Mathematics previous year question of Straight Line for NDA
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previous year question of Straight Line for NDA

Previous Year Straight Line Question Of NDA - set 1

Previous Year Straight Line Question Of NDA
Q 2723080841

The incentre of the triangle with vertices `A(1, sqrt3), B(0, 0)` and `C(2, 0)` is
NDA Paper 1 2017
(A)

`( 1 , sqrt3/2)`

(B)

`(2/3 , 1/sqrt3)`

(C)

`(2/3 , sqrt3/2)`

(D)

`(1 , 1/sqrt3)`

Solution:

`AB= sqrt((1-0)^2 + (sqrt 3-0)^2) =2`

`AC= sqrt((1-2)^2+ ( sqrt 3 -0)^2)=2`

`BC= sqrt((2-0)^2 +(0-0)^2) = 2`

`AB=BC=CA`

Hence `ABC` is equilateral triangle.

`=>` incentre = centroid
Correct Answer is `=>` (D) `(1 , 1/sqrt3)`
Q 2186312277

Consider the lines `y = 3x, y = 6x` and `y= 9`

What is the area of the triangle formed by these lines?
NDA Paper 1 2016
(A)

`27/4` sq units

(B)

`27/2` sq units

(C)

`19/4` sq units

(D)

`19/2`sq units

Solution:

Given, lines are

`y=3x` ..........(i)

`y=6x` ...............(ii)

and `y =9` ...........(iii)

On solving Eqs. (i) and (iii), we get

`x=3, y =9`

:.Coordinates of `A= (3, 9)`

On solving Eqs. (ii) and (iii), we get

` x=3/2, y=9`

:. Coordinates of `B = (3/2, 9)`

Now, `AB = sqrt ((3-3/2)^(2) + (9 - 9)^(2))`

` = sqrt((3/2)^(2) + 0^(2)) = sqrt(9/4) = 3/2 `

:. Area of ` DeltaOAB =1/2 xx AB xx AM`

[as area of triangle `= 1/2 xx base xx height]`

`= 1/2 xx 3/2 xx 9`

` = 27/4` sq units
Correct Answer is `=>` (A) `27/4` sq units
Q 2116412370

Consider the lines `y = 3x, y = 6x` and `y= 9`

The centroid of the triangle is at which one of the
following points?
NDA Paper 1 2016
(A)

`(3, 6)`

(B)

`(3/2 , 6)`

(C)

`(3, 3)`

(D)

`(3/2 , 9)`

Solution:

We have, three vertices of `Delta ABC` are `O(0, 0), A(3, 9)` and

`B (3/2 , 9)` Let `C(x,y)` be the centroid of the triangle.

Then `x = ( 0 + 3 + 3/2 )/3 = 3/2` and `y = (0 + 9 + 9)/3 = 6`

Coordinates of the centroid is `( 3/2, 9)`.
Correct Answer is `=>` (B) `(3/2 , 6)`
Q 2116234170

Consider a parallelogram whose vertices are `A(1, 2), B(4, y), C(x, 6)` ard `D(3, 5)` taken in order.

What is the value of `AC^( 2) - BD^( 2)?`
NDA Paper 1 2016
(A)

`25`

(B)

`30`

(C)

`36`

(D)

`40`

Solution:

Since, In a parallelogram diagonals bisect each other.

:. Mid-point of `AC` =Mid-point of `DB`

`=> ((x+ 1)/2 , (6 + 2)/2) = ((4+3)/2 , (y+ 5)/2) `

` => (x+1)/2 = 7/2 => x = 6`

and `y+5=6+2 => y=3`

Coordinates of `O` are `(7/2 , 8/2)` i.e. `(7/ 2, 4)`

Now, `AC^(2) = (6- 1)^(2) + (6- 2)^(2) = 25 + 16 = 41`

and `BD^(2) = (4-3)^(2) + (3 -5)^(2) = 1 + 4=5`

`:. AC^(2) - BD^(2) = 41 -5 = 36`
Correct Answer is `=>` (C) `36`
Q 2116334270

Consider a parallelogram whose
vertices are `A(1, 2), B(4, y) C(x, 6)` ard `D(3, 5)` taken in order.

What is the point of intersection of the diagonals?
NDA Paper 1 2016
(A)

`(7/2 , 4)`

(B)

`(3 , 4)`

(C)

`(7/2 , 5)`

(D)

`(3 , 5)`

Solution:

Since, in a parallelogram diagonals bisect each other.

:. Mid-point of `AC` =Mid-point of `DB`

`=> ((x+ 1)/2 , (6 + 2)/2) = ((4+3)/2 + (y+ 5)/2) `

` => (x+1)/2 = 7/2 => x = 6`

and `y+5=6+2 => y=3`

`:.` Coordinates of `O` are `(7/2 , 8/2)` i.e. `(7/ 2, 4)`
Correct Answer is `=>` (A) `(7/2 , 4)`
Q 2126678571

A straight line intersects `x` and `y` axes at `P` and `Q`,
respectively. If `(3, 5)` is the middle point of `PQ`, then what
is the area of the triangle `OPQ?`
NDA Paper 1 2016
(A)

`12` sq units

(B)

`15` sq units

(C)

`20` sq units

(D)

`30` sq units

Solution:

Let coordinates of `P` and `Q` are `(x, 0)` and `(0, y)`,
respectively.

`:.` Mid-point of `PQ` is `(x + 0)/2 =3 => x =6`

and `(0+y)/2 = 5 =>y=10 `

`:.` Area of `DeltaPOQ = 1/2 xx OP xx OQ`

` = 1/2 xx 6 xx 10 = 30` sq units
Correct Answer is `=>` (D) `30` sq units
Q 2701580428

An equilateral triangle has one vertex at (0,0) and another at `(3 , sqrt 3)` .What are the coordinates of the third vertex ?
NDA Paper 1 2016
(A)

`(0, 2 sqrt3 )` only

(B)

`(3, -sqrt3 )` only

(C)

`(0,2sqrt3 )` or `( 3 , -sqrt3)`

(D)

Neither `(0,2sqrt3 )` nor `(3, -sqrt3)`

Solution:

Two vertices of an equilateral triangle are `(0, 0)` and `(3, √3)`.

Let the third vertex of the equilaterla triangle be `(x, y)`

Distance between ` (0, 0)` and `(x, y) =` Distance between` (0, 0)` and `(3, √3) =` Distance between `(x, y)` and `(3, √3)`

`√(x^2 + y^2) = √[(x - 3)^2 + (y - √3)^2]`

`x^2 + y^2 = 12`

`x^2 + 9 - 6x + y^2 + 3 - 2√3y = 12`

`24 - 6x - 2√3y = 12`

`- 6x - 2√3y = - 12`

`3x + √3y = 6`

`x = (6 - √3y) / 3`

`⇒ [(6 - √3y)/3]2 + y^2 = 12`

`⇒ (36 + 3y^2 - 12√3y) / 9 + y^2 = 12`

`⇒ 36 + 3y^2 - 12√3y + 9y^2 = 108`

`⇒ - 12√3y + 12y^2 - 72 = 0`

`⇒ -√3y + y^2 - 6 = 0`

`⇒ (y - 2√3)(y + √3) = 0`

`⇒ y = 2√3` or `- √3`

If `y = 2√3, x = (6 - 6) / 3 = 0`

If `y = -√3, x = (6 + 3) / 3 = 3`

So, the third vertex of the equilateral triangle `= (0, 2√3) ` or `(3, -√3)`.
Correct Answer is `=>` (C) `(0,2sqrt3 )` or `( 3 , -sqrt3)`
Q 2731680522

What is the equation of the right bisector of the line segment joining `(1,1)` and `(2, 3)` ?
NDA Paper 1 2016
(A)

`2 x + 4 y -11 =0`

(B)

`2x -4 y - 5 =0`

(C)

`2x - 4 y -11 =0`

(D)

`x -y +1 =0`

Solution:

mid point

`((1+2)/2 , (1+3)/2)`

`(3/2 , 2)`

Slop of `AB = (3-1)/(2-1) =2`

Slop of right bisector `= -1/2`

So right bisector will the line passing through `(3/2 , 2)` with slop of `-1/2`

`y-2 = -1/2 (x-3/2)`

`4y -8 =-2x +3`

`2x +4y -11 =0`
Correct Answer is `=>` (A) `2 x + 4 y -11 =0`
Q 2761880725

If the point (a,a) lies between the lines `| x + y | =2`, then which one of the following is correct ?
NDA Paper 1 2016
(A)

`| a | < 2 `

(B)

`| a | < sqrt2`

(C)

`|a | < 1`

(D)

` |a | < 1/sqrt2`

Solution:

`| x+ y| = 2`

`=> ` line s are `x+y = 2 , x+y = -2`

`(a , a) ` will lie on

` y = x`


`OA = OB = | (1 * 0+ 0*1 pm 2)/sqrt(1+1) | = sqrt2`

to line point in between, the `bot`distance from (0,0) to (a,a) should be less then `bot ` distance between (0,0) to (1,1) as (1,1) is point of intersection of `y = x` and `x+y=2`

`sqrt 2 |a| < OA => |a| < sqrt2`

`|a| < 1`
Correct Answer is `=>` (C) `|a | < 1`
Q 2761080825

If `(a , b)` is at unit distance from the line `8x+6y+1 = 0` then which of the following conditions are correct ?

1. `3a-4b-4 = 0`

2. `8a+6b+11 = 0`

3. `8a+6b-9 = 0`

Select the correct answer using the code given below :
NDA Paper 1 2016
(A)

1 and 2 only

(B)

2 and 3 only

(C)

1 and 3 only

(D)

1 , 2 and 3

Solution:

Distance of
`(a,b)` from `8 x + 6y +1 =0` is `=1`

`| (8a + 6b +1)/( sqrt (8^2 + 6^2) ) | =1`

`=18 a + 6b +11 = 10`

`8a + 6b -9 =0` or `8a + 6b +11 =0`
Correct Answer is `=>` (B) 2 and 3 only
Q 2156191074

Consider the two lines `x + y + 1 = 0` and `3x + 2y + 1 = 0`

What is the equation of the line passing through the
point of intersection of the given lines and parallel to
X-axis?
NDA Paper 1 2016
(A)

`y + 1 = 0`

(B)

`y- 1 = 0`

(C)

`y- 2 = 0`

(D)

`y + 2 = 0`

Solution:

Given lines are `x + y + 1 = 0` ..........(i)

and `3x + 2y + 1 =0` ...........(ii)

Point of intersection of these two lines are `(1, - 2)`.

`:.` Required equation of line` y = - 2` or `y + 2 = 0`
Correct Answer is `=>` (D) `y + 2 = 0`
Q 2186291177

Consider the two lines `x + y + 1 = 0` and `3x + 2y + 1 = 0`

What is the equation of the line passing through the
point of intersection of the given lines and parallel to
Y-axis?
NDA Paper 1 2016
(A)

`x + 1 = 0`

(B)

`x -1 = 0`

(C)

`x- 2 = 0`

(D)

`x + 2 = 0`

Solution:

Given lines are `x + y + 1 = 0` .........(i)

and `3x + 2y + 1 = 0` ......(ii)

Equation of line parallel to Y-axis
Correct Answer is `=>` (B) `x -1 = 0`
Q 2341680523

(a, 2b) is the mid-point of the line segment joining the points (10, -6) and (k, 4). If a- 2b = 7, then what is the value of k?
NDA Paper 1 2016
(A)

2

(B)

3

(C)

4

(D)

5

Solution:

Since, (a, 2b) is the mid-point of the line segment joining the points - and (k, 4), therefore we have

`(a, 2b ) = ( (10 +k)/2, (-6 +4)/2)`

`=> (a, 2 b) = ((10 +k)/2 , -1)`

`=> = (10 +k)/2 ` and ` 2b = -1`

`=> a = (10 + k )/2 ` and `b =-1/2`

Also , it is given that ` 1 - 2b = 7`

`=> (10 +k)/2 -2 (-1/2) =7`

`=> 10 + k +2 = 14`

`=> k =2`
Correct Answer is `=>` (A) 2
Q 2117212180

If `a`, `b` and `c` are the position vectors of the vertices of an
equilateral triangle whose orthocentre is at the origin,
then which one of the following is correct ?
NDA Paper 1 2016
(A)

`a + b + c = 0`

(B)

`a + b + c =` unit vector

(C)

`a + b + c `

(D)

`a = b + c`

Solution:

Given vertices of equilateral triangle are `a, b` and `c`.

`:.` Orthocentre of an equilateral triangle `= (a+b+c)/3`

Since, orthocentre of triangle is at the origin.

So, `(a+b+c)/3 = 0 => a + b + c = 0`

[`∵` in a equilateral triangle, centroid and orthocentre coincide]
Correct Answer is `=>` (A) `a + b + c = 0`
Q 2107512488

What is the acute angle between the lines represented by
the equations `y - sqrt(3)x - 5 = 0` and `sqrt(3)y - x + 6 = 0`?
NDA Paper 1 2016
(A)

`30^0`

(B)

`45^0`

(C)

`60^0`

(D)

`75^0`

Solution:

Given lines are

`y - sqrt(3)x - 5 = 0` and ` sqrt(3)y - x + 6 = 0`

`:. y = sqrt(3) x + 5` .......(i)

and ` sqrt(3)y = x - 6 => y = x/sqrt(3) - (6/sqrt(3))` ......(ii)

From Eqs. (i) and (ii), we have

` m_1 = sqrt(3)` and `m_2 = 1/sqrt(3)`

Hence, acute angle `tan theta = | (m_1 - m_2)/(1 + m_1 +m_2) |`

` = | ( sqrt(3) - 1/sqrt(3) )/( 1 + sqrt(3) xx 1/sqrt(3) )| = | ((3 -1 )/sqrt(3))/2 | = | (2/sqrt(3))/2 |`

` => tan theta = 1/sqrt(3)`

`=> tan theta = tan 30^0`

`=> theta = 30^0`
Correct Answer is `=>` (A) `30^0`
Q 1628180001

Consider the ` Delta ABC` with
vertices `A ( -2, 3), B (2, 1)` and `C (1, 2)`.

What is the circumcentre of the ` angle ABC?`
NDA Paper 1 2015
(A)

`(-2, -2)`

(B)

`(2, 2)`

(C)

`(-2, 2)`

(D)

`( 2, -2)`

Solution:

Let `P (x, y)` is the circumcentre of the `angleABC`.

`:. AP^2 = PB^2`

`=> (x + 2)^2 + (y- 3)^2 (x - 2)^2 + (y- 1)^2`

`=> x^ 2 + 4 + 4x + y^2 + 9 - 6y =`

`x^ 2 + 4 - 4x + y^2 + 1- 2y`

`=> 4x - 6y + 13 = '4 - 4x + 1 - 2y`

`=> 8x - 4y + 8 = 0`

`=> 2x - y + 2 = 0`... (i)

Also, `AP^2 = PC^2`

`=> (x + 2)^2 + (y- 3)^2 = (x - 1)^2 + (y- 2)^2`

`=> x^ 2 + 4 + 4x + y^2 + 9 - 6y`

`= x^ 2 + 1 - 2x + y^2 + 4 - 4y`

`=> 4x - 6y + 13 = -2x - 4y + 5`

`=> 6x - 2y + 8 = 0`

`=> 3x-y + 4= 0` ... (ii)

On subtracting Eq. (ii) from Eq (i), we get

`- x - 2 = 0`

`=> x = -2`

`:. y = 3 x (-2) + 4 = - 2`

Hence, the required circumcentre is `(-2, -2)`.
Correct Answer is `=>` (A) `(-2, -2)`
Q 1638180002

Consider the ` Delta ABC` with
vertices `A ( -2, 3), B (2, 1)` and `C (1, 2)`.

What is the centroid of the ` Delta ABC ?`
NDA Paper 1 2015
(A)

` ( 1/3 ,1)`

(B)

` ( 1/3 , 2)`

(C)

` ( 1 ,2/3)`

(D)

` ( 1/2 , 3)`

Solution:

Given, vertices of a triangle are `A(-2, 3), B(2, 1)`
and `C(1,2)`.

Then centroid of ` Delta ABC = ( (x_1 + x_2 + x_3)/3 ,( y_1 + y_2 + y_3)/3)`

Centroid of the `Delta ABC = (( -2 + 2 +1 )/3 ,( 3 + 1 + 2)/3 )= ( 1/3 ,2)`
Correct Answer is `=>` (B) ` ( 1/3 , 2)`
Q 1648180003

Consider the ` Delta ABC` with
vertices `A ( -2, 3), B (2, 1)` and `C (1, 2)`.

What is the foot of the altitude from the vertex A of
the triangle `ABC?`
NDA Paper 1 2015
(A)

`( 1, 4)`

(B)

`( -1, 3)`

(C)

`( -2, 4)`

(D)

`( -1, 4)`

Solution:

Let D be the foot of altitude from `A` in `Delta ABC` and

`D= (x, y)`

Slope of `BC = (1-2)/(2-1) = - 1`

`:. m = -1`

Also, slope of `AD` is `(( y_1 - 3)/ ( x_1 +2))`

But ` ( y_1 - 3)/ ( x_1 +2) . - 1 = -1 `

` => y_1 - 3 = x_1 + 2 => y_1 - x_1 = 5`

From the given points, only `(-1, 4)` satisfies this equation

Hence, the required foot of altitude is `(-1, 4)`.
Correct Answer is `=>` (D) `( -1, 4)`
Q 2211223129

The area of the figure formed by the lines
`ax + by + c = 0, ax - by + c = 0, ax + by - c = 0` and
`ax - by - c = 0` is
NDA Paper 1 2015
(A)

` c^2/(ab)`

(B)

` (2c^2)/(ab)`

(C)

` c^2/(2ab)`

(D)

` c^2/(4ab)`

Solution:

The figure formed by the lines

`ax + by + c = 0, ax - by + c = 0, ax + by - c = 0` and

`ax - by - c = 0` is shown below

Clearly, `ABCD` is a parallelogram and its area

`= 1/2 xx` Product of diagonals

` = 1/2 xx (2c)/b xx (2c)/a xx (2c^2)/(ab)` sq units
Correct Answer is `=>` (B) ` (2c^2)/(ab)`
Q 2231423322

If a line is perpendicular to the line `5x - y = 0` and forms
a triangle of area `5 sq` units with coordinate axes, then its
equation is
NDA Paper 1 2015
(A)

`x + 5y pm 5 sqrt(2) = 0`

(B)

`x - 5y pm 5 sqrt(2) = 0`

(C)

`5x + y pm 5 sqrt(2) = 0`

(D)

`5x - y pm 5 sqrt(2) = 0`

Solution:

Equation of the line perpendicular to `5x - y = 0`

is `x + 5y = lamda`.

` lamda^2 = 50`

Now, area of `Delta AOB = 5` sq units

`=> 1/2 lamda xx lamda/5 = 5`

` => lamda= pm 5 sqrt(2) `

Equation of the lines are `x + 5y = pm 5 sqrt(2)`

i.e. `x + 5y pm 5sqrt(2) = 0`.
Correct Answer is `=>` (A) `x + 5y pm 5 sqrt(2) = 0`
Q 1668180005

A line passes through `(2, 2)` and is perpendicular to
the line `3x + y = 3`, its `y` -intercept is
NDA Paper 1 2015
(A)

` 3/4 `

(B)

`4/3`

(C)

`1/3`

(D)

`3`

Solution:

Given equation of line is `y = 3- 3x`

`=> m = -3`

:. Slope of the line perpendicular to the line `y = 3- 3x`,
is

` m' = (-1)/(-3) = 1/3`

:. Required equation of the line is

` y - 2 = 1/3 (x -2)`

`=> 3y- 6 = x - 2 => 3y - x = 4`

`=> x/(-4) + y/(4/3) =1`

So , `y` Intercept is `4/ 3`.
Correct Answer is `=>` (B) `4/3`
Q 1688180007

The perpendicular distance between the straight
lines `6x + 8 y + 15 = 0` and `3x + 4y + 9 = 0` is
NDA Paper 1 2015
(A)

`3/2` units

(B)

`3/(10)` units

(C)

`3/4` units

(D)

`2/7` units

Solution:

We have,

`6x + 8y + 15 = 0`

and `6x + 8y + 18 = 0`

:. Perpendicular distance between them is

` | (18 - 15)/( sqrt(36) + 64) |`

` = 3/(10) unit`
Correct Answer is `=>` (B) `3/(10)` units
Q 2271723626

The three lines `4x + 4y = 1,8x - 3y = 2, y = 0` are
NDA Paper 1 2015
(A)

the sides of an isosceles triangle

(B)

concurrent

(C)

mutually perpendicular

(D)

the sides of an equilateral triangle

Solution:

Since, the point of intersection of `y = 0` and `4x + 4y = 1`

is `( 1/4 , 0)` and `( 1/4 , 0)` lies on the line `8x- 3 y = 2`.

Hence, the given lines are concurrent.
Correct Answer is `=>` (B) concurrent
Q 2221023821

The line `3x + 4 y - 24 = 0` intersects the `X`-axis at `A` and
`Y`-axis at `B`. Then, the circurncentre of the ` Delta OAB`, where `O`
is the origin, is
NDA Paper 1 2015
(A)

`(2, 3)`

(B)

`(3, 3)`

(C)

`(4, 3)`

(D)

None of these

Solution:

Given `3x + 4 y - 24 = 0`

Its `x`-intercept `A= (8, 0)`

and `y`-intercept `B = (0, 6)`

Since, `AOB` is a right angled triangle.

So, the mid-point of hypotenuse is the circumcentre.

Hence,circumcentre `= (4, 3)`

`[ ∵(8+0)/2 , (0+6)/2 = (4,3) ]`
Correct Answer is `=>` (C) `(4, 3)`
Q 1618280100

The length of perpendicular from the origin to a
line is `5` units and the line makes an angle `120^0`
with the positive direction of `X`-axis. The equation
of the line is
NDA Paper 1 2015
(A)

`x + sqrt(3) y = 5`

(B)

`sqrt(3) y + y =10`

(C)

`sqrt(3) x - y =10`

(D)

None of these

Solution:

`xcosalpha + ysinalpha = p`

Here `alpha = 30^0 ` `p= 5`

`x(sqrt3/2) + y (1/2) = 5`

`=> xsqrt3 + y = 10`

Alternatively :


Let the required line intercept `X`-axis at a.

Slope of the line is

`m = tan120^0 =- sqrt(3)`

Also, `sin 60^0 = 5/a`

`=> sqrt(3)/2 = 5/a`

`:. a = (10)/sqrt(3)`

:. Equation of line passing through `(a, 0)` and having

slope `-sqrt(3)` is

`y - y = m(x - x_1)`

`=> y- 0 = - sqrt(3) ( x - (10)/sqrt(3) )`

`=> y = - sqrt(3) x + 10`

`=> sqrt(3x) + y = 10`
Correct Answer is `=>` (B) `sqrt(3) y + y =10`
Q 1628280101

The equation of the line joining the origin to the
point of intersection of the lines `x/a + y/b = 1` and
`x/b + y/a = 1`
NDA Paper 1 2015
(A)

`x- y =0`

(B)

`x + y =0`

(C)

`x = 0`

(D)

`y = 0`

Solution:

We have,

`x/a + y/b = 1` .........(1)

and `x/b + y/a = 1` ..........(2)

`=> bx + ay = ab`

`=> ax+ by= ab`

`:. bx + ay = ax + by`

` => x (b -a)= y (b - a)`

`=> x/y = 1 => x = y`

`:. ` Equation the line is `x - y = 0`

Now, from Eq. (i), we get

`:. y/a + y/b = 1`

` y = (ab)/(a+b) `

and ` x = (ab)/(a+b) `

:. Equation of line Joining onigin and `( (ab)/(a+b) , (ab)/(a+b) )` .is

`y - 0 = 1 (x- 0)`

`=> y = x`
Correct Answer is `=>` (A) `x- y =0`
Q 2301291128

The product of the perpendiculars from the two points
(± 4,0) to the line `3x cos phi + 5y sin phi = 15` is
NDA Paper 1 2015
(A)

25

(B)

16

(C)

9

(D)

8

Solution:

Given, `3x cos phi + 5y sin phi = 15`

`:. ` Distance from the point `(x_1, y_1 )` to ax+ by + c = 0

`(-|ax_1 + by_1 + c|)/sqrt(a^2 +b^2)`

Lengths of perpendicular from the point (± 4, 0),

`p_1 = |(12 cos phi -15)/sqrt(9 cos phi + 25 sin^2 phi)| ` .........(i)

and `p_2 = | (-12 cos phi - 15)/sqrt(9 cos phi + 25 sin^2 phi)| ` ..................(ii)

On multiplying Eqs. (i) and (ii), we get

`p_1p_2 = | ((12 cos phi)^2 - (15)^2)/(9 cos ^2 phi + 25 sin^2 phi) | = | (144 cos ^2 phi -225)/(9 + 16 sin^2 phi)|`

`|(144 (1 - sin^2 phi) -225)/(9 +16 sin^2 phi) | = | (-9 ( 9 + 16 sin^ phi))/(9 + 16 sin^2 phi) | = 9`
Correct Answer is `=>` (C) 9
Q 2231534422

Two straight lines passing through the point `A(3, 2)` cut
the line `2y = x + 3` and `X`- axis perpendicularly at `p` and `Q`,
respectively. The equation of the line `PQ` is
NDA Paper 1 2015
(A)

`7x + y - 21 = 0`

(B)

`x + 7y + 21 = 0`

(C)

`2x + y - 8 = 0`

(D)

`x + 2y + 8 = 0`

Solution:

Given, `2y = x + 3 ....... (i)`

A line passes through `A(3, 2)` and having slope `-2`.

[since, line is perpendicular to line having slope `1/2`]

`:. y - 2 = - 2 (x- 3)`

[∵ point-slope form, `y - y_1 = m (x- x_1)]`

`=> 2x + y - 8 = 0 .... (ii)`

On solving Eqs. (i) and (ii), we get

`P = ((13)/5 , (14)/5)`

Now, a line passes through `A(3, 2)` and having slope `1/0`.

[since, line is perpendicular to `X`-axis]

`:. y - 2 = 1/0 (x-3) => x = 3`

Coordinate at point `Q = (3, 0)`

Now, equation of line `PQ` is `y- (14)/5 = (0- (14)/5)/( 3 - (13)/5) ( x - (13)/5)`

`=> 7x + y - 21 = 0`
Correct Answer is `=>` (A) `7x + y - 21 = 0`
Q 1688280107

From the point `P(3, -1, 11)`,
a perpendicular is drawn on the line `L` given by the
equation `x/2 = (y-2)/3 = (z-3)/4 `. Let `Q` be the foot of the
perpendicular.

What is the length of the line segment `PQ?`
NDA Paper 1 2015
(A)

`sqrt(47)` unit

(B)

` 7` unit

(C)

`sqrt(53)` unit

(D)

` 8` unit

Solution:

Length of the line segment `PQ`, when `Q = (2, 5, 7)`

`:. PQ = sqrt((2 - 3)^2 + (5 + 1)^2 + (7 - 11)^2)`

`= sqrt(1 + 36 + 16) = sqrt(53)` units
Correct Answer is `=>` (C) `sqrt(53)` unit
Q 2281134927

`ABCD` is a parallelogram and `P` is the point of
intersection of the diagonals. If `O` is the origin, then
`OA+OB+OC+OD` is equal to
NDA Paper 1 2015
(A)

`4 OP`

(B)

`2 OP`

(C)

`OP`

(D)

Null vector

Solution:

We know that `P` will be the mid-point of `AC` and `BD`.

`:. OA + OC =2 OP` .........(i)

and `OB + OD = 2 OP` ..............(ii)

On adding Eqs. (i) and (ii), we get

`OA + OB+ OC + OD = 4 OP`
Correct Answer is `=>` (A) `4 OP`
Q 1732045832

Consider the following points
1. `(0, 5)`
2. `(2, -1)`
3. `(3, -4)`

Which of the above lie on the line `3x + y = 5` and at a
distance `sqrt(10)` from `(1, 2)`?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

1 and 2

(D)

1, 2 and 3

Solution:

Given, equation of line

` 3x + y = 5`

` => x/(5//3) + y/5 = 1` ..............(1)

Let `S = 3x + y - 5 = 0` ..........(2)

1. At point `(0, 5)`

` S_((0.5)) = 3(0) + 5 - 5`

` = 0 + 0 = 0`

So, the point `(0, 5)` lie on the given line.

and distance between `(0, 5)` and `(1, 2)`

` = sqrt ( (1- 0)^2 + (2- 5)^2)`

` = sqrt( 1^2 + (-3)^2)`

` [ ∵` by distance formula, `d = sqrt ( (x_2 - x_1 )^2 + (y_ 2 - y_1)^2 ) `

` = sqrt( 1 + 9) = sqrt(10)`

2. At point `(2, -1)`

` S_((2 ,-1)) = 3(2) + (-1) - 5`

` = 6 - 1 - 5 = 6 - 6= 0 `

So, the point `(2, -1)` also lie on the given line

and distance between `(2, - 1 )` and `(1 , 2)`

` = sqrt ( (1-2)^2 + (2+ 1)^2)`

` = sqrt((-1)^2 + (3)^2)`

` = sqrt( 1 + 9) = sqrt(10)`

3 . At point `(3, - 4)`

`S _((3, -4)) = 3(3) + (- 4) - 5`

` = 9 - 4 - 5 = 9 - 9 = 0 `

So, the point `(3, - 4)` also lie on the given line

and distance between `(3, - 4)` and `(1, 2)`

` = sqrt((1- 3)^2 + (2 + 4)^2 = sqrt( (-2)^2 + (6)^2)`

` = sqrt( 4+ 36) = sqrt(40) =2sqrt(10)`

Hence, points `(0, 5)` and `(2, -1)` lie on the line `3x + y = 5`

and at a distance `sqrt(10)` from `(1, 2)`.
Correct Answer is `=>` (C) 1 and 2
Q 1762045835

What is the equation of the line through `(1, 2)`, so that
the segment of the line intercepted between the axes is
bisected at this point?
NDA Paper 1 2014
(A)

`2x - y = 4`

(B)

`2x - y + 4 = 0`

(C)

`2x + y = 4`

(D)

`2x + y + 4 = 0`

Solution:

Let the required equation be,

` x/a + y/b = 1` .........(1)

Whose intercept on `X` and `Y`-axes are `a` and `b`, respectively.

Since, the Eq. (i) is passes through the point `(1, 2)`.

`:. 1/a + 2/b = 1` ......... (2)

Also, given that the line segment is bisected by the point

`(1, 2)`. i.e., `(1, 2)` is the middle point of the line segment

passes through the points `(a, 0)` and `(0, b)`.

`:. (1,2) = ( (a+0)/2 , (0+b)/2 )`

` => a/2 = 1`

` => a = 2` and `b/2 = 2 => b = 4`

Put the value of a and bin Eq. (1), we get

` x/2 + y/4 = 1 => 2x + y = 4`

which is the required equation.
Correct Answer is `=>` (C) `2x + y = 4`
Q 1752145934

What is the equation of straight line passing through the
point `(4, 3)` and making equal intercepts on the
coordinate axes?
NDA Paper 1 2014
(A)

`x + y = 7`

(B)

`3x + 4y = 7`

(C)

`x- y = 1`

(D)

None of these

Solution:

Let the required equation in intercept form

` x/a + y/b = 1` ...........(1)

where a and bare the intercepts on `X` and `Y`-axes,

respectively.

Given that, the straight line (i) making equal intercept on

the coordinate axes.

`:. a = b`

From Eq. (i),

` x/a + y/a = 1`

` => x + y = a` ...........(2)

Also. the straight line passing through the point `(4, 3)`.

`:. 4 + 3 = a`

` => a= 7`

Now, put the value of a in Eq. (2), we get

` x + y = 7`

which is the required equation.
Correct Answer is `=>` (A) `x + y = 7`
Q 1762145935

What is the equation of the line midway between the
lines `3x- 4y + 12 = 0` and `3x- 4y = 6?`
NDA Paper 1 2014
(A)

`3x - 4y - 9 = 0`

(B)

`3x - 4y + 9 = 0`

(C)

`3x - 4y - 3 = 0`

(D)

`3x - 4y + 3 = 0`

Solution:

The given equation of lines are,

` 3x - 4y + 12 = 0` ............(1)

and `3x - 4y = 6`

or `3x - 4y - 6 = 0` ..........(2)

On comparing with `ax + by + c = 0` , we get

`a_1 = a_2 = 3 , b_1 = b_2 = - 4` and `c_1 = 12, c_2 = -6`

So, the equatiton of line mid way between the given lines is

` 3x - 4y + ( (c_1 + c_2))/2 = 0`

` => 3x - 4y + ((12 - 6))/2 = 0`

` => 3x- 4y + 3 = 0`

which is the required equation.
Correct Answer is `=>` (D) `3x - 4y + 3 = 0`
Q 1753167044

The vertices of a `Delta ABC` are `A(2, 3, 1)`,
`B( -2, 2, 0)` and `C(0, 1, -1)`.

What is the magnitude of the line joining mid-points of
the sides `AC` and `BC`?
NDA Paper 1 2014
(A)

`1/sqrt(2)` unit

(B)

`1` unit

(C)

`3/sqrt(2)` units

(D)

`2` units

Solution:

Given that vertices of a triangle are,

let `(x_1 , y_1 , z_1) = A (2, 3, 1)`,

`(x_2 ,y_2 , z_2 ) = B (- 2, 2, 0)`,

and `(x_3, y_3, z_3) = C (0, 1, -1)`,

Mid-point of

` AC = ( (2 + 0)/2 , (3+1)/2 ,(1-1)/2)`

`P = (1, 2, 0)` and mid-point of

` BC = ( (-2+0)/2 , (2 + 1)/2 ,(0-1)/2 )`

` Q = (- 1 , 3//2, -1//2)`

Now, magnitude of the line joining `P` and `Q =| PQ|`

` = sqrt( (1 +1)^2 + (2 - 3//2)^2 + (0 + 1//2)^2 )`

` = sqrt( (2)^2 + (1//2)^2 + (1//2)^2)`

` sqrt(4 + 1/4 + 1/4 ) = sqrt(4 + 1/2 )`

` sqrt(9//2) = 3/sqrt(2) `units
Correct Answer is `=>` (C) `3/sqrt(2)` units
Q 2377334286

If the three vertices of the parallelogram `ABCD` are `A ( 1, a), B (3, a)` and `C (2, b)`, then `D` is equal
to?
NDA Paper 1 2013
(A)

`(3, b)`

(B)

`(0, b)`

(C)

`(4, b)`

(D)

`(5, b)`

Solution:

We know that, in parallelogram, diagonals bisect each other.

`:.` Mid-point of `BD =` Mid-point of `AC`

`=> ( (x+3)/2 , (y+a) /2)= (3/2, (a+b)/2)`

`=> (x+3)/2 =3/2 =>x= 0`

and `(y+a)/2 = (a+b)/2 => y = b`

So, the coordinate of point `D` is `(0, b)`.
Correct Answer is `=>` (B) `(0, b)`
Q 2377434386

What is the equation of the line which passes
through `(4, -5)` and is perpendicular to `3x + 4y + 5 =0`?
NDA Paper 1 2013
(A)

`4x- 3y- 31 = 0`

(B)

`3x- 4y- 41 = 0`

(C)

`4x + 3y- 1 = 0`

(D)

`3x + 4y + 8 = 0`

Solution:

Since, the required line is perpendicular to the line

`3x + 4y + 5 = 0`.

So . the slope of required line is `[-1/(-3/4)] = 4/3`

Also. required line passing through the point `(4,- 5)`. Then. its
equation

`(y+5) = 4/3 (x-4)`

`=> 3y +15 = 4x -16 => 4x-3y=31`
Correct Answer is `=>` (A) `4x- 3y- 31 = 0`
Q 2317534480

For what value of `k` are the two straight lines `3x + 4 y = 1` and `4x + 3 y + 2k = 0` equidistant from the point `(1, 1)`?
NDA Paper 1 2013
(A)

`1/2`

(B)

`2`

(C)

`-2`

(D)

`-1/2`

Solution:

Perpendicular distance of the line `3x + 4 y - 1 = 0` from

the point `(1, 1) =` Perpendicular distance of the line

`4x + 3y + 2k = 0` from the point `(1, 1)`

`=> (|3 xx 1 + 4 xx 1-1 | )/(sqrt (9+16) ) = ( |4 xx 1 +3 xx 1 +2k |)/(sqrt (16+9) )`

`=> (|3+4-1 |)/5 = ( |4+3+2k | )/5`

`=> 6= 7+2k`

`=>2k = -1`

`:. k = -1/2`
Correct Answer is `=>` (D) `-1/2`
Q 2377534486

`A` point `P` moves such that its distances from `( 1, 2)` and `(-2, 3)` are equal. Then, the locus of `P` is
NDA Paper 1 2013
(A)

straight line

(B)

parabola

(C)

ellipse

(D)

hyperbola

Solution:

Let the coordinates of point `P` is `(h, k)`.

Now, by given condition

Distance between `(h, k)` and `(1, 2) =` Distance between `(h, k)` and `(-2, 3)`


`=> sqrt ( (h-1)^2 + (k-2)^2 ) = sqrt ( (h+2)^2 + (k-3)^2 )`

`=> h^2 + 1 - 2h + k^2 + 4 - 4k = h^2 + 4 + 4h + k^2 + 9 6k `

`=> - 2h - 4k + 5 = 4h - 6k + 13`

`=> 6h - 2k + 8 = 0`

`=> 3h - k + 4 = 0`

So, the locus of `P` is `3x- y + 4 = 0`, which represent a straight line.
Correct Answer is `=>` (A) straight line
Q 2317134989

The equation of the locus of a point which is equidistant from the axes is
NDA Paper 1 2013
(A)

`Y=2x`

(B)

`x=2y`

(C)

`y = pm x`

(D)

`2y +x = 0`

Solution:

A point which is equidistant from the axes means a line
which is equally inclined with the axes and passes through the
.origin.


i.e., `m = 45^(circ)` and `135^(circ)`

`:.` Required equations are

`y =tan 45^(circ) x` and `y = tan 135^(circ) x`

`y = x` and `y=- x`

So, the combined equation is `y = pm x`.
Correct Answer is `=>` (C) `y = pm x`
Q 2327445381

What angle does the line segment joining `(5, 2)`
and `(6, -15)` subtend at `(0, 0)`?
NDA Paper 1 2013
(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/2`

(D)

`(3 pi)/4`

Solution:

Slope of `OA (m_1) = (2-0)/(5-0) = 2/5`

and slope of `OB (m_2) = (-15-0)/(6-0) = -5/2`

`:. m_1 * m_2 = 2/5 xx -5/2 = -1`

i.e., angle between `OA` and `OB` is `pi/2`

Hence, the line segment `AB` subtend right angle at origin `O`.
Correct Answer is `=>` (C) `pi/2`
Q 2317645580

What is the equation to the straight line passing
through `(5,- 2)` and `(-4, 7)`?
NDA Paper 1 2013
(A)

`5x-2y= 4`

(B)

`-4x + 7 y = 9`

(C)

`x+ y= 3`

(D)

`x-y = -1`

Solution:

Equation of straight line which passes through the
points `(5,- 2)` and `(- 4, 7)` is

`(y- y_1 ) = (y_2 -y_1)/(x_2 -x_1) (x- x_1)`

`=> (y+2) = (7+2)/(-4-5) (x-5)`

`=> y+2 = 9/-9 (x-5)`

`=> y+2 = -x +5`

`:. x+y =3`
Correct Answer is `=>` (C) `x+ y= 3`

Set 2

Problem set 2
Q 2317645589

What is the angle between the lines `x + y = 1` and
`x-y= 1 ?`
NDA Paper 1 2013
(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/3`

(D)

`pi/2`

Solution:

Given equation of lines is

`x+y=1 => y =-x+1` .....................(i)

and `x - y == 1 => y = x - 1`................(ii)

`:.` Slope of line (i) is `m_1 =- 1`

and slope of line (ii) is `m_2 = 1`

`:. m_1 * m_2 = (- 1) (1) =- 1`

Thus, the angle between both lines is `pi/2`
Correct Answer is `=>` (D) `pi/2`
Q 2307045888

The centroid of the triangle With vertices

`(2, 3), (-2,- 5)` and `(3, 5)` is at

NDA Paper 1 2013
(A)

`(1,1)`

(B)

`(2,-1)`

(C)

`(1,-1)`

(D)

`(1,2)`

Solution:

Let the vertices of the triangle are,

`P(x_1, y_1) -> (2, 3)`

`Q(x_2, Y_2 ) -> (-2, -5)`

and `R (x_3, y_3) -> (3, 5)`

`:.` Centroid of `Delta PQR`

`= ( (x_1 +x_2 +x_3)/3, (y_1+y_2 +y_3)/3)`

`=( (2-2 +3)/3, (3-5+5)/3)= (1,1)`
Correct Answer is `=>` (A) `(1,1)`
Q 2377145986

The equation of the line, the reciprocals of whose
intercepts on the axes are `m` and `n`, is given by
NDA Paper 1 2013
(A)

`nx + my= mn`

(B)

`mx + ny = 1`

(C)

`mx + ny = mn`

(D)

`mx- ny = 1 `

Solution:

We know that, the equation of straight l'ine in intercept
form is

`x/a + y/b =1` ................(i)

By given condition, `a = 1/m ` and `b =1/n`

On putting the values of `a + b` in Eq. (i), we get

`x/(1/m) + y/(1/n) =1 => mx+ny =1`

which is the required equation of straight line.
Correct Answer is `=>` (B) `mx + ny = 1`
Q 2307156088

What is the equation of a straight fine which
passes through `( 3, 4)` and the sum of whose `x` and `y`
intercepts is `14`?
Paper 1 2013
(A)

`4x + 3y =24`

(B)

`x+ y = 7`

(C)

`4x- 3y= 0`

(D)

`3x + 4y=25`

Solution:

The equation of line in intercept form is

`x/a+ y/b =1` ..................(i)

where a and b are length of intercepts of the line with `X` and
`Y`-axes, respectively.

Now, by given condition,

Sum of `x` and `y`-intercepts `= 4`

`=> a + b = 14` ... (ii)

Since, the line (i) passes through the point `(3, 4)` then

`3/a +4/b =1`

`=> 3/a+ 4/(14-a) =1`, [from Eq. (ii) ]


`=>3(14-a)+ 4a= a(14-a)`

`=> 42-3a+4a=14a-a^2`

`=> a^2- 13 a + 42= 0`.

`=> a^2-6a-7a+ 42 =0`

`=> a(a- 6)-7(a-6) = 0`

`=> (a- 6)(a-7) = 0` ·

`=> a= 6,7`

`=> b = 8,7`

Hence, the required equation of straight line

`x/6 + y/8 =1` (when, `a= 6` and `b=8`)

`=> 8x+6y -48 =0`

`=> 4x +3y -24 =0`

and `x/7 +y/7 =1` (when,`a=b=7`)

`=> x+y =7`
Correct Answer is `=>` (B) `x+ y = 7`
Q 2317156089

What is the equation of a straight fine which
passes through `( 3, 4)` and the sum of whose `x` and `y`
intercepts is `14` ?
Paper 1 2013
(A)

`4x + 3y =24`

(B)

`x+ y = 7`

(C)

`4x- 3y= 0`

(D)

`3x + 4y=25`

Solution:

The equation of line in intercept form is

`x/a+ y/b =1` ..................(i)

where a and b are length of intercepts of the line with `X` and
`Y`-axes, respectively.

Now, by given condition,

Sum of `x` and `y`-intercepts `= 4`

`=> a + b = 14` ... (ii)

Since, the line (i) passes through the point `(3, 4)` then

`3/a +4/b =1`

`=> 3/a+ 4/(14-a) =1`, [from Eq. (ii) ]


`=>3(14-a)+ 4a= a(14-a)`

`=> 42-3a+4a=14a-a^2`

`=> a^2- 13 a + 42= 0`.

`=> a^2-6a-7a+ 42 =0`

`=> a(a- 6)-7(a-6) = 0`

`=> (a- 6)(a-7) = 0` ·

`=> a= 6,7`

`=> b = 8,7`

Hence, the required equation of straight line

`x/6 + y/8 =1` (when, `a= 6` and `b=8`)

`=> 8x+6y -48 =0`

`=> 4x +3y -24 =0`

and `x/7 +y/7 =1` (when,`a=b=7`)

`=> x+y =7`
Correct Answer is `=>` (B) `x+ y = 7`
Q 2317256180

What is the equation of a straight line which
passes through `( 3, 4)` and the sum of whose `x` and `y`
intercepts is `14`?
NDA Paper 1 2013
(A)

`4x + 3y =24`

(B)

`x+ y = 7`

(C)

`4x- 3y= 0`

(D)

`3x + 4y=25`

Solution:

The equation of line in intercept form is

`x/a+ y/b =1` ..................(i)

where a and b are length of intercepts of the line with `X` and
`Y`-axes, respectively.

Now, by given condition,

Sum of `x` and `y`-intercepts `= 4`

`=> a + b = 14` ... (ii)

Since, the line (i) passes through the point `(3, 4)` then

`3/a +4/b =1`

`=> 3/a+ 4/(14-a) =1`, [from Eq. (ii) ]


`=>3(14-a)+ 4a= a(14-a)`

`=> 42-3a+4a=14a-a^2`

`=> a^2- 13 a + 42= 0`.

`=> a^2-6a-7a+ 42 =0`

`=> a(a- 6)-7(a-6) = 0`

`=> (a- 6)(a-7) = 0` ·

`=> a= 6,7`

`=> b = 8,7`

Hence, the required equation of straight line

`x/6 + y/8 =1` (when, `a= 6` and `b=8`)

`=> 8x+6y -48 =0`

`=> 4x +3y -24 =0`

and `x/7 +y/7 =1` (when,`a=b=7`)

`=> x+y =7`
Correct Answer is `=>` (B) `x+ y = 7`
Q 2357456384

The point whose abscissa is equal to its ordinate
and which is equidistant from `A (-1 0)` and `B (0,5)`
is
NDA Paper 1 2013
(A)

`(1,1)`

(B)

`(2,2)`

(C)

`(-2,-2)`

(D)

`(3,3)`

Solution:

Let the point whose abscissa is equal to its ordinate is
`P(x,x)`.

Its distance from `A(-1, 0)` and `B(0,5)` is equal.

i.e., `PA= PB`

`=> PA^2=PB^2`

`=> (x+1)^2 + (x-0)^2 = (x-0)^2 + (x-5)^2`

`=> x^2+ 1 + 2x + x^2= x^2+x^2+25-10x`

`=> 1 + 2x = 25 - 10x`

`=> 12x= 24`

`:. x= 2`

So, the required point is `P(x,x)= P(2, 2)`.
Correct Answer is `=>` (B) `(2,2)`
Q 2357556484

What is area of the triangle whose vertices are
`(3, 0), (0, 4)` and `(3, 4)`?
NDA Paper 1 2013
(A)

`6` sq units

(B)

`7.5` sq units

(C)

`9` sq units

(D)

`12` sq units

Solution:

Let the vertices of the triangle are

`(x_1, y_1 ) -> (3, 0)`

`(x_2, y_2 ) -> (0, 4)`

`(x_3, y_3 ) -> (3, 4)`

Now, area of triangle `= 1/2 | | (x_1, y_1,1), (x_2,y_2,1), (x_3, y_3 ,1) | |`

`=1/2 | | (3,0,1),(0,4,1),(3,4,1) | |`

`=1/2 | [ 3 (4-4) -0 +1 (0-12) ] |`

`=1/2 |[0-0-12] | = 1/2 xx 12`

`=6 ` units
Correct Answer is `=>` (A) `6` sq units
Q 2317656580

A straight line passes through the poits `(5, 0)` and `(0, 3)`. The length of the perpendicular from
the point `(4, 4)` on the line is
NDA Paper 1 2013
(A)

`sqrt(17)/2`

(B)

`sqrt (17/2)`

(C)

`15/sqrt(34)`

(D)

`17/2`

Solution:

A line which passes through the points `( 5, 0)` and `(0, 3)` is

`(y-0)=(3-0)/(0-5) (x+5)`

`=> -5y = 3x -15`

`=> 3x+5y -15 =0` ..............(i)


Now, length of the perpendicular from the point ( 4, 4( on the
line (i) is

`= ( | 3(4) +5(4) -15 |)/(sqrt ( (3)^2 + (5)^2 ) ) = ( |12+20 -15 | )/(sqrt (9+25) )`

`=17/sqrt(34) = sqrt(17/2)`
Correct Answer is `=>` (B) `sqrt (17/2)`
Q 2317756680

What is the inclination of the line `sqrt (3)x - y = 0`?
NDA Paper 1 2013
(A)

`30^(circ)`

(B)

`60^(circ)`

(C)

`135^(circ)`

(D)

`150^(circ)`

Solution:

Given equation of line,

`sqrt(3)x-y -1 =0`

`=> y= sqrt (3)x -1`

On comparing with `y= mx + c`, we get

`m = sqrt(3)` `( :. m = tan theta )`

`=> tan theta = sqrt(3) = tan 60^(circ)`

`=> theta = 60^(circ)`

So, nclination of the given line is `60^(circ)`.
Correct Answer is `=>` (B) `60^(circ)`
Q 2387756687

If two straight line paths are representee!. by the
equations `2x - y = 2` and `-4x + 2 y = 6`, then the
paths will
NDA Paper 1 2013
(A)

cross each other at one point

(B)

not cross each other

(C)

cross each other at two points

(D)

cross each other at infinitely many points

Solution:

Given equation of straight lines,

`2x- y-2=0` ..................(i)

and `-4x+ 2y-6= 0` .................(ii)

On comparing with `ax + by+ c = 0`, we get

`a_1 =2, b_1 =-1` and `c_1 =-2`

and `a_2= -4 ,b_2 =2`


and `c_2 = -6`

Now, `a_1/a_2 = -1/2 ,b_1/b_2=-1/2` and `c_1/c_2 =1/3`

Since, `a_1/a_2 = b_1/b_2 ne c_1/c_2`

Hence, both straight lines are parallel i.e., they never cross each
other.
Correct Answer is `=>` (D) cross each other at infinitely many points
Q 2317756689

What is the area of the triangle bounded! by the
side `x =0 ,y =0` and `x + y =2`?
NDA Paper 1 2013
(A)

1 sq unit

(B)

2 sq units

(C)

4 sq units

(D)

8 sq units

Solution:

Given sides are

`x= 0, y=0` and `x+ y= 2`

`:.` Required area of `Delta ABC =1/2 xx OA xx OB`

`=1/2 xx 2 xx 2 =2 ` sq units
Correct Answer is `=>` (B) 2 sq units
Q 2377156986

The points `(5, 1 ), ( 1, -1)` and `(11, 4)` are
NDA Paper 1 2012
(A)

collinear

(B)

vertices of right angled triangle

(C)

vertices of equilateral triangle

(D)

vertices of an isosceles triangle

Solution:

The points `(5, 1), (1,- 1)` and `(11, 4)` are collinear, if

`x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0`

Here,`x_1 =5,y_1 =1,x_2 =1,y_2 =-1,x_3 =11 ,y_3 =4`

From Eq. (i),

`5 (- 1- 4) + 1 (4- 1) + 11(1 + 1) = 0`

`=> 5(-5)+1(3)+22=0`

`=> -25 +3 +22 =0`

`=> 0 =0`

Hence, the points `(5, 1), (1, -1)` and `(11, 4)` satisfy the given Eq. (i).
So, they are collinear.
Correct Answer is `=>` (A) collinear
Q 2367856785

The locus of a point equidistant from three
collinear points is
NDA Paper 1 2012
(A)

a straight line

(B)

a pair of points

(C)

a point

(D)

the null set

Solution:

Let the three points `A(3, 1), 8 (12,- 2)` and `C(O, 2)` are
coli near and the point `P(h, k)` are equidistant from these points `A,
B` and `C`.

Now, ` PA^2 = PB^2 = PC^2`

`=> (h- 3)^2 + (k- 1)^2 = (h- 12)^2 = (h- 10)^2 + (k- 2)^2`

`=> h^2 + k^2 = 6h- 2k + 10 = h^2 + k^2 - 24h + 4k + 148`

`= h^2 + k^2 - 4k + 4`

Taking first and third, we get `3h- k = - 3` ... (i)

Taking second and third, we get `3h- k = 18` ... (ii)

Since, Eqs. (i) and (ii) are two parallel lines.

Hence, the locus will be a null set.
Correct Answer is `=>` (D) the null set
Q 2317156980

The equation to the locus of a point which is
always equidistant from the points `(1, 0)` and
`(0, -2)` is
NDA Paper 1 2012
(A)

`2x + 4y + 3 = 0`

(B)

`4x + 2y + 3 = 0`

(C)

`2x + 4y- 3 = 0`

(D)

`4x + 2y- 3 = 0`

Solution:

Let `A(1, 0)` and `B(0,- 2)` are two given points and let

`P(h, k)` be any variable point, then
According to tho question,

`PA = PB`

`=> PA^2 = PB^2`

`=> (h - 1)^2 + (k -0)^2 = (h - 0)^2 + (k + 2)^2`

`=> h^2 + 1 - 2h + k^2 = h^2 + k^2 + 4 + 4k`

`=> 4k + 2h + 3= 0`

`=> 2h + 4k + 3 = 0`

Hence, locus of point `P (h, k)` is `2x + 4y + 3 = 0`.
Correct Answer is `=>` (A) `2x + 4y + 3 = 0`
Q 2307156988

The points `(5, 1 ), ( 1, -1)` and `(11, 4)` are
NDA Paper 1 2012
(A)

collinear

(B)

vertices of right angled triangle

(C)

vertices of equilateral triangle

(D)

vertices of an isosceles triangle

Solution:

The points `(5, 1), (1,- 1)` and `(11, 4)` are collinear, if

`x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0`

Here,`x_1 =5,y_1 =1,x_2 =1,y_2 =-1,x_3 =11 ,y_3 =4`

From Eq. (i),

`5 (- 1- 4) + 1 (4- 1) + 11(1 + 1) = 0`

`=> 5(-5)+1(3)+22=0`

`=> -25 +3 +22 =0`

`=> 0 =0`

Hence, the points `(5, 1), (1, -1)` and `(11, 4)` satisfy the given Eq. (i).
So, they are collinear.
Correct Answer is `=>` (A) collinear
Q 2357167084

What is the perpendicular distance between the
parallel lines `3x + 4y =9` and `9x + 12y + 28 =0`?
NDA Paper 1 2012
(A)

`7/3` units

(B)

`8/3` units

(C)

`10/3` units

(D)

`11/3` units

Solution:

We know that, if two parallel lines `ax + by + c_1= 0` and


`ax +by +c_2 =0` , then the distance between them is

`| (c_2 -c_1)/ (sqrt (a^2 +b^2) )|`

For lines `3x + 4 y - 9 = 0` ...............(i)

and `9x + 12 y + 28 = 0`

`=> 3x +4y +28/3 =0` ..................(ii)

`:.` Distance between them `= | (28/3 +9 )/(sqrt(9+16)) | = |55/3 xx 1/5 | = 11/3` units
Correct Answer is `=>` (D) `11/3` units
Q 2347267183

If `p, q, r` and `s` be the distances from origin of the points `(2, 6), (3, 4), (4, 5)` and `(-2, 5)` respectively,
then which one of the following is a whole number?
NDA Paper 1 2012
(A)

`p`

(B)

`q`

(C)

`r`

(D)

`s`

Solution:

Distance between `(2, 6)` and `(0, 0)`

`p = sqrt(36 +4) = sqrt(40)`

Distance between `(3, 4)` and `(0, 0)`

`q= sqrt(9+16) =5`

Distance between `(4, 5)` and `(0, 0)`

`r= sqrt (16 +25) = sqrt(41)`

Distance between `(- 2, 5)` and `(0, 0)`

`s= sqrt((-2)^2 + (5)^2) = sqrt(4+25) = sqrt (29)`

Clearly, the value of `q` is a whole number.
Correct Answer is `=>` (B) `q`
Q 2317367289

From the point `( 4, 3)` a perpendicular is drapped
on the `X`-axis as well as on the `Y`-axis. If the lengths of perpendiculars are `p` and `q`, respectiveb~ then
which one of the follovring is correct?
NDA Paper 1 2012
(A)

`p =q`

(B)

`3p =4q`

(C)

`4p =3q`

(D)

`p+q =5`

Solution:

From adjoining figure,

`P=3` and `q=4`

`:. 4p= 3q`
Correct Answer is `=>` (C) `4p =3q`
Q 2357467384

What is the value of `lambda`, if the straight line `(2x + 3 y + 4) + lambda (6x - y + I 2) = 0` is parallel to
`Y`-axis?
NDA Paper 1 2012
(A)

`3`

(B)

`-6`

(C)

`4`

(D)

`-3`

Solution:

Given, `(2x + 3y + 4) + lambda (6x - y + 12) = 0`

`=> 2x + 6 lambda x + 3y - lambda y + 4 + 12 lambda = 0`

`=> 2x (3 lambda + 1) + y(3 - lambda )+ 4 + 12 lambda = 0` ... (i)

Since, line (i) is parallel to `Y`-axis.

Since, the coefficient of `y` must be zero.

`:. 3 - lambda = 0 => lambda =3`
Correct Answer is `=>` (A) `3`
Q 2327767681

The line `y = 0` divides the line joining the points
`(3, -5)` and `(-4, 7)` in the ratio
NDA Paper 1 2012
(A)

`3:4`

(B)

`4:5`

(C)

`5:7`

(D)

`7:9`

Solution:

Let the line `y = 0` divides the line joining the points
`(3,- 5)` and `(-4, 7)` in the ratio `n : m`, then

By internal section formula,

`x= (m(3) +n(-4) )/(m+n), = (3m -4n)/(m+n)`

`= y = (m(-5) +n (7))/(m+n) = (-5m + 7n)/(m+n)`

Given, `y= 0 => (-5m +7n)/(m+n) =0`

`=> 5m =7n => m/n = 7/5 => n/m =5/7`

`:. n:m = 5:7`
Correct Answer is `=>` (C) `5:7`
Q 2317767689

The equation of a straight line which makes an
angle `45^(circ)` with the `X`-axis with `y`-intercept
`101` units is
NDA Paper 1 2012
(A)

`10x + 101y = 1`

(B)

`101x + y = 1`

(C)

`x + y- 101 = 0`

(D)

`x- y + 101 = 0`

Solution:

We know that, if the line making an angle `theta` with the
positive direction of `X`-axis with `y`- intercept asc. Then, equation of
the line is `y = mx + c = tan theta * x + e`

`:. theta = 45^(circ)` and `c = 101` units

`:. y =tan 45^(circ) * x + 101 => y = 1 * x + 101 => x- y + 101 = 0`
Correct Answer is `=>` (D) `x- y + 101 = 0`
Q 2377867786

If the points `(2, 4), (2, 6)` and `(2 + sqrt (3) , k)` are the
vertices of an equilateral triangle, then what is the
value of `k`?
NDA Paper 1 2012
(A)

`6`

(B)

`5`

(C)

`-3`

(D)

`1`

Solution:

Let `A= (2, 4), B = (2, 6)` and `C = (2 + sqrt( 3 ) ,k)`

`AB = sqrt ( (2-2)^2 + (6-4)^2 ) = sqrt( (2)^2 ) =2`

`BC = sqrt ( (2+ sqrt (3) -2)^2 + (k-6)^2 ) = sqrt(3+ (k-6)^2)`

`= sqrt (3+ k^2 +36 -12k)= sqrt (k^2 -12k +39)`

`AC = sqrt ( (2+ sqrt (3) -2)^2 + (k-4)^2)`

`AC =sqrt(3+k^2 +16 -8k) =sqrt (k^2 -8k +19)`

Since, `ABC` is an equilateral triangle.

`:. AB =BC =CA =>AB^2 =BC^2 =CA^2`

`=> underset (I) (4) = underset (II) (k^2 -12k +39) = underset(III) ( k^2 -8k + 19)`

Taking II and Ill,

`k^2 -12k +39 = k^2 -8k +19`

`=>-12k +39 +8k -19 =0 => -4k +20 = 0`

`=> -4k = -20 =>k =5`
Correct Answer is `=>` (B) `5`
Q 2377378286

What is the equation of line passing thrqugh
`(0, 1)` and making an angle with the `Y`-axis equal
to the inclination of the line `x - y = 4` with `X`-axis?
NDA Paper 1 2012
(A)

`y = x + 1`

(B)

`x = y + 1`

(C)

`2x = y + 2`

(D)

None of these

Solution:

Since, the line passes through the point `(0,1)` and
making an angle with `Y`-axis which is equivalent to the slope of the
line `y = x- 4`.

i.e., `theta = 45^(circ) =>tan theta = 1 =m`

`:.` Equation of line is

`(y-1) =m (x-0) = 1(x)`

`=> y = x+1`
Correct Answer is `=>` (A) `y = x + 1`
Q 2367478385

What is the distance between the lines
`3x + 4y = 9` and `6x + 8 y = 18?`
NDA Paper 1 2012
(A)

`0`

(B)

`3` units

(C)

`9` units

(D)

`18` units

Solution:

Since, both equation of lines are same that means both
lines are coincident to each other

`3x + 4y = 9` ... (i)

and `6x +By= 18`

`=> 3x + 4y= 9` ... (ii)

So, distance between two coincident line is zero.
Correct Answer is `=>` (A) `0`
Q 2347578483

What is the perimeter of the triangle with vertices
`A(-4, 2), B(0, -1)` and `C(3, 3)?`
NDA Paper 1 2012
(A)

`7 + 3 sqrt (2)`

(B)

`10+5 sqrt(2)`

(C)

`11 + 6 sqrt(2)`

(D)

`5+ sqrt (2)`

Solution:

Given that, the vertices of the triangle are
`A(-4, 2), B(0,-1)` and `C(3, 3)`.

`:.` Length of `AB = sqrt ( (-4-0)^2 + (2+1)^2 ) = sqrt(16 xx 9) = sqrt(25) = 5`

Length of `BC = sqrt ( (0-3)^2 + (-1-3)^2) = sqrt(9+16) =sqrt (25) =5`

Length of `CA = sqrt ( (3+4)^2 + (3-2)^2) =sqrt(49+1) = sqrt(50) = 5 sqrt(2)`

`:.` Perimeter of `Delta ABC = (AB +BC +CA)`

`= (5+5 +5 sqrt(2)) = (10 + 5 sqrt (2))`
Correct Answer is `=>` (B) `10+5 sqrt(2)`
Q 2317678580

If the mid-point between the points `(a+b, a- b)`
and `(-a, b)` lies on the line `ax+ by=k`, what is the
value of `k`?
NDA Paper 1 2012
(A)

`a/b`

(B)

`a+b`

(C)

`ab`

(D)

`a-b`

Solution:

Given that, mid-point of the point `(a + b, a - b )` and
`(-a, b)` lies on tile line `ax+ by= k`

`:.` Mid -point` = [ ( (a+b) + (-a) )/2, ( (a-b) +b )/2] =[b/2, a/2]`

which lies on the line `ax + by = k`.

`=> a*b/2 + b*a/2 = k => k = ab .`
Correct Answer is `=>` (C) `ab`
Q 2347778683

The acute angle which the perpendicular from
origin on the line `7 x-3 y=4` makes with the `x`-axis
is
NDA Paper 1 2012
(A)

zero

(B)

positive but not `pi / 4`

(C)

negative but not `pi/4`

(D)

`pi/4`

Solution:

The equation of perpendicular line to `7x - 3y = 4` is

Slop of this line `m = 7/3` angle with x-axis `tan^(-1)(7/3)`


Slop of perpendicular line the given line `= - 1/m = -3/7`

angle ` = tan^(-1)(-3/7) = - tan^(-1)(3/7)`
Correct Answer is `=>` (C) negative but not `pi/4`
Q 2317078880

What is the perpendicular distance of the point
`(x,y)` from `X`-axis?
NDA Paper 1 2012
(A)

`x`

(B)

`y`

(C)

`|x|`

(D)

`|y|`

Solution:

Tile perpendicular distance of the point `(x_1 ,y_1)` to line

`ax + by + c = 0` is

`p= ( |ax_1 + by_1 +c | )/(sqrt (a^2 + b^2) )`

The equation of `X`-axis is `y = 0`.

`:. p = ( | y | )/(sqrt ((1)^2 ) ) = |y |`
Correct Answer is `=>` (D) `|y|`
Q 2367180085

For what value of `k`, are the lines `x + 2y -9 =0`
and `kx + 4y + 5 = 0` parallel?
NDA Paper 1 2011
(A)

`2`

(B)

`-1`

(C)

`1`

(D)

`0`

Solution:

If two lines `a_1 x + b_1 y + c_1 =0` and

`a_2 x+b_2 y + c_2 =0` are parallel, then `a_1/a_2 =b_1/b_2 =c_1/c_2`

Then, for the lines `x + 2y- 9 = 0` and `kx + 4y + 5 =0`

`1/k = 2/4 = -9/5`

Taking the first two parts,

`k=2`
Correct Answer is `=>` (A) `2`
Q 2367280185

What is the equation of a line parallel to `X-` axis at
a distance of `5` units below `X`-axis?
NDA Paper 1 2011
(A)

`x=5`

(B)

`x= -5`

(C)

`y = 5`

(D)

`y=-5`

Solution:

The equation of the line parallel tp `X-` axis at a distance
of `5` units below `X`-axis, i.e., at `(0, - 5)` is `y = - 5`.
Correct Answer is `=>` (D) `y=-5`
Q 2387380287

What are the coordinates of the foot of the
perpendicular from the point `(2, 3)` on the line
`x+ y- 11 = 0`?
NDA Paper 1 2011
(A)

`(2,9)`

(B)

`(5,6)`

(C)

`(-5, 6)`

(D)

`(6,5)`

Solution:

Let the point of foot of the
perpendicular be `M(x_1, y_1)`.

Now, `PM bot AB`

Slope of line `AB` is

`m_1 = -1`

and slope of line `MP` is

`m_2= (y_1 -3)/(x_1 -2)`

`=>m_1m_2 = -1`

`=> -1 ((y_ -3)/(x_1 -2)) = -1`

`=> y_1 -3 =x_1 -2 => x_1 -y_1 +1 =0` ..........(i)

Since, the point `M` lies on the line `AB`, then

`x_1 + y_1 -11 =0`

On solving Eqs. (i) and (ii), we get

`2x_1 -10 =0 => x_1 =5` and `y_1 =6`


So, required foot of the perpendicular `M` is `(5, 6)`.
Correct Answer is `=>` (B) `(5,6)`
Q 2367480385

If `(p, q)` be the point on the `X`-axis equidistant
from the points `(1, 2)` and `(2, 3)`, then which one
of the following is correct ?
NDA Paper 1 2011
(A)

`p = 0, q = 4`

(B)

`p = 4, q = 0`

(C)

`P= 3/2,q = 0`

(D)

`P= 1,q = 0`

Solution:

Let `A(p, q)` be the point on the `X`-axis which is
equidistant from the points `B(1, 2)` and `C(2, 3)`, then

`AB= AC`

`=> AB^2 = AC^2`

`=> (p-1)^2 + (q -2)^2 = (p- 2)^2 + (q - 3)^2`

`=> p^2 + 1- 2p + q^2 + 4- 4q = p^2 + 4- 4p + q^2 + 9- 6q`

`=> 2p + 2q = 8`

`=> p + q = 4` ... (i)

Since, the value of `p = 4` and `q = 0` satisfies the Eq. (i) and on the

`X`-axis `q` must be zero, so `(p, q) = (4, 0)`.
Correct Answer is `=>` (B) `p = 4, q = 0`
Q 2307480388

If `p` is the length of the perpendicular drawn from
the origin to the line `x/a + y/b =1`, then which one of

the following is correct?
NDA Paper 1 2011
(A)

`1/p^2 = 1/a^2 + 1/b^2`

(B)

`1/p^2 =1/a^2 -1/b^2`

(C)

`1/p = 1/a + 1/b`

(D)

`1/p= 1/a -1/b`

Solution:

Given equation of line is `x/a + y/b =1` ............(i)

Then, the length of perpendicular drawn from the origin to the line

`(i) =p` (given)

`=> ( |1/a xx 0 +1/b xx 0-1 | )/(sqrt ( (1/a)^2 + (1/b)^2 ) ) =p`

`=> (| -1| )/( sqrt ( 1/a^2 +1/b^2) ) =p`

`=> 1/p= sqrt (1/a^2 + 1/b^2)`

`:. 1/p^2 = 1/a^2 + 1/b^2`
Correct Answer is `=>` (A) `1/p^2 = 1/a^2 + 1/b^2`
Q 2387780687

What is the locus of a point which moves
equidistant from the coordinate axes?
NDA Paper 1 2011
(A)

`x pm y = 0`

(B)

`x+ 2y =0`

(C)

`2x + y =0`

(D)

None of these

Solution:

Since, the lines `y = x` and `y = - x` lie at the same
distances in coordinate axes.

`:. y = pm x => x pm y =0`

So, `x pm y = 0` is the locus of a point which moves equidistant
from the coordinate axes.
Correct Answer is `=>` (A) `x pm y = 0`
Q 2317880780

What is the equation of the line joining the origin
with the point of intersection of the lines
`4x + 3 y = 12` and `3x + 4y = 12`?
NDA Paper 1 2011
(A)

`x+y =1`

(B)

`x-y = 1`

(C)

`3y = 4x`

(D)

`x =y`

Solution:

The equation of given lines are

`4x+ 3y =12`, ................(i)
and `3x+ 4y =12` .................(ii)

`=> x= 12/7` and `y = 12/7`

Since, point of intersection of the given line is `(12/7 ,12/7)`

`:.` Equation of line passing through `(0, 0)` and `(12/7 ,12/7)` is

`y-0 = (12/7 -0)/(12/7 -0) (x-0)`

`:. y =x`
Correct Answer is `=>` (D) `x =y`
Q 2307880788

If the sum of the squares of the distances of the
point `(x,y)` from the points `(a, 0)` and `(-a, 0)` be
`2b^2` , then which one of the following is correct?
NDA Paper 1 2011
(A)

`x^2 + a^2 = b^2 + y^2`

(B)

`x^2 + a^2 = 2b^2 -y^2`

(C)

`x^2 -a^2 = b^2 +y^2`

(D)

`x^2 + a^2 = b^2- y^2`

Solution:

Let the sum of the distances of the point `P(x, y)` from
the points `A(a, 0)` and `B(-a, 0)` be `2b^2`.

`:. PA^2 + PB^2 = 2b^2`

`(x - a)^2 + (y- 0)^2 + (x + a)^2 + (y- 0)^2 = 2b^2`

`=> x^2 + a^2- 2ax + y^2 + x^2 + a^2 + 2ax + y^2 = 2b^2`

`=> x^2 + a^2 + y^2 = b^2`

`=> x^2 + a^2 = b^2 - y^2`
Correct Answer is `=>` (D) `x^2 + a^2 = b^2- y^2`
Q 2337080882

If the line `mx + ny = 1` passes through the points
`( 1, 2)` and `(2, 1 )`, then what is the value of `m`?
NDA Paper 1 2011
(A)

`1`

(B)

`3`

(C)

`1/2`

(D)

`1/3`

Solution:

Since, the line `mx + ny = 1` passes through `(1, 2)` and `(2,1)`

`:. m+2n =1,` ............(i)

and `2m + n =1` ..............(ii)

From Eqs. (i) and (ii), `m = n`

From Eq. (i), `m + 2m= 1`

`:. m =1/3`
Correct Answer is `=>` (D) `1/3`
Q 2307080888

What is the equation of the line passing through
`(2,- 3)`, and parallel to `Y`-axis?
NDA Paper 1 2011
(A)

`y= -3`

(B)

`y =2`

(C)

`x=2`

(D)

`x= -3`

Solution:

The equation of line passing through `(2, - 3)` and
parallel to`Y`-axis is `(y + 3) =tan 90 (x- 2) => x - 2 = 0`

`=> x = 2`.
Correct Answer is `=>` (C) `x=2`
Q 2317080889

What is the equation of the line passing through
`(2,- 3)`, and parallel to `Y`-axis?
NDA Paper 1 2011
(A)

`y= -3`

(B)

`y =2`

(C)

`x=2`

(D)

`x= -3`

Solution:

The equation of line passing through `(2, - 3)` and
parallel to`Y`-axis is `(y + 3) =tan 90 (x- 2) => x- 2 = 0`

`=> x = 2`.
Correct Answer is `=>` (C) `x=2`
Q 2367180985

What is the locus of the point which is at a
distance of `8` units to the left of `Y`-axis?
NDA Paper 1 2011
(A)

`X= 8`

(B)

`Y=8`

(C)

`X= - 8`

(D)

`Y = -8`

Solution:

Required locus is `X = - 8` which is at a distance of
`8` units to the left of `Y`-axis.
Correct Answer is `=>` (C) `X= - 8`

Problem Set 3

Set 3
Q 2387180987

Two straight lines `x-3y -2 =0` and
`2x- 6y- 6 =0`
NDA Paper 1 2011
(A)

never intersect

(B)

intersect at a single point

(C)

intersect at infinite number of points

(D)

intersect at more than one points (but finite number of points)

Solution:

Here, `a_1 = 1, a_2 = 2, b_1 =- 3, b_2 =- 6, c_1= -2` and `c_2 = -6`

`:. a_1/a_2 =1/2 , b_1/b_2 = 1/2, c_1/c_2 =1/3`

`=> a_1/a_2 =b_1/b_2 ne c_1/c_2`

So, both the straight lines never intersect because both lines are
parallel to each other.
Correct Answer is `=>` (A) never intersect
Q 2317191080

If `(a, 0), (0, b)` and `(1, 1)` are collinear, then what
is `(a + b - ab)` equal to?
NDA Paper 1 2011
(A)

`2`

(B)

`1`

(C)

`0`

(D)

`-1`

Solution:

Since, `(a, 0), (0, b)` and `(1, 1)` are collinear

`:. | (a,0,1), (0,b,1), (1,1,1) | =0`

`=> a(b-1) +1 (0-b) =0`

`=> ab -a -b =0 =>a+b -ab =0`
Correct Answer is `=>` (C) `0`
Q 2347191083

What is the equation to the straight line joining
the origin to the point of intersection of the lines

`x/a + y/b =1` and `x/b + y/a = 1`?


NDA Paper 1 2010
(A)

`x+y =0`

(B)

`x+y +1= 0`

(C)

`x-y = 0`

(D)

`x+y +2 =0`

Solution:

We know that, the equation of the straight line passing
through the intersection point of two lines `x/a + y/b =1` and

`x/b + y/a =1`, is

`(x/a + y/b -1) + lambda (x/b + y/a -1) = 0` ..............(i)

Since. this line pass through the origin.

`:. (0- 1) + lambda (0- 1) = 0 => lambda = -1`

On putting the value of `lambda` in Eq. (i), we get

`(x/a + y/b -1) -1 (x/b +y/a -1) =0`

`=> x/a + y/b -1 -x/b -y/a+1 =0`

`=> x (1/a -1/b) -y (1/a -1/b) =0`

`:. x-y =0`
Correct Answer is `=>` (C) `x-y = 0`
Q 2387191087

If the straight lines `x-2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`, then what is the value of `k` ?
Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`1/2`

(D)

`-1/2`

Solution:

Since, the straight lines `x- 2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`

Since, the point `(1,1/2)` will satisfy the equation `kx + y = 1`.

`:. k* 1 +1/2 =1 => k =1/2`
Correct Answer is `=>` (C) `1/2`
Q 2317191089

If the straight lines `x-2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`, then what is the value of `k` ?
Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`1/2`

(D)

`-1/2`

Solution:

Since, the straight lines `x- 2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`

Since, the point `(1,1/2)` will satisfy the equation `kx + y = 1`.

`:. k* 1 +1/2 =1 => k =1/2`
Correct Answer is `=>` (C) `1/2`
Q 2327291181

If the straight lines `x-2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`, then what is the value of `k` ?
Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`1/2`

(D)

`-1/2`

Solution:

Since, the straight lines `x- 2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`

Since, the point `(1,1/2)` will satisfy the equation `kx + y = 1`.

`:. k* 1 +1/2 =1 => k =1/2`
Correct Answer is `=>` (C) `1/2`
Q 2337291182

If the straight lines `x-2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`, then what is the value of `k` ?
NDA Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`1/2`

(D)

`-1/2`

Solution:

Since, the straight lines `x- 2y = 0` and `kx + y = 1`

intersect at the point `(1,1/2)`

Since, the point `(1,1/2)` will satisfy the equation `kx + y = 1`.

`:. k* 1 +1/2 =1 => k =1/2`
Correct Answer is `=>` (C) `1/2`
Q 2357291184

If a point `P` moves such that the difference of its
distances from two given points `(c, 0)` and `(- c, 0)`
is constant, then what is the locus of the point `P`?
NDA Paper 1 2010
(A)

Circle

(B)

Ellipse

(C)

Hyperbola

(D)

Parabola

Solution:

We know that, if `P` is such that the difference of its
distances from two given points is constant, then locus of `P` is a
hyperbola, which is the definition of hyperbola.
Correct Answer is `=>` (C) Hyperbola
Q 2377291186

What is the slope of the line perpendicular to the
line `x/4 + y/3 =1`?
NDA Paper 1 2010
(A)

`3/4`

(B)

`-3/4`

(C)

`-4/3`

(D)

`4/3`

Solution:

Since, the slope of the line `x/4 + y/3 =1` is `-3/4`

`:.` Slope of the line perpendicular to this line `= - (-1/(3/4)) =4/3`
Correct Answer is `=>` (D) `4/3`
Q 2317291189

If the area of a triangle with vertices `(- 3, 0), (3, 0)`
and `(0, k)` is `9` sq units, then what is the value of `k` ?
NDA Paper 1 2010
(A)

`3`

(B)

`6`

(C)

`9`

(D)

`12`

Solution:

Let the vertices of the `Delta ABC` be `A(-3, 0), B(3, 0)` and `C(0,k)`

`:.` Area of `Delta ABC = 1/2 | (-3,0,1), (3,0,1), (0,k,1) |`

`=> 9 =1/2 {-3 (-k) +1 (3k) }` (given)

`=> 18 = k + 3k`

`:. k =18/6 = 3`
Correct Answer is `=>` (A) `3`
Q 2327391281

What is the locus of points, the difference of whose distances from two points being constant?
NDA Paper 1 2010
(A)

Pair of straight lines

(B)

An ellipse

(C)

A hyperbola

(D)

A parabola

Solution:

We know that, the locus of the points, the difference of
whose distances from two points being constant, is a hyperbola.
(by definition of hyperbola)
Correct Answer is `=>` (C) A hyperbola
Q 2347391283

What is the image of the point `( 1, 2)` on the line

`3x + 4y -1 = 0`?
NDA Paper 1 2010
(A)

`(-7/5, -6/5)`

(B)

`(7/8 ,1/2)`

(C)

`(7/8, -1/2)`

(D)

`(-7/5 , 1/2)`

Solution:

Let `(alpha , beta)` be the image of point `(1, 2)` w.r.t. line

`3x + 4y -1 = 0`.

So, `( (alpha +1)/2 , (beta +2)/2)` will be on the line `3x + 4y- 1 = 0`.

`=> 3 ((alpha +1)/2) +4 ((beta +2)/2) -1 =0`

`=> 3 alpha +3 + 4 beta + 8 -2 =0`

`=> 3 alpha +4 beta + 9 =0`


which is satisfied by `(-7/5, -6/5)`

Thus, the image of point `(1, 2)` is `(-7/5,-6/5)`
Correct Answer is `=>` (A) `(-7/5, -6/5)`
Q 2357391284

If the lines `3 y + 4x = 1, y = x + 5` and `5 y + bx =3`
are concurrent, then what is the value of `b`?
NDA Paper 1 2010
(A)

`1`

(B)

`3`

(C)

`6`

(D)

`0`

Solution:

The equation of given lines are

`3y + 4x = 1` ................(i)

`y= x+ 5`.............(ii)

and `5y + bx = 3`

On solving Eqs. (i) and (ii), we get

`x = -2` and `y = 3`

If there lines are concurrent, then these values must satisfy the
third equation,

`15-2b=3`

`=> 2b = 12`

` :. b=6`
Correct Answer is `=>` (C) `6`
Q 2367391285

If `(- 5, 4)` divides the line segment between the
coordinate axes in the ratio `1 : 2`, then what is its
equation?
NDA Paper 1 2010
(A)

`8x + 5y + 20 = 0`

(B)

`5x + 8y- 7 = 0`

(C)

`8x - 5y + 60 = 0`

(D)

`5x- 8y+ 57= 0`

Solution:

Let `A(a, 0)` and `B(0, b)` be two points on respective

coordinate axes and `(-5, 4)` divides `AB` in the ratio `1 :2`.

`:. -5= (1 xx 0 +2 xx a)/3`

`=> a= -15/2`

and `4= (1 xx b +2 xx 0)/3`

`=> b =12`

Hence, the equation of line joining `(-15/2,0)` and `(0,12)` is

`(y-0) = (12-0)/(0+ 15/2)* (x+ 15/2)`

`=> y= 4/5 (2x +15)`

`=> 5y = (8x +60)`

`:. 8x -5y + 60 =0`
Correct Answer is `=>` (C) `8x - 5y + 60 = 0`
Q 2377391286

What is the product of the perpendiculars from

the two points ` (pm sqrt (b^2 -a^2 , 0) )` to the line

`ax cos phi + by sin phi = ab ?`


NDA Paper 1 2009
(A)

`a^2`

(B)

`b^2`

(C)

`ab`

(D)

`a/b`

Solution:

Given that, `ax cos phi + by sin phi - ab =0`

At point `(sqrt (b^2 -a^2) )`,

`d_1 = | (a sqrt (b^2 -a^2) cos phi -ab)/ (sqrt (a^2 cos^2 phi + b^2 sin^2 phi) ) |`

At point `(- sqrt (b^2 -a^2 ),0 )`,

`d_2 = | (-a sqrt (b^2 -a^2) cos phi - ab )/ (sqrt (a^2 cos^2 phi + b^2 sin^2 phi) ) |`

`:. d_1 d_2 = | - ( [a^2 (b^2 -a^2) cos^2 phi - a^2 b^2 ] )/(a^2 cos^2 phi + b^2 sin^2 phi) |`

`= | - (a^2 (-b^2 sin^2 phi - a^2 cos^2 phi))/ (a^2 cos^2 phi + b^2 sin^2 phi) |`

`= a^2`
Correct Answer is `=>` (A) `a^2`
Q 2387391287

The middle point of the segment of the straight
line joining the points `(p, q)` and `(q, - p)` is
`(r /2, s /2)`. What is the length of the segment?
NDA Paper 1 2009
(A)

`[(s^2 +r^2)^(1/2)]/2`

(B)

`[ (s^2 +r^2)^(1/2)]/4`

(C)

`(s^2 +r^2)^(1/2)`

(D)

`s+r`

Solution:

Mid-point of `(p, q)` and `( q,- p)` is `((p+q)/2 , (q+p)/2)` which

is give `(r/2 ,s/2)`

`:. (p+q)/2 = r/2` and `(q-p)/2 =s/2`

Now, length of segment `= sqrt ( (p-q)^2 + (q+p)^2)`

`= sqrt (s^2 + r^2)`
Correct Answer is `=>` (C) `(s^2 +r^2)^(1/2)`
Q 2388312207

What is the locus of a point which is equidistant
from the point `(m + n, n - m)` and the point
`(m- n, n + m)`?
NDA Paper 1 2009
(A)

`mx = ny`

(B)

`nx =-my`

(C)

`nx =my`

(D)

`mx=-ny`

Solution:

Let the locus,of the point be `(x_1, y_1)`.

`:. sqrt ( [ x_1 - (m+n)]^2 + [y_1 - (n-m) ]^2 )`

`= sqrt ( [x_1 - (m-n) ]^2 + [y_1 - (n+m) ]^2 )`

`=> x_1^2 + (m+n)^2 -2x_1 (m+n) + y_1^2 + (n-m)^2 -2y_1 (n-m)`

`=x_1^2 + (m-n)^2 -2x_1 (m-n) + y_1^2 + (n+m)^2 -2y_1 (n+m)`

`=> 2x_1 (m-n-m-n) + 2y_1 (n+m -n +m) =0`

`=> -4x_1 n + 4 y_1 m =0`

`=> my_1 =n x_1`

Hence, the locus of the point is `nx = my`.
Correct Answer is `=>` (C) `nx =my`
Q 2318412300

What is the area of the triangle formed by the
lines `y-x = 0, y + x = 0, x = c`?
NDA Paper 1 2009
(A)

`c/2`

(B)

`c^2`

(C)

`2c^2`

(D)

`c^2/2`

Solution:

Required area `= 2` area `(Delta AOD)`

`= 2 xx 1/2 xx OD xx AD = c^2``
Correct Answer is `=>` (B) `c^2`
Q 2338412302

If `x cos theta + y sin theta = 2` is perpendicular to the line
`x - y = 3`, then what is one of the values of `theta`?
NDA Paper 1 2009
(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/2`

(D)

`pi/3`

Solution:

Since, slope of line `x cos theta + y sin theta = 2` is `- cot theta` and
slope of line `x - y = 3` is `1`.

Also, these lines are perpendicular to each other.

`:. (-cot theta ) (1) = -1`

`=> cot theta = 1 = cot pi/4`

`:. theta = pi/4`
Correct Answer is `=>` (B) `pi/4`
Q 2388412307

What is the foot of the perpendicular from the
point `(2, 3)` on the line `x + y -11 = 0` ?
NDA Paper 1 2009
(A)

`(1,10)`

(B)

`(5,6)`

(C)

`(6,5)`

(D)

`(7,4)`

Solution:

The equation of line perpendicular to the given line

`x + y - 11 = 0` .. (i)

is `-x + y+ lambda = 0` ... (ii)

Since, this equation passes through `(2, 3)`.

`:. -2 + 3 +lambda = 0`

`lambda = -1`

From Eq. (ii),

`-x + y- 1 = 0 => y = x + 1`

From Eq. (i),

`x+y+1-11=0`

`=> 2x = 10`

`=> x = 5` and `y = 5 + 1 = 6`

Hence, coordinates of foot of perpendicular from `(2, 3)` to given
line is `(5, 6)`.
Correct Answer is `=>` (B) `(5,6)`
Q 2318412309

Consider the following statements

I. The equation to a straight line paral.!el to the
axis of `X` is `y = d`, where `d` is a constant.

II. The equation to the axis of `X` is `x = 0`.

Which if the statement(s) given above is/are correct?
NDA Paper 1 2009
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

We know that, the equation of `X`-axis is `y = 0`.

Thus, only Statement I is correct.
Correct Answer is `=>` (A) Only I
Q 2368512405

The points `(1, 3, 4), (-1, 6, 10), (-7, 4, 7)` and
`(-5, 1, 1)` are vertices of a
NDA Paper 1 2009
(A)

rhombus

(B)

rectangle

(C)

parallelogram

(D)

square

Solution:

Let the coordinates of the points `A, B, C` and `D` be
`(1, 3, 4), (-1, 6, 10), (-7, 4, 7)` and `(-5, 1, 1)`, respectively.

`:. AB = sqrt ((-1-1)^2 + (6-3)^2 + (10-4)^2 ) = sqrt (4+9 +36) =7`

`BC = sqrt ( (-7+1)^2 + (4-6)^2 + (7-10)^2 ) =sqrt (36 +4 +9 ) =7`

`DA = sqrt ((1+5)^2 + (3-1)^2 + (4-1)^2) = sqrt (36 +4+9) =7`

`AC = sqrt ( (-7-1)^2 + (4-3)^2 + (7-4)^2) = sqrt (64 +1+9) =sqrt (74)`

and `CD = sqrt ((-5+7)^2 + (1-4)^2 + (1-7)^2 ) = sqrt (4+9 +36) = 7`

`BD =sqrt ( (-5+1)^2 + (1-6)^2 + (1-10)^2) = sqrt (16 +25 +81)`

`= sqrt (122)`

`:. AB =BC =CD =DA`

But `BD ne AC`

So, the points `A, B, C` and `D` form a rhombus.
Correct Answer is `=>` (A) rhombus
Q 2328612501

What does the equation `x^3 y + xy^3 - xy = 0`
represent?
NDA Paper 1 2008
(A)

A pair of straight lines only

(B)

A pair of straight lines and a circle

(C)

A rectangular hyperbola only

(D)

A rectangular hyperbola and a circle

Solution:

Given equation,

`x^3 y + xy^3 - xy = 0`

`=> xy(x^2 + y^2) = xy`

`=> xy(x^2+ y^2 -1) = 0`

`=> x^2 + y^2 = 1, xy = 0`

Thus, the equation represents a pair of straight lines and a circle.
Correct Answer is `=>` (B) A pair of straight lines and a circle
Q 2368612505

If the point of intersection of the two lines
`2x + 3 y + 4 = 0` and `4x + 3 y + 2 = 0` is at a
distance `d` from origin, then what is the value of `d`?
NDA Paper 1 2008
(A)

`sqrt(2)`

(B)

`sqrt(3)`

(C)

`sqrt(5)`

(D)

`sqrt(7)`

Solution:

Given equations of two lines as

`2x + 3y+ 4= 0`, ... (i)

and `4x + 3y+ 2 = 0` ... (ii)

On solving Eqs (i) and (ii), the coordinates of the intersecting
point are `(1,-2)`


Now `sqrt ( (0-1)^2 + {0-(-2) }^2 ) = d`

`d = sqrt (1+4) = sqrt (5)`
Correct Answer is `=>` (C) `sqrt(5)`
Q 2318612509

The line through the points `(4, 3)` and `(2,5)`
cutsoff intercepts of lengths `lambda` and `mu` on axes.
Which one of the following is correct?
NDA Paper 1 2008
(A)

`lambda > mu`

(B)

`lambda < mu`

(C)

`lambda > -mu`

(D)

`lambda = mu`

Solution:

Let the equation of the line be

`x/lambda + y/mu =1`

Since, it passes through `(4, 3)` and `(2, 5)`.

`:. 4/lambda + 3/mu =1` .............(i)

and `2/lambda +3/mu =1` ...............(ii)

On solving Eqs. (i) and (ii), we get

`mu = lambda =7`
Correct Answer is `=>` (D) `lambda = mu`
Q 2388723607

What is the locus of a point which is equidistant
from the points `(a+ b, a- b)` and `(b- a, a+b)?`
NDA Paper 1 2008
(A)

`bx - ay = 0`

(B)

`bx + ay = 0`

(C)

`-ax+ by= 0`

(D)

`ax+ by= 0`

Solution:

let the coordinates of the point be `(x, y)`.

`:. sqrt ( {x- (a+b ) }^2 + {y- (a-b) }^2 )`

`= sqrt ( {x- (b-a) }^2 - {y- (a+b) }^2)`

On squaring both side, we get
`x^2 + (a+ b)^2 -2x(a + b)+ y^2 + (a- b)^2 - 2y(a- b)`

`= x^2 + (b- a)^2 -2x(b- a)+ y^2 + (a+ b)^2 -2y(a + b )`

`=> 2x(a +b)+ 2y(a- b) = 2x(b- a)+ 2y(a + b)`

`=> x{(a + b)- (b- a)}+ y{(a- b)- (a+ b)}= 0`

`=> 2ax + (-2by) = 0`

`=> ax -by =0`

`:. -ax + by =0`
Correct Answer is `=>` (C) `-ax+ by= 0`
Q 2318823700

If `A (2, 3), B ( 1, 4), C (0,- 2)` and `D (x,y)` are
vertices of a parallelogram, then what is the
of `(x,y)`?
NDA Paper 1 2008
(A)

`(1,-3)`

(B)

`(2,4)`

(C)

`(1,1)`

(D)

`(0,0)`

Solution:

Given, `A(2, 3), B(1, 4), C(0, -2)` and `D(x,y)` are the
vertices of a parallelogram. We know that, the mid-points of
diagonals of a parallelogram are same.

Mid-point of `AC= ( (0+2)/2 , (3-2)/2)`

Mid-point of `BD = ( (x+1)/2, (y+4)/2)`

`:. (2+0)/2 = (1+x)/2` and `(3-2)/2 = (4+y)/2`

`=> x =1` and `y =-3`

`:. ` Value of `(x,y) = (1,-3)`
Correct Answer is `=>` (A) `(1,-3)`
Q 2348334203

If `O` be the origin and `A(x_1,y_1), B(x_2,y_2)` are two

points , then what is `(OA) (OB) cos angle AOB`?
NDA Paper 1 NoYear
(A)

`x_1^2 + x_2^2`

(B)

`y_1^2 + y_2^2`

(C)

`x_1y_1 +x_2y_2`

(D)

`x_1y_1 - x_2y_2`

Solution:

`OA = sqrt ((x_1 - 0)^2 + ( y_1 - 0)^2 ) = sqrt ( x_1^2 +y_1^2)`

and `AB = sqrt ( (x_1 - x_2)^2 + (y_1 - y_2 )^2 )`

by cosine law ,

`cos angle AOB = (OA^2 + OB^2 -AB^2)/(2OA * OB)`

`=> OA*OB* cos angle AOB =1/2 {x_1^2 +y_1^2 + x_2^2 + y_2^2`

`- x_1^2 +x_2^2 - 2 x_1x_2 ) - (y_1^2 + y_2^2 - 2 y_1y_2)}`

`=1/2 { 2x_1x_2 + 2y_1y_2 } `

`= ( x_1x_2 + y_1y_2)`
Correct Answer is `=>` (C) `x_1y_1 +x_2y_2`
Q 2358334204

If `O` be the origin and `A(x_1,y_1), B(x_2,y_2)` are two

points , then what is `(OA) (OB) cos angle AOB`?
NDA Paper 1 2008
(A)

`x_1^2 + x_2^2`

(B)

`y_1^2 + y_2^2`

(C)

`x_1y_1 + x_2y_2`

(D)

`x_1y_1 - x_2y_2`

Solution:

`OA = sqrt ((x_1 - 0)^2 + ( y_1 - 0)^2 ) = sqrt ( x_1^2 +y_1^2)`

and `AB = sqrt ( (x_1 - x_2)^2 + (y_1 - y_2 )^2 )`

by cosine law ,

`cos angle AOB = (OA^2 + OB^2 -AB^2)/(2OA * OB)`

`=> OA*OB* cos angle AOB =1/2 {x_1^2 +y_1^2 + x_2^2 + y_2^2`

`- x_1^2 +x_2^2 - 2 x_1x_2 ) - (y_1^2 + y_2^2 - 2 y_1y_2)}`

`=1/2 { 2x_1x_2 + 2y_1y_2 } `

`= ( x_1x_2 + y_1y_2)`
Correct Answer is `=>` (C) `x_1y_1 + x_2y_2`
Q 2318434300

If `O` be the origin and `A(x_1,y_1), B(x_2,y_2)` are two

points , then what is `(OA) (OB) cos angle AOB ?`
NDA Paper 1 2008
(A)

`x_1^2 + x_2^2`

(B)

`y_1^2 + y_2^2`

(C)

`x_1x_2 + y_1 y_2`

(D)

`x_1y_1 +x_2y_2`

Solution:

`OA = sqrt ( (x_1 -0 )^2 + (y_1 - 0)^2 ) = sqrt(x_1^2+ y_1^2)`,

and `AB =sqrt( (x_1 +x_2)^2 + (y_1 -y_2)^2)`

By cosine law,

`cos angle AOB = (OA^2 + OB^2 - AB^2)/(2OA * OB)`

`=> OA* OB* cos angle AOB =1/2 { x_1^2 + y_1^2 +x_2^2 +y_2^2`

`- (x_1^2 + x_2^2 -2x_1x_2)-(y_1^2 +y_2^2 -2y_1y_2)}`

`=1/2 {2x_1x_2 +2y_1y_2 } = (x_1x_2 + y_1y_2)`
Correct Answer is `=>` (C) `x_1x_2 + y_1 y_2`
Q 2358534404

If `(a, b), (c, d)` and `(a- c, b- d)` are collinear, then
which one of the following is correct?
NDA Paper 1 2008
(A)

`bc - ad = 0`

(B)

`ab - cd = 0`

(C)

`bc + ad = 0`

(D)

`ab + cd = 0`

Solution:

Since, `(a, b), (c, d)` and `{(a- c), (b- d)}` are collinear,
then the area of the triangle should be zero

`:. Delta =1/2 {x_1 (y_2- y_3) + x_2 (y_3 -y_1) +x_3 (y_1 - y_2) } =0`

`:. a (d-b+ d) +c (b- d -b) + (a-c) (b-d) =0`

`=> 2ad -ab -cd +ab -ad -bc +cd =0`

`=> ad- bc=0`

`=> bc -ad=0`
Correct Answer is `=>` (A) `bc - ad = 0`
Q 2338734602

The coordinates of `P` and `Q` are `(- 3, 4)` and `(2, 1 )`,
respectively. If `PQ` is extended to `R` such that
`PR = 2QR` then what are the coordinates of `R?`
NDA Paper 1 2007
(A)

`(3,7)`

(B)

`(2,4)`

(C)

`(-1/2,5/2)`

(D)

`(7,-2)`

Solution:

Since, the coordinates of `P` and `Q` are `( -3, 4)` and `(2, 1)`,
respectively.

Let coordinates of `R` be `(x, y)`.

Also, `PR = 2QR`

`=> m:n=2:1`

By external section formula,

`:. x= (2 xx 2 +1 xx 3)/(2-1) = 4+3 =7`

and `y = (2 xx 1 -1 xx 4)/(2-1)= 2-4 =-2`
Correct Answer is `=>` (D) `(7,-2)`
Q 2338034802

If the points with the coordinates `(a, ma),
{b, (m + 1)b}, {c, (m + 2) c}` are collinear, then
which one of the following is correct?
NDA Paper 1 2007
(A)

`a, b` and `c` are in arithmetic progression for all `m`

(B)

`a, b` and `c` are in geometric progression for all `m`

(C)

`a, b` and `c` are in harmonic progression for all `m`

(D)

`a, b` and `c` are in arithmetic progression only for `m= 1`

Solution:

Since, the points `A(a, ma), B [b, (m + 1) b]` and

`C [c, (m + 2)c]` are collinear area of the triangle should be zero


formed by these points.

`Delta = 1/2 [x_1 (y_2-y_3) +x_2 (y_3-y_1) +x_3 (y_1 -y_2) ]`

`:. a{(m + 1)b- (m + 2)c} + b{(m + 2)c- ma}`

`+ c{ma- (m + 1)b} = 0`

`=> mab +ab -mac -2ac +mbc +2bc`

`-mab +mac -mbc -bc =0`

`=> ab -2ac +2bc -bc =0`

`=> ab + bc =2ac`

`=> b = (2ac)/(a+c)`

Hence, `a, b` and `c` are in harmonic progression for all `m`
Correct Answer is `=>` (C) `a, b` and `c` are in harmonic progression for all `m`
Q 2308245108

Which one of the following points on the line
`2x -3y =5` is equidistant from `( 1, 2)` and `( 3, 4)`?
NDA Paper 1 2007
(A)

`(7,3)`

(B)

`(4,1)`

(C)

`(1,-1)`

(D)

`(-2,-3)`

Solution:

Let, the point `P(x_1, y_1)` be equidistant from points

`A(1, 2)` and `B{3, 4)`.


`:. PA = PB \ \ => \ \ PA^2 = PB^2`

`=> (1-x_1)^2 + (2-y_1)^2 = (3-x_1)^2 + (4-y_1)^2`

`=> 1 + x^2 - 2x_1 + 4 + y_1^2 - 4y_1 = 9 + x_1^2 - 6x_1 + 16 + y_1^2 - 8y_1`

`=> x_1 +y_1 =5` ... (i)

Since, the point `P(x_1, y_1)` lies on `2x- 3y = 5`

`:. 2x_1 - 3y_1 = 5` ... (ii)

On solving Eqs. (i) and (ii), we get

`x_ 1 = 4` and `y_1 = 1`

So, the coordinates of `P` are `(4, 1)`.
Correct Answer is `=>` (B) `(4,1)`
Q 2308645508

What is the acute angle between the lines
`Ax+ By= A+ B` and `A(x-y) + B(x + y) = 2B`?
NDA Paper 1 2007
(A)

`45^(circ)`

(B)

`tan^(-1) (A/(sqrt (A^2 +B^2) ))`

(C)

`tan^(-1) (B/(sqrt (A^2 +B^2`) ))`

(D)

`60^(circ)`

Solution:

The given equations of lines be

`Ax+ By= A+ B`, ............(i)

and `A(x- y) + B(x- y) = 2B`

`=> (A + B)x + (B - A)y = 2B` ... (ii)

Let slope of line (i) is `m_1`.

`=> m_1 = -A/B`

And the slope of line (ii) is `m_2`.

`=> m_2 = - (A+B)/(B-A)`

Let `theta` be the angle between both the lines, then

`:. tan theta = | (m_1 -m_2)/ (1+ m_1m_2) |`

`:. tan theta = | (-A/B + ((A+B)/(B-A)) )/ (1+A/B ((A+B)/(B-A)) ) | = | (-A (B-A) +B (A+B) )/(B (B-A) +A (A+B) ) |`

`= tan^(-1) | (AB +B^2 -AB +A^2)/(A^2 +AB + B^2 -AB) | = tan^(-1) | (A^2 + B^2)/(A^2 + B^2) |`

`= tan^(-1) (1) = 45^(circ)`
Correct Answer is `=>` (A) `45^(circ)`
Q 2388845707

If `p` be the length of the perpendicular from the
origin on the straight line `x + 2by = 2p`, then what
is the value of `b`?
NDA Paper 1 2007
(A)

`1/p`

(B)

`p`

(C)

`1/2`

(D)

`sqrt(3) /2`

Solution:

The given equation of straight line is

`x + 2by- 2p = 0`.............(i)

Length of perpendicular from origin to line (i) `= p`

`:. | (0+0 -2p)/(sqrt (1+4b^2) ) | =p => (2p)/(sqrt(1+4b^2)) =p`

`=> 4= 1+4b^2`

`=> 4b^2 =3 => b= sqrt(3)/2`
Correct Answer is `=>` (D) `sqrt(3) /2`
Q 2328356201

In what ratio does the line `y - x + 2 = 0` cut the
line joining `(3, -1)` and `(8, 9)`?
NDA Paper 1 2007
(A)

`2:3`

(B)

`3:2`

(C)

`3:-2`

(D)

`1:2`

Solution:

Let `C` be the point which divides the join of `(3, - 1)` and
`(8, 9)` in the ratio `lambda : 1`.

Then, the coordinates of `C` are

`((lambda xx 8 +1 xx 3)/(lambda +1), (9 xx lambda -1 xx 1) /(lambda +1)) = ((8 lambda +3)/(lambda +1) , (9 lambda -1)/(lambda +1))`

(by internal section formula)

Since, this point also lies on `y- x + 2 = 0`.

`:. (9 lambda -1)/(lambda +1) - (8 lambda +3)/(lambda +1) +2 =0`

`=> 9 lambda -1 -8 lambda -3+ 2 lambda +2 =0`

`=> 3 lambda -2 =0 => lambda = 2/3`

`:.` Required ratio `= 2 : 3`
Correct Answer is `=>` (A) `2:3`
Q 2308556408

The points `(2, -2), (8, 4), (4, 6)` and `(-1, 1)` in
order are the vertices of which one of the following
quadrilaterals?
NDA Paper 1 2007
(A)

Square

(B)

Rhombus

(C)

Rectangle (but not square)

(D)

Trapezium

Solution:

Let `A(2, -2), 8(8, 4),C(4, 6)` and `D(-1, 1)` are in order.

`AB = sqrt ((8-2)^2 + (4+2)^2) = sqrt(36+36) = sqrt(72)`

`BC = sqrt ((4-8)^2 + (6-4)^2) =sqrt(16+4) = sqrt(20)`

`CD = sqrt ((-1-4)^2 + (1-6)^2 ) = sqrt(25+25) = sqrt (50)`

and `DA = sqrt ((2+1)^2 + (2-1)^2) =sqrt (9+1) = sqrt (10)`

`:. AB ne BC ne CD ne DA`

So, the quadrilateral `ABCD` is a trapezium.
Correct Answer is `=>` (D) Trapezium
Q 2378656506

If `p` be the length of the perpendicular from the
origin on the straight line `ax + by = p` and `b = sqrt(3)/2`,

then what is the angle between the perpendicular
and the positive direction of `X`-axis?
NDA Paper 1 2007
(A)

`30^(circ)`

(B)

`45^(circ)`

(C)

`60^(circ)`

(D)

`90^(circ)`

Solution:

Since, `p` is the length of perpendicular from the origin
on the straight line `ax + by - p = 0`.

`:. | (0+0 -p)/(sqrt (a^2 +b^2) ) | =p => 1 = sqrt(a^2 +b^2)`

`=> 1=a^2 +3/4` `( :. b = sqrt(3)/2)`

`=> a^2 = 1/4 => a =1/2`

Thus, equation of straight line is

`1/2 x + sqrt(3)/2 y = p`

`=> x cos 60^(circ) + y sin 60^(circ) = p`

Hence, the required angle is `60^(circ)`, which is tile equation of line in
normal form, where `60^(circ)` be the angle between the perpendicular
and the positive direction of `X`-axis.
Correct Answer is `=>` (C) `60^(circ)`
Q 2338756602

The straight line `ax+ by + c = 0` and the
coordinate axes form an isosceles triangle under
which one of the following conditions?
NDA Paper 1 2007
(A)

`|a| = |b |`

(B)

`|a| = |c|`

(C)

`|b| =|c|`

(D)

None of these

Solution:

The straight line `ax + by + c = 0` and the coordinate
axes form an isosceles triangle.

`ax+by = -c`

`=> x/(-c/a) + y/(-c/b) =1` (intercept form)

If `-c/a = -c/b`

i.e., !intercept on `X`-axis = Intercept on `Y`-axis

`=> 1/a = 1/b => | a| = |b |`
Correct Answer is `=>` (A) `|a| = |b |`
 
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