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Must Do Problems of Straight Line for NDA

Must Do Problems For NDA

Must Do Problems For NDA
Q 2835534462

If `(a cos theta_1 , a sin theta_1 ), (a cos theta_2, a sin theta_2)` and
`(a cos theta_3, a sin theta_3)` represents the vertices of an
equilateral triangle inscribed in a circle, then
consider the following statements.

I. `cos theta_1 +cos theta_2 + cos theta_3 = 0`

II.` sin theta_1 + sin theta_2 + sin theta_3 = 0`

Which of the above statement(s) is/are correct?



(A)

Only I

(B)

Only I

(C)

Both I and II

(D)

Neither I nor II

Solution:

Vertices
`(a cos theta_1 , a sin theta_1 ),(a cos theta_2 ,a sin theta_2 )`

and `(a cos theta_3 , a sin theta_3)` are equidistant
from origin (0, 0). Hence, the origin is
circumccntre (centroid) of circnmcirclc.
Therefore, the coordinates of centroid sre

`( ( a (cos theta_1 + cos theta_2 + cos theta_3 ) )/3 , (a (sin theta_1 + sin theta_2 + sin theta_3) )/3 ) `

But as the centroid is the origin (0, 0),

therefore `cos theta_1 + cos theta_2 + cos theta_3 = 0`

and `sin theta_1 + sin theta_2 + sin theta_3 = 0`

Hence, both statements arc correct.
Correct Answer is `=>` (C) Both I and II
Q 2805434368

ABC is a triangle formed by the lines xy = 0 and
x + y = 1. Consider the following statements

I. Orthocentre of `Delta ABC` is at the origin.

II. Circumcentre of `Delta ABC` is at the point
(1/ 2, 1/ 2).
Which of the above statement(s) is I are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

None of these

Solution:

The equations of the sides of `Delta ABC`
arc x = 0, y = 0 and x + y = 1 Clcarlv,
`Delta ABC` is right angled triatngle with right
angle at the origin and coordinates the
end-points of hypotenuse as ( 1, 0) and
( 0, 1). So, orthocentre of the triangle
is at the origin and cirumccntre at
the mid-point of its hypotenuse, i.e. at
(1/2, 1 /2).

So, both statements are correct.
Correct Answer is `=>` (C) Both I and II
Q 2815434360

The two points (2, 1) and (3,- 1) with respect to
the line 3x- 5y + 7 = 0

(A)

on the line

(B)

on same side of the line

(C)

on opposite side of the line

(D)

None of these

Solution:

The equations of the sides of `Delta ABC`
arc x = 0, y = 0 and x + y = 1 Clcarlv,
`Delta ABC` is right angled triatngle with right
angle at the origin and coordinates the
end-points of hypotenuse as ( 1, 0) and
( 0, 1). So, orthocentre of the triangle
is at the origin and cirumccntre at
the mid-point of its hypotenuse, i.e. at
(1/2, 1 /2).

So, both statements are correct.
Correct Answer is `=>` (B) on same side of the line
Q 2815123969

If `x cos theta + y sin theta = 2` is perpendicular to the line
x- y = 3, then what is one of the value of `theta` ?

(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 arc perpendicular to
each other.

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

`=> cot theta = 1 = cot pi/4 => theta= pi/4`
Correct Answer is `=>` (B) `pi/4`
Q 2855123964

What is the locus of a point which is equidistant
from the point (m + n, n- m) and the point
(m- n, n + m)?

(A)

mx=ny

(B)

nx=-my

(C)

nx=my

(D)

mx=-ny

Solution:

Let the coordinate of the moving
point P be (h, k) .

Then, `[ b - (m+n) ]^2 + [ k - (n-m) ]^2`

`= [b - (m-n) ]^2 + [ k - (n + m ) ]^2`

`=> b^2 + (m+n)^2 -2 b (m+n) +k^2`

`+ (n -m )^2 -2k ( n-m ) = b^2 + ( m-n )^2`

`-2b (m-n) + k^2 + (n+m)^2 -2 k (m+n)`

`=> -2 [b (m+n) + k (n-m) ]`

`= -2 [ b (m-n) + k (m+n ) ]`

`=> mb + nb + nk - mk`

`= mb -nb + mk + nk`

`=> 2nb = 2mk => nb = mk `

`:.` Required locus is `nx = my`
Correct Answer is `=>` (C) nx=my
Q 2805023868

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

(A)

8x + 5y + 20 = 0

(B)

5x+ 8y-7=0

(C)

8x- 5y + 60 = 0

(D)

5x- 8y + 57 = 0

Solution:

Let A (a, O) 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, equation of line joining

`(-15/2 ,0)` and `(0,12)` is

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

`=> 8x -5y + 60 = 0`
Correct Answer is `=>` (C) 8x- 5y + 60 = 0
Q 2815723669

The equation of straight line passing through the
point of intersection of the straight line
3x- y + 2 = 0 and 5x- 2y + 7 = 0 and having
infinite slope is

(A)

X= 2

(B)

X+ y = 3

(C)

X= 3

(D)

X= 4

Solution:

Required line should be

`(3x y+2)+λ (5x-2y+7)=0` ... (i)

`=> (3+5λ)x-(2λ+ 1)y+(2+ Tλ) = 0`

`=> y = ( 3+5 λ)/(2λ+1 ) x + (2+ 7 λ)/( 2λ +1)` .........(ii)

As, the Eq. (ii) has infinite slope,

`2 λ+1=0=> λ=-1/2`

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

`X =3`
Correct Answer is `=>` (C) X= 3
Q 2815523469

The diagonals of a quadrilateral ABCD are along
the lines x + 3y = 4 and Gx- 2y = 7. Then, ABCD
must be a

(A)

rectangle

(B)

parallelogram

(C)

cyclic quadrilateral

(D)

rhombus

Solution:

Slope of line x + 3 y = 4 is `m_1 = -1/3`

Slope of line 6x- 2y = 7 is `m_2 = (-6)/(-2) = 3`

`:. m_1 m_2 = -1`

`:.` The given diagonals arc perpendicular
to each other.
Thus, ABCD is a rhombus.
Correct Answer is `=>` (D) rhombus
Q 2815523460

The area of a triangle is 5 and two of its vertices
are A (2, 1), B(3,- 2). Then, the third vertex, m
Ist quadrant which lies on the line y = x + 3 is

(A)

(7/2,13/2)

(B)

(5/2,5/2)

(C)

(3/2 , 3/2)

(D)

(0,0)

Solution:

Let the third vertex be (p, q), then
q=p+3 ... (i)

Also, `Delta = | 5 | = pm 5`

`:. q+3 p-7 = pm 10`

`:. 3p + q = 17` ............(ii)

and `3p + q = -3` ..........(iii)

On solving Eq. (i) with Eq. (ii) and Eq.

(iii) we get `(7 /2, 13/2) ` and `(-3/2 , 3/2)`

`:.` Required vertex `= (7/2, 13/ 2)`
Correct Answer is `=>` (A) (7/2,13/2)
Q 2845423363

A point P (b, k) lies on the straight line
x + y + 1 = 0 and is at a distance 5 units from the
origin. If h is negative, then h is equal to

(A)

-3

(B)

3

(C)

-4

(D)

4

Solution:

Since, the point ( b, k) lies on

`x+y+1=0=> b + k + 1 = 0`

and `b^2 + k^2 = 25`

`(-1-k)^2 +k^2 =25`

`=> 2k^2 + 2k- 24 = 0`

`=> k^2 + k- 12 = 0`

`=> k = -4 ` or k = 3 [rejected as `k < 0` ]

`:. b = -1-(-4)= 3`
Correct Answer is `=>` (B) 3
Q 2815323269

Foot of perpendicular drawn from (0, 5) to the
line 3x - 4y - 5 = 0 is

(A)

(1, 3)

(B)

(2, 3)

(C)

(3, 2)

(D)

(3, 1)

Solution:

Equation of perpendicular line from

3x - 4y 5 = 0 ... (i)

is 4x + 3v + c = 0

Since it passes through (0, 5).

`:.` c = -15

`=> 4x + 3y - 15 = 0` ... (ii)

On solving Eqs. (i) and (ii), we get (3, 1)
Correct Answer is `=>` (D) (3, 1)
Q 2805112968

The equation of line parallel to the line
2x + 3y + 5 = 0 and sum of whose intercepts on
the axes is 15 is

(A)

2x + 3y = 15

(B)

3x-2y+ 2 = 0

(C)

3x-2y+ 8 = 0

(D)

2x+ 3y-18=0

Solution:

Equation of line parallel to

`2x + 3 y + 5 = 0` is `2x + 3 y = lambda`


`=> x/(lambda/2) + y/(lambda/3) = 1`

Given, `lambda/2 + lambda/3 = 15 => lambda = 18`

`:.` Required line is 2x + 3y = 18
Correct Answer is `=>` (D) 2x+ 3y-18=0
Q 2864723655

The distane between the lines 4x + 3y = 11 and
8x + 6y = 15 is

(A)

7/2

(B)

7/3

(C)

7/5

(D)

7/10

Solution:

Given lines arc 4x + 3 y = 11

and 4x + 3y = 15/2.

Distance between them

`= | (11 -15/2)/(sqrt (16+9 ) ) | = | 7/(2 xx 5) | = 7/10`
Correct Answer is `=>` (D) 7/10
Q 2854723654

The area of quadrilateral ABCD whose vertices in
order are `A (1, 1), B(7,- 3), C(12, 2)` and `D(7, 21)` is

(A)

66 sq units

(B)

132 sq units

(C)

124 sq units

(D)

86.5 sq units

Solution:

Area of quadrilateral ABCD

`=1/2 | ( x_1 - x_3 , y_1 - y_3 ), ( x_2 - x_4 , y_2 - y_4 ), ( x_2 - x_4 , y_2 - y_4 ) |`

`=1/2 | ( (1-12) , (1-2) ), ( ( 7-7 ), ( -3 - 21) ) |`

`=1/2 | (-11 , -1 ), ( 0 , -24) | = 1/2 (264 - 0 )`

`= 132` sq units.
Correct Answer is `=>` (B) 132 sq units
Q 2844723653

If `t_1 ne t_2` and the points `A( a, 0), B(at_(1)^2 , 2at_1 )` and
`C(at_(2)^2 , 2at_2)` are collinear, then `t_1 t_2` is equal to

(A)

1

(B)

2

(C)

-1

(D)

-2

Solution:

`Delta = 1/2 | (a , 0 ,1 ), ( at^2 , 2at_1 ,1 ), ( at_(2)^2 , 2at_(2) ,1) |`

` = 1/2 xx (2a) xx a xx | (1,0,1 ), ( t_(1)^2 , t_1 , 1) , ( t_(2)^2 , t_2 ,1 ) |`

`:. Delta = 0 => (t_1 - t_2) + (t_(1)^2 t_2 - t_(2)^2 t_1) = 0`

`=> (t_1 -t_2) +t_1 t_2(t_1 - t_2)= 0`

`=> (t_1 - t_2) (1+ t_1 t_2) = 0`

`=> t_1 t_2 = -1`
Correct Answer is `=>` (C) -1
Q 2824723651

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?

(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 given `(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 2834623552

The co-ordinates of incentre of `Delta ABC` with
vertices A(O, 6), B(S, 12) and C(S, 0) is

(A)

`(16/3 , 0 )`

(B)

`(8,11)`

(C)

`(-4,3 )`

(D)

`(5, 6 )`

Solution:

`a= BC = sqrt (0^2 + (12- 0)^2 ) = 12`

`b =AC = sqrt ( (0-8)^2 + (6-0)^2 ) =10`

`c= AB = sqrt (8^2 + 6^2) =10 `

Incentre is

`( (ax_1 + bx_2 + cx_3 )/( a+b +c) , ( ay_1 + by_2 + cy_3 )/( a+b +c) )`

i.e., `( (12 xx 0 + 10 xx 8 + 10 xx 8 )/(12+10+10) , (12 xx 6 + 10 xx 12 +10 xx 0 )/(12 + 10 + 10 ) )`

`= (160/32 , 192/32) = (5,6)`
Correct Answer is `=>` (D) `(5, 6 )`
Q 2884323257

The coordinates of the middle points of the sides
of a triangle are (4, 2), (8, 3) and (2, 2), then find
the coordinates of its centroid are

(A)

`(3, 7/3 )`

(B)

`(3,3)`

(C)

`(4,3 )`

(D)

None of these

Solution:

Centroid of `Delta ABC` coincide with the
centroid of triangle formed by
mid-points of AB, BC and CA.

`:. ` Required coordinates

` ≡ ( (4+3+2)/3 , (2+3+2)/3) ≡ (3, 7/3)`
Correct Answer is `=>` (A) `(3, 7/3 )`
Q 2815345260

Given two points
A (-2, 0) and 8 (0, 4), M is a point with coordinates
`(x, x), x ge 0`. P divides the joining of A and 8 in the
ratio 2 : 1. C and D are the mid-points of BM and
AM, respectively.
Area of the `Delta AMB` is minimum, if the coordinates
of M are

(A)

`(1,1)`

(B)

`(0,0 )`

(C)

`(2,2)`

(D)

`(3,3)`

Solution:

Area of the `Delta AMB = 1/2 | (x,x,1) ,( -2, 0 ,1 ), ( 0,4, 1) |`

`= |1/2 (-4x +2x - 8) | = | - (x+4) |`

which is minimum for x = 0 and thus
the coordinates of M are (0, 0).
Correct Answer is `=>` (B) `(0,0 )`
Q 2815345260

Given two points
A (-2, 0) and 8 (0, 4), M is a point with coordinates
`(x, x), x ge 0`. P divides the joining of A and 8 in the
ratio 2 : 1. C and D are the mid-points of BM and
AM, respectively.
Ratio of the areas of the `Delta's APM` and `BPM` is

(A)

2:1

(B)

1:2

(C)

2:3

(D)

1:3

Solution:

As P divides AB in the ratio 2 : 1.
The base of the `Delta's APM` and `BPM` are
in the ratio `2 : 1` and the length of the
perpendicular from the vertex M on the
base is same. So, the ratio of the areas of
the `Delta APM` and `Delta BPM ` is also 2 : 1.
Correct Answer is `=>` (A) 2:1
Q 2865134965

The lines

`L_1 : 4x - 3y + 7 = 0` and `L_2 : 3x- 4y + 14 = 0`, intersect
the line `L_3 : x + y = 0` at P and Q, respectively. The
bisectors of the acute angle between `L_1` and `L_2`
intersect `L_3` at R.
The equation of the bisector of acute angle is

(A)

x + y + 3 = 0

(B)

x- y- 3 = 0

(C)

x- y + 3 = 0

(D)

3x - y- 7 = 3

Solution:

The equations of lines `L_1` and `L_2` by

making constant term positive, are

`4x- 3y + 7 = 0` ... (i)

and `3x - 4y + 14 = 0` ... (ii)

`·: 4 xx 3 +( -3)(- 4) = 24 > 0`

i.e. `a_1 a_2 + b_1 b_2 > 0`

So, the bisector of the acute angle is
given by

`(4x -3y + 7 )/( sqrt ( 4^2 + (-3)^2 ) ) = - (3x -4y + 14 )/(sqrt (3^2 + (-4)^2 ) )`

`=> 4x - 3y + 7 = -3x + 4y - 14`

`=> x- y + 3 = 0`
Correct Answer is `=>` (C) x- y + 3 = 0
Q 2865134965

The lines

`L_1 : 4x - 3y + 7 = 0` and `L_2 : 3x- 4y + 14 = 0`, intersect
the line `L_3 : x + y = 0` at P and Q, respectively. The
bisectors of the acute angle between `L_1` and `L_2`
intersect `L_3` at R.
The ratio PR :RQ equals to

(A)

`2 sqrt 2 : sqrt 5`

(B)

`2:1`

(C)

`1:1`

(D)

`sqrt 5 : sqrt 2`

Solution:

Let O be the points of intersection of
lines `L_1` and `L_2` .

Solving Eqs. (i) and (ii), we get

` O ≡ (2,5)`

Equation of `L_3` is x + y = 0 ... (iii)

Solving Eqs. (i) and (iii), we get P=( -1, 1)

Solving Eqs. (ii) and (iii), we get

`Q ≡ (- 2, 2)`

`:. OP = sqrt ( (2+1)^2 + (5-1)^2 ) = 5`

and `OQ = sqrt ( (2+2)^2 + (5-2)^2 ) = 5 `

`:. OP = sqrt ( (2+1)^2 + (5-1)^2 ) = 5`

and `OQ = sqrt ( (2+2)^2 + (5-2)^2 ) = 5`

`:.` In any triangle, bisector of an angle
divides the triangle into two similar
triangles

`:. (PR)/(RQ) = (OP)/(OQ) = 5/5 = 1/1 = 1: 1`
Correct Answer is `=>` (C) `1:1`
Q 2865134965

The lines

`L_1 : 4x - 3y + 7 = 0` and `L_2 : 3x- 4y + 14 = 0`, intersect
the line `L_3 : x + y = 0` at P and Q, respectively. The
bisectors of the acute angle between `L_1` and `L_2`
intersect `L_3` at R.
Area of triangle formed by lines `L_1` , `L_2` and `L_3` is

(A)

13/2 sq units

(B)

7/2 sq units

(C)

9/2 sq units

(D)

8 sq units

Solution:

Area of `Delta OPQ = 1/2 | (1,5,1), ( -1,1,1) ,( -2,2,1) |`

`=1/2 | [ -2 -5 +0 ] | = 7/2 ` sq units
Correct Answer is `=>` (B) 7/2 sq units
Q 2875834766

The equation of an
altitude of an equilateral triangle is `sqrt 3 x + y = 2 sqrt 3`
and one of the vertices is `(3, sqrt 3)` .
The possible number of triangles is

(A)

1

(B)

2

(C)

3

(D)

4

Solution:

Two equilateral triangles ABC and

`A' BC` arc possible with one vertices

`(3, sqrt 3)` and `AD` as an altitude.
Correct Answer is `=>` (B) 2
Q 2875834766

The equation of an
altitude of an equilateral triangle is `sqrt 3 x + y = 2 sqrt 3`
and one of the vertices is `(3, sqrt 3)` .
The area of equilateral triangle is (in sq units)

(A)

`3 sqrt 3`

(B)

`sqrt 3`

(C)

`6 sqrt 3`

(D)

`2 sqrt 3 `

Solution:

Two equilateral triangles ABC and

`A' BC` arc possible with one vertices

`(3, sqrt 3)` and `AD` as an altitude.
Correct Answer is `=>` (A) `3 sqrt 3`
Q 2855834764

Consider the line segment

`L : x sec theta + y tan theta = p`.
The locus of mid-point of the portion of the line L
intercepted between the axes is

(A)

`p^2/(4x^2) - p^2/(4 y^2) =1`

(B)

`p^2/(4x^2) - p^2/(4y^2) = 1`

(C)

`p^2/(2x^2) - p^2/(2y^2) =1`

(D)

None of these

Solution:

The equation of the given line is

`x/ (p cos theta) + y/(p cot theta) = 1` ...........(i)

Let the line (i) cuts X-axis and Y-axis at
the points A and B respectively, then

`A ≡ (p cos theta , 0)` and `B = (0, p cot theta )`

If `P(x, y)` is the mid-point of AB ; then

`2x = p cos theta` and `2 y = p cot theta`

`:. sec theta = p/(2x)` and `tan theta = p/(2y)`

Since `sec^2 theta - tan^2 theta = 1`

`=> p^2/(4x^2) - p^2/(4y^2) = 1`
Correct Answer is `=>` (A) `p^2/(4x^2) - p^2/(4 y^2) =1`
Q 2855834764

Consider the line segment

`L : x sec theta + y tan theta = p`.
The equation of line perpendicular to line L and
passing through `(p cot theta , 2 p cos^3 theta)` is

(A)

`x tan theta - y sec theta + p cos 2 theta = 0`

(B)

`x tan theta - y sec theta - p cos2 theta = 0`

(C)

`x tan theta + y sec theta + p sin 2 theta = 0`

(D)

None of the above

Solution:

Equation of line perpendicular to
line L is `x tan theta - y sec theta + k = 0`

Since, this line passes through

`(p cot theta , 2 p cos^3 theta)`

`:. p cot theta * tan theta - 2 p cos^3 theta * sec theta + k = 0`

`=> p -2 p cos^2 theta + k = 0`

`=> k = p (2 cos^2 theta -1) = p cos 2 theta`

`:.` Required equation is

`x tan theta - y sec theta + p cos 2 theta = 0`
Correct Answer is `=>` (A) `x tan theta - y sec theta + p cos 2 theta = 0`
Q 2815634569

Let ABCD be a parallelogram whose equations for the consecutive
sides AB and AD are 4x + 5y = 0 and 7x + 2y = 0. The
equation of one of the diagonal is 11x + 7y = 9.
The equation of the other diagonal is

(A)

x + y =0

(B)

x- y = 0

(C)

2x + y = 0

(D)

x-3y = 0

Solution:

Equations for sides AB and AD are

`4x + 5y = 0` ...........(i)


and `7x +2y = 0` .........(ii)

The point of intersection of AB and AD
is A( 0, 0), clearly A( 0, 0) does not lie on
diagonal

` 11 x + 7y = 9 `... (iii)

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

`B ≡ ( 5/3 , (-4)/3 ) `

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

`D ≡ ( -2/3 , 7 /3 )`

Since, H is the middle point of BD.

`:. H = ( ( 5/3 - 2/3 )/2 , ( -4/3 + 7/3)/2 ) ≡ (1/2 , 1/2 )`


Equation of diagonal AC is

`y - 0 = ( 0- 1/2 )/( 0 - 1/2) (x-0 ) =>x -y = 0`
Correct Answer is `=>` (B) x- y = 0
Q 2815634569

Let ABCD be a parallelogram whose equations for the consecutive
sides AB and AD are 4x + 5y = 0 and 7x + 2y = 0. The
equation of one of the diagonal is 11x + 7y = 9.
Area of parallelogram ABCD is

(A)

8 sq units

(B)

7/2 sq units

(C)

3 sq units

(D)

6 sq units

Solution:

Area of `Delta ABD = 1/2 | ( 0 ,0 ,1 ), ( 5/3 , -4/3 , 1 ), ( -2/3 , 7/3 , 1 ) |`

`=1/2 [35/9 - 8/9] = 3/2` sq units

`:. ` Area of ` | |` gm ABCD `=2 xx ar (Delta ABD)`

`= 2 xx 3/2 = 3` sq units
Correct Answer is `=>` (C) 3 sq units

Advance Practice Problems for NDA

Q 1560356215

The point on the line `3x + 4y = 5` which is equidistant from `(1,2)` and `(3, 4)` is
BITSAT 2009
(A)

`(7, - 4)`

(B)

`(15, -10)`

(C)

`(1/7, 8/7)`

(D)

`(0, 5/4)`

Solution:

Let point `(x_1, y_1)` be on the line `3x + 4y = 5`.

`3x_1 + 4y_1 = 5 ………….. (i)`

Also, `(x_1 - 1)^2 + (y_1 – 2)^2 = (x_1 – 3)^2 + (y_1 – 4)^2`

`=> x_1^2 + y_1^2 - 2x_1 – 4y_1 + 5= x_1 ^2+ y_1^2 - 6x_1 – 8 y_1 + 25 `

` 4x_1 + 4y_1 = 20 ……….. (ii)`

On solving Eqs. (i) and (ii), we get, `x_1 = 15, y_1 = -10`
Correct Answer is `=>` (B) `(15, -10)`
Q 2489645517

The equation `x^3 - yx^2 + x - y = 0` represents
WBJEE 2016
(A)

a hyperbola and two straight lines

(B)

a straight line

(C)

a parabola and two straight lines

(D)

a straight line and a circle

Solution:

Given equation is

`x^3 - y x^2 +x -y =0`

`=> x^2 (x-y) +(x-y) =0`

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

Now, `x^2 +1 ne 0`

and `x-y =0`

`:. x =y`

So, the equation represents a straight line.
Correct Answer is `=>` (B) a straight line
Q 1502191038

Let `A = (3, -4), B(1, 2)` and `P = (2k -1, 2k + 1) ` is a variable point such that `PA + PB` is the minimum. Then `k` is :
BITSAT 2013
(A)

`\ frac{7}{9}`

(B)

`0`

(C)

`\ frac{7}{8}`

(D)

None of these

Solution:

Since we need the minimum value of `PA + PB`, point `P` must lie on the line joining `A` and `B`.

Equating the slopes, we get `\ frac{2k + 1 - 2}{2k - 1 - 1} = \ frac{-4 - 2}{3 - 1}`

`\Rightarrow \ frac{2k - 1}{2k - 2} = \ frac{-6}{2}`

`\Rightarrow 2k - 1 = -3(2k - 2)`

i.e. `2k - 1 = -6k + 6`

`\Rightarrow 8k = 7`

`\Rightarrow k = \ frac{7}{8}`
Correct Answer is `=>` (C) `\ frac{7}{8}`
Q 2559791614

Given four lines with equations, `x + 2y - 3 = 0, 2x + 3y - 4 = 0`, `3x + 4y- 5 = 0, 4x + 5y- 6 = 0` These lines are
BCECE Mains 2015
(A)

concurrent

(B)

the sides of a quadrilateral

(C)

the sides of a parallelogram

(D)

the sides of a square

Solution:

Given equations of lines are

`x + 2y - 3 = 0` .........(i)

`2x + 3y - 4 = 0` ..........(ii)

`3x + 4y- 5 = 0` .........(iii)

and `4x + 5y- 6 = 0` .........(iv)

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

`x = -1 , y = 2`

From Eq. (iii), we get

`3(-1) + 4(2) - 5 = 0`

`=> 0 = 0`

From Eq. (iv),we get

`4(-1) + 5(2)- 6 = 0`

`=> 0 = 0`

Hence, given lines are concurrent.
Correct Answer is `=>` (A) concurrent
Q 2458880704

The points `(-a,-b),(a,b),(0,0)`and
`(a^2 , ab), a ne 0, b ne 0` are always
WBJEE 2016
(A)

collinear

(B)

vertices of a parallelogram

(C)

vertices of a rectangle

(D)

lie on a circle

Solution:

Let the four points be `A(- a,- b), B(a, b), C(0, 0)` and
`D(a^2 , ab)`

If `A, B` and `C` are collinear.

Then, `| (-a,-b,1), (a,b,1), (0,0,1) | = 0`

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

`=> ab + ab = 0`

Hence, `A, B` and `C` ae collinear.

Again, if `B, C` and `D` are collinear.

Then, `|(a,b,1),(0,0,1) , (a^2 ,ab ,1) | =0`

`=> a(0 - ab)- b(0 - a^2) + 1(0) = 0`

`=> - a^2b + a^2b = 0`

Hence, `B, C` and `D` are collinear.

So, the points `A, B ,C` and `D` are always collinear.
Correct Answer is `=>` (A) collinear
Q 2583745647

If the three points `(3q, 0), (0, 3p)` and `(1, 1)` are collinear, then which one is true?
WBJEE 2010
(A)

`1/p+1/q = 1`

(B)

`1/p+1/q = -1`

(C)

`1/p+1/q = 3`

(D)

`1/p+3/q = 1`

Solution:

`A(3q, 0), B(0,3p), C (1,1)` are collinear.

`therefore` slope of AC =slope of BC

`=> (1-0)/(1-3q) = (1-3p)/(1-00`


`=> 1/(1-3q) = (1-3p)/1`


`=> 1 = (1-3p)(1-3q)`


`=> 1 = 1- 3q- 3p + 9pq`

`=> 3p + 3q = 9pq`


`1/q+1/p = 3`
Correct Answer is `=>` (C) `1/p+1/q = 3`
Q 2449380213

The equation of the straight line passing through the point ( 4, 3) and making intercepts on the coordinate axes whose sum is 1, is
BCECE Stage 1 2012
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

Let a and b intercepts on the coordinate axes

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

Equation of line is `x/a+y/b = 1`

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


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


Since, this line passes through ( 4, 3).

`therefore 4/a - 3/(a+1) = 1 => a+4 = a^2+a`

`=> a^2 = 4 => a = pm2`


`therefore` Equation of line is `x/2-y/3 = 1` or `x/(-2)+y/1 = 1` [From Eq. (i)]
Correct Answer is `=>` (D) `x/2-y/3 = 1` and `x/(-2)+y/1 = 1`
Q 2560645515

Point `R (h, k)` divides a line segment between the axes in the ratio `1 : 2`. Find equation of the line.
BCECE Stage 1 2013
(A)

`2kx + hy = 3hk`

(B)

`2kx + hy = 2hk`

(C)

`2kx - hy = 3hk`

(D)

None of the above

Solution:

Let the equation of line `AB` is

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

Let a point `R (h, k)` divide line `AB` in the ratio

`1:2`. By using internally ratio,

`R(h, k) = ( ( 1x_2 + 2x_1)/(1 + 2) , (1y_2 + 2y_1)/(1 + 2) )`

` ∵ h = (1 xx 0 + 2 xx a)/( 1+ 2) , k = ( 1 xx b + 2 xx 0)/( 1+ 2)`

[`∵ P(x, y)` divide the line `A (x_1 , y_1)` and `B (x_2 . y_2)`

in the ratio `m : n` internally]

` :. P(x, y) = ( (nx_2 + mx_2)/(n + m) , (ny_2 + my_1)/(n + m) )`

` => h = (2a)/3 , k = b/3`

` => a = (3h)/2 , b = 3k`

On putting the values of a and bin Eq. (i), we

get

` => ( 2 kx + hy)/(3 hk) = 1`

` 2kx + 17y = 3 hk`
Correct Answer is `=>` (A) `2kx + hy = 3hk`
Q 1533680542

The equation of the base `BC` of an equilateral triangle `ABC` is `x+y=2` and `A` is `(2, -1)`. The length of the side of the triangle is:
BITSAT 2012
(A)

`\sqrt 2`

(B)

` ( \frac{3}{2} )^{1/2}`

(C)

`( \frac{1}{2} )^{1/2}`

(D)

`( \frac{2}{3} )^{1/2}`

Solution:

Length of altitude through `A` to side `BC` is

`L= |\ frac{2-1-2}{\sqrt{1^2+1^2}}\ |=\ frac{1}{\sqrt 2}`

Hence, length of side of equilateral triangle is always equal to `\ frac2{\sqrt3}\times\text{Length of Altitude}=\ frac{2}{\sqrt{3}}L= (\ frac{2}{3} )^{1/2}`
Correct Answer is `=>` (D) `( \frac{2}{3} )^{1/2}`
Q 2532345232

The straight lines `x + y = 0, 5x + y = 4` and `x + 5y = 4` form
WBJEE 2014
(A)

an isosceles triangle

(B)

an equilateral triangle

(C)

a scalene triangle

(D)

a right angled triangle

Solution:

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

`5x+y=4` ........ (ii)

and `x + 5y = 4` ........... (iii)

On solving above equations pairwise, we get

`A(1,-1) , B(2/3, 2/3), C(-1,1)`

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

`= sqrt(1/9+25/9)`

`=sqrt(26/9)`

`sqrt26/3`

and `CA.= sqrt((1 + 1)^2 + ( -1 - 1)^2)`

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

Here, we see that `AB = BC`

Hence, given lines form an isosceles triangle
Correct Answer is `=>` (A) an isosceles triangle
Q 2552512434

Let `P (2, - 3) , Q ( -2, 1)` be the vertices of the `Delta PQR`. If the centroid of ` Delta PQR` lies on the line `2x + 3y = 1`, then the locus of `R` is
WBJEE 2012
(A)

`2x + 3y = 9`

(B)

`2x - 3y = 7`

(C)

`3x + 2y = 5`

(D)

`3x - 2y = 5`

Solution:

Let the vertices of `R` be `(h, k)`.

`:.` Centroid of triangle is given by

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

i.e., `( (2+h-2)/3 ,( -3+1+k)/3 )`

i.e., `( h/3 , (- 2+ k)/3 )`

Since, the point `( h/3 , (- 2+ k)/3 )` lies on the line

`2x + 3y = 1`, therefore the point `( h/3 , (- 2+ k)/3 )`

must be satisfy the equation of the given line.

i.e., `2(h/3) + 3 ((- 2 + k)/3) = 1`

`=> ( 2h - 6 + 3k)/3 = 1`

`=> 2h - 6 + 3k = 3`

`=> 2h + 3k = 9`

Replace `h = x` and `k = y`, we get the required

locus of `R`.

i.e., `2x + 3y = 9`
Correct Answer is `=>` (A) `2x + 3y = 9`
Q 1520578411

The equation of the straight line perpendicular to the straight line `3X+ 2Y = 0` and passing through. the point of intersection of the lines `X + 3Y - 1 = 0` and `X - 2Y + 4 = 0` is
BITSAT 2009
(A)

`2X - 3Y + 1 = 0`

(B)

`2X - 3Y + 3 = 0`

(C)

`2X - 3Y + 5 = 0`

(D)

`2X - 3Y + 7 = 0`

Solution:

The point of intersection of lines

`X + 3Y - 1 = 0` and `X - 2Y + 4 = 0` is `(-2, 1)`.

Let equation of line perpendicular to the given line is

`2X - 3Y + lamda = 0`.

Since, it passes through `(-2, 1).`

`2(-2) - 3(1) + lamda = 0`

` lamda = 7`

Required line is `2X - 3Y + 7 = 0`
Correct Answer is `=>` (D) `2X - 3Y + 7 = 0`
Q 2520601511

If `(a, a^2 )` falls inside the angle made by the lines `y = x/2, x > 0` and `y = 3x, x > 0`, then a belongs to
BCECE Stage 1 2014
(A)

`(3, oo)`

(B)

`(1/2, 3)`

(C)

`(-3, -1/2)`

(D)

`(0, 1/2)`

Solution:

Since, the given point `(a, a^2 )` lies inside the angle between two lines

Then , `a^2 -a/2 > 0 ` and `a^2 - 3a < 0`

`=> a in (-oo,0) uu (1/2 , oo)` and `a in (0,3)`

`=> a in (1/2 , 3)`
Correct Answer is `=>` (B) `(1/2, 3)`
Q 2513745649

The equations `y = pm sqrt3x , y = 1` are the sides of
WBJEE 2010
(A)

an equilateral triangle

(B)

a right angled triangle

(C)

an isosceles triangle

(D)

an obtuse angled triangle

Solution:

`y = tan60^0 x , y = -tan60^0 x`

`y =1` (equilateral)
Correct Answer is `=>` (A) an equilateral triangle
Q 2446801773

If `p` is the perpendicular from origin to the line `x/a +y/b =1` then
UPSEE 2010
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

Given line is

`x/a + y/b =1`

`=> bx+ay=ab`

Thus, the length of perpendicular from origin to
the line

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

`p=(b xx 0 +a xx 0-ab)/(sqrt (b^2 +a^2))`

According to question,

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

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

`=> 1/p^2 =1/a^2+1/b^2`
Correct Answer is `=>` (A) `1/p^2 =1/a^2 +1/b^2`
Q 1570434316

If the sum of the distances of a point `P` from two perpendicular lines in a plane is `1`, then the locus of `P` is a
BITSAT 2008
(A)

rhombus

(B)

circle

(C)

straight line

(D)

pair of straight lines

Solution:

The sum of the distance of a point P from two perpendicular lines in a plane is 1, then the locus of P is a rhombus
Correct Answer is `=>` (A) rhombus
Q 2670423316

A straight line L through the point `(3,- 2)` is inclined at an angle `60°` to the line `sqrt3x + y = 1`. If L also intersects the X-axis, then the equation of L is
BCECE Mains 2015
(A)

`y + sqrt3 x + 2 - 3 sqrt3 = 0`

(B)

`y - sqrt3 x + 2 + 3 sqrt3 = 0`

(C)

`sqrt3 y - x + 3 + 2 sqrt3 = 0`

(D)

`sqrt3 y + x - 3 + 2 sqrt3 = 0`

Solution:

A straight line passing through P and making an

angle of `alpha = 60°`, is given by

` ( y - y_1)/(x - x_1) = tan ( theta pm alpha)`

`=> sqrt3 x + y = 1`

`=> y =- sqrt3 x + 1`, then `tan theta =- sqrt3`

`=> (y +2)/(x - 3) = (tan theta ± tan alpha)/( 1 + tan theta tan alpha)`

` => (y+ 2)/(x-3) = ( - sqrt3 ± sqrt3)/(-1 (- sqrt3) (sqrt 3))`

`=> y + 2 = 0` and `= (y + 2)/(x - 3) = (-2 sqrt3)/(1 - 3) = sqrt3`

`=> y + 2 = sqrt3 x - 3sqrt3`

[ neglecting `y + 2 = 0` as it does not intersect X-axis]
Q 2500501418

Let A(2, -3) and B(-2, 1) be vertices of a MBQ. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line
BCECE Stage 1 2014
(A)

2x+ 3y=9

(B)

2x - 3y=7

(C)

3x + 2y = 5

(D)

3x - 2 y = 3

Solution:

Let (x, y) be the coordinates of vertex C and `(x_ 1 , y_ 1 )` be the coordinates of centroid of the triangle.

`:. x_1 = (x+2 -2 )/3` and `y_1 = (y -3 +1) /3`

`=> x_ = x/3` and `y_1 = (y -2)/3`

Since, the centroid lies on the line `2 x + 3y =1`

`:. 2 x_1 + 3 y_1 =1`

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

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

`=> 2x + 3y =9`

which is the required locus of vertex C
Correct Answer is `=>` (A) 2x+ 3y=9
Q 2582867737

The incentre of an equilateral triangle is `(1,1)` and the equation of one side is `3x + 4y + 3 = 0`. Then, the equation of the circumcircle of the triangle is
WBJEE 2012
(A)

`x^2 + y^2 - 2x - 2y - 2 = 0`

(B)

`x^2 + y^2 - 2x - 2y - 14 = 0`

(C)

`x^2 + y^2 - 2x - 2y + 2 = 0`

(D)

`x^2 + y^2 - 2x - 2y + 14 = 0`

Solution:

Since, triangle is equilateral therefore incentre

`(1, 1)` lies on the centroid of the `Delta ABC`.

`:.GD =` Length of perpendicular from the point

`G(1, 1)` to the line `3x + 4y + 3 = 0`

` = (3(1)+ 4(1)+ 3 )/sqrt(3^2 + 4^2) = 2`

`AG = 2GD = 4`

`:.` Equation of circumcircle with centre at `(1, 1)`

and radius `= 4` units

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

`=> x^2 + y^2 - 2x - 2y - 14 = 0`
Correct Answer is `=>` (B) `x^2 + y^2 - 2x - 2y - 14 = 0`
 
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