Mathematics Tricks & Tips of Straight Line for NDA-1

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Tricks & Tips of Straight Line for NDA-1

Section Formula

If we're supposed to find out ratio in which a point divides line joining two point or a line divides line joining two points

we have to assume ratio `lambda :1` and not `m: n` to find dividing ratio

If lambda turns out to be -ve `=>` external bisection

If lambda turns out to be +ve `=>` internal bisection

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 2743180943

What is the ratio in which the point `C (-2/7 , -20/7)` divides the line joining the points `A(-2, - 2)` and `B(2, -4)?`
NDA Paper 1 2017

(A)

`1 : 3`

(B)

`3 : 4`

(C)

` 1 : 2`

(D)

`2 : 3`

Solution:

Let ratio `lambda : 1`

`(-2)/7 =(2 lambda -2)/(lambda+1)`

`lambda=3/4`

Ratio `3/4 :1 = 3: 4`

Correct Answer is `=>` (B) `3 : 4`

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 2602545438

A line perpendicular to the line segment joining the points `(1, 0)` and ` (2, 3)` divides it in the ratio `1 : n`. Then, the equation of line is

(A)

`x (n+1) + 3 (n+1) =n + 11`

(B)

`x (n+1) + (3 n+1) =n+11`

(C)

`x(n+1)-3 (n+1) y =n+11`

(D)

`x(n+1) -(3n +1)y =n+11`

Solution:

Let the given points are `A(1, 0)` and `B(2. 3)` . Let the line `PQ` divide `AB` in the ratio `1 : n` at `R`.

Using section formula for internal division

`R= ((1 xx x_2 + n x_1)/(1+n) , ( 1xx y_2 + n y_1)/(1+n))`

`= ((1 xx 2 + n xx 1)/(1+n) , (1 xx 3 + n xx0)/(1+n))`

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

Also, `PO bot AB`

Let slope of line `PO` is `m.`

`:. ` Slope of line `PO xx` Slope of line `AB = - 1`

`=> m xx (y_2-y_1)/(x_2-x_1) =-1`

`=> m xx (3-0)/(2-0)=-1`

`=> m xx 3 =-1`

`=> m=-1/3`

Now , equation of the line `PQ` by using

`y-y_1 =m(x-x_1)`

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

`=> (3 (n+1) y-9)/(1+n) =(-x (n+1) +(n+2))/(n+1)`

`=> 3 (n + 1)y- 9 =- x (n + 1) + (n + 2)`

`=> x(n + 1) + 3 (n + 1)y = n + 2 + 9`

`=> x(n + 1) + 3 (n + 1) y = n + 11`

Correct Answer is `=>` (A) `x (n+1) + 3 (n+1) =n + 11`

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)

(B)

(C)

(D)

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 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 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`

Collinearity of Three points

There are three ways to prove

(i) By area of triangle formed by these point `Delta =0`

(ii) Slop of `AB= ` Slope of `BC=` Slope of `CA`

(iii) `AB + BC = AC`

`AC + CB = AB`

`BA + AC = BC`

`=>` Always use first method to prove collinearity

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 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 1243891743

If `a, b, c` are in `A.P` then the points `(a, x), (b, y)` and `(c, z)` are collinear if

(This question may have multiple correct answers)

(A) `x, y, z` are in `G.P.`

(B) `x, y, z` are in `H.P.`

(D) `x = y = z`

Solution:

The given point are collinear if

`0 = |(a,x,1), (b,y,1), (c,z,1)|`

`[(R_2→,R_2−R_1), (R_3→,R_3−R_2)]`

`|(a, x, 1), (b-a, y-x, 0), (c-b, z-y,0)|=0`

or if `z - y = x - y` [`a, b, c` being in `A.P` so `b-a=c-b` ]

or if `2y = x + z`

or, if `x = z = y`.

The correct answer is: `x, y, z` are in `A.P., x = y = z`.

Correct Answer is `=>` (C)

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`

Inclination and slope

A line in a coordinate plane forms two angles with the x-axis, which are supplementary.

The angle (say) `θ` made by the line `l` with positive direction of `x`-axis and measured anti clockwise is called the inclination of the line. Obviously `0° ≤ θ ≤ 180°`

Slope (Gradient) of a line :

.(i) lf the inclination of a line (i.e. non vertical line) is `theta` and `theta ne pi/2` ,then the slope of a line is defined to be `tan theta`

(ii) `0° > theta < 180° (theta ne 90^o)`

(iii) If `theta=0` then line is parallel to x-axis
lf `theta=90°` then line is perpendicular to `x`-axis or parallel to `y`-axis

(iv) lf `A(x_1, y_1)` & `B (x_2, y_2), x_1 ne x_2` are points on a straight line then the slope `m` of the line is given by

`tan theta = m=((y_2-y_1))/((x_2-x_1))` (from fig)

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)`

Angle between two Lines :

The angle 0 between the lines having slopes `m_1` and `m_2` is given by

`tan theta = | (m_2 - m_1)/( 1 + m_1m_2)|`

1. Parallelism of Lines :

If two lines of slopes `m_1` and `m_2` are parallel , then the angle `theta` between them is 0°.

`:. tan theta = tan 0° = 0 => (m_2 - m_1 )/ ( 1 + m_1 m_2) = 0`

`=> m_1 = m_2`

Thus, when two lines are parallel, their slopes are equal.

`=>` The equation of line parallel to `ax + by + c = 0` is given `ax+by +lambda = 0`

2. Perpendicularity of Two Lines :

If two lines of slopes `m_1` and `m_2` are perpendicular, then the angle e between them is 90°.

`cot theta = 0`

`=> (1 + m_1m_2 )/( m_1 - m_2 ) = 0 => m_1 * m_2 = -1`

Thus, when two lines are perpendicular, the product of their slopes is -1.

If m is the slope of a line, then the slope of a line perpendicular to it is `( -1/m)`

`=>` The equation of line parallel to `ax + by + c = 0` is given `bx-ay + lambda = 0`

Q 2743191043

What is the acute angle between the pair of straight lines `sqrt2x+ sqrt3y = 1` and `sqrt3x +sqrt2y = 2?`
NDA Paper 1 2017

(A)

`tan^(-1) (1/(2 sqrt6))`

(B)

`tan^(-1) (1/sqrt2)`

(C)

`tan^(-1) (3)`

(D)

`tan^(-1) (1/sqrt3)`

Solution:

`m_1 =(- sqrt 2)/(sqrt 3)`

`m_2= (- sqrt 3)/(sqrt 2)`

`theta= tan^(-1) |(m_1 -m_2)/(1+ m_1m_2)|`

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

`tan^(-1) (1/(2 sqrt 6))`

Correct Answer is `=>` (A) `tan^(-1) (1/(2 sqrt6))`

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 2713180949

What is the equation of the straight line parallel to `2x + 3y + 1 = 0` and passes through the point `(-1, 2)?`
NDA Paper 1 2017

(A)

`2x+3y-4 = 0`

(B)

`2x+3y-5=0`

(C)

`x + y-1 = 0`

(D)

`3x-2y + 7 = 0`

Solution:

Required Equation of line

`2x+3y + lambda=0`

it passes through `(-1 ,2)`

`-2 +6 + lambda=0`

`lambda = 4`

Line `2x+3 y -4=0`

Correct Answer is `=>` (A) `2x+3y-4 = 0`

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 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 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 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 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`

Concurrent line

Three lines are said to be concurrent if they pass through a common point, i.e. they meet at a point.

`a_1 x + b_1 y + c_1 = 0` .............`(i)`

`a_2 x + b_2 y + c_2 = 0` ..................`(ii)`

`a_3 x + b_3 y + c_3 = 0` .................`(iii)`

then

` => | ( a_1, b_1, c_1), (a_2, b_2, c_2 ), ( a_3, b_3 , c_3)| =0`

or

Find the point of intersection of two lines, this intersection must satisfy third line then lines will be concurrent lines.

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

Locus

`text(Equation to Locus of a Point :)`

The equation to the locus of a point is the relation which is satisfied by the coordinates of every point on the locus of the point.

`text(Note : Steps to find locus of a point.)`

`text(Step I :)` Assume the coordinates of the point say `(h, k)` whose locus is to be determined.

`text(Step II :)` Write the given condition in mathematical form involving `h, k`.

`text(Step III :)` Eliminate the variable `(s)`, if any.

`text(Step IV : )` Replace `h` by `x` and `k` by `y` in the result obtained in step `III.`

The equation so obtained is the locus of the point which moves under some condition `(s)`.

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 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`