Question

P

is the point

(- 1, 2)

. A variable line through

P

cuts the

x

and

y

axes at

A

and

B

, respectively.

Q

is the point on

A B

such that

P A, P Q, P B

are

H . P

. Show that the locus of

Q

is the line

y = 2 x

.

Accepted Answer

A line passing through a point $(- 1, 2)$ and inclined at an angle $x$ with the positive direction of $y$ -axis and at a distance $A$ from the point $B$ is $\frac{x - x_{1}}{\cos θ} = \frac{y - y_{1}}{\sin θ} = r$ .

Thus, a line passing through $P (- 1, 2)$ which is inclined at an angle $θ$ with the positive direction of $x$ -axis and at a distance $r$ from the point $(x, y)$ is $\frac{x + 1}{\cos θ} = \frac{y - 2}{\sin θ} = r$

$\Rightarrow x + 1 = r \cos θ, y - 2 = r \sin θ$

$\Rightarrow x = - 1 + r \cos θ, y = 2 + r \sin θ \dots (i)$

Thus, any point on the line through $P$ can be taken as $(- 1 + r \cos θ, 2 + r \sin θ)$ , where $r$ is the distance of the point from $P$ .

This line cuts the $x$ and $y$ -axes are $A$ and $B$ and let $Q \equiv (h, k)$ be a point whose locus is to be determined.

Also, let the points $P$ and $B$ are at distances $r_{1}, r_{2}$ and $P A, P Q, P B$ respectively from the point $P$ .

Then, using the equation $(i)$ , the co-ordinates of $A, Q$ and $B$ are respectively, $(- 1 + r_{1} \cos θ, 2 + r_{1} \sin θ), (- 1 + r_{2} \cos θ, 2 + r_{2} \sin θ)$ and $(- 1 + r_{3} \cos θ, 2 + r_{3} \sin θ)$ .

As, the point $A$ is on the $x$ -axis, hence its $y$ -co-ordinate is zero, i.e., $2 + r_{1} \sin θ = 0$

$\Rightarrow r_{1} = - \frac{2}{\sin θ} \dots (ii)$

Also, the point $B$ is on the $y$ -axis, hence its $x$ -co-ordinate is zero, i.e., $- 1 + r_{3} \cos θ = 0$

$\Rightarrow r_{3} = \frac{1}{\cos θ} \dots (iii)$

Also, let the co-ordinate of the point $Q \equiv (h, k) = (- 1 + r_{2} \cos θ, 2 + r_{2} \sin θ)$

$\Rightarrow h = - 1 + r_{2} \cos θ, k = 2 + r_{2} \sin θ$

$\Rightarrow \frac{h + 1}{r_{2}} = \cos θ, \frac{k - 2}{r_{2}} = \sin θ \dots (iv)$

Now, given $P A, P Q, P B$ are in $H . P .$ , hence, we have $\frac{1}{P A} + \frac{1}{P B} = \frac{2}{P Q}$

On putting the values from the equations $(ii)$ and $(iii)$ , we get $- \frac{\sin θ}{2} + \cos θ = \frac{2}{r_{2}}$

Now, putting the values from the equation $(iv)$ , we get $- \frac{(\frac{k - 2}{r_{2}})}{2} + (\frac{h + 1}{r_{2}}) = \frac{2}{r_{2}}$

$\Rightarrow \frac{2 - k}{2 r_{2}} + \frac{h + 1}{r_{2}} = \frac{2}{r_{2}}$

$\Rightarrow \frac{2 - k}{2} + h + 1 = 2$

$\Rightarrow 2 - k + 2 h + 2 = 4$

$\Rightarrow 2 h = k$ .

Thus, the required locus is obtained by replacing $(h, k)$ by $(x, y)$ , hence, the locus is $2 x = y$ or $y = 2 x$ .

$P$ is the point $(- 1, 2)$ . A variable line through $P$ cuts the $x$ and $y$ axes at $A$ and $B$ , respectively. $Q$ is the point on $A B$ such that $P A, P Q, P B$ are $H . P$ . Show that the locus of $Q$ is the line $y = 2 x$ .

Important Questions on Straight Line

Learn and Prepare for any exam you want !

P is the point -1,2. A variable line through P cuts the x and y axes at A and B, respectively. Q is the point on AB such that PA, PQ, PB are H.P. Show that the locus of Q is the line y=2x.

Important Questions on Straight Line

Learn and Prepare for any exam you want !

$P$ is the point $(- 1, 2)$ . A variable line through $P$ cuts the $x$ and $y$ axes at $A$ and $B$ , respectively. $Q$ is the point on $A B$ such that $P A, P Q, P B$ are $H . P$ . Show that the locus of $Q$ is the line $y = 2 x$ .