Question

The locus of the image of the point $(2,3)$ in the line
$(2x-3y+4)+k(x-2y+3)=0,k\in R$ is a

• A. Circle of radius $\sqrt{2}$
• B. Circle of radius $\sqrt{3}$
• C. Straight line parallel to $x$-axis
• D. Straight line parallel to $y$-axis

Answer 1

The correct option is A Circle of radius $\sqrt{2}$

First we need to find the point of intersection of the line $2x-3y+4=0$ and $x-2y+3=0$.

Now, the line $(2x-3y+4)+k(x-2y+3)=0$ is the perpendicular bisector of the line joining points $P(2,3)$ and $P^{\prime }(h,k)$. Now, $AP=A{P}^{\prime },$

The given line is $(2x-3y+4)+k(x-2y+3)=0,k\in R$ --- ( 1 )

This line will pass through the point of intersection of the llines

$2x-3y+4=0$ ----- ( 2 )

And $x-2y+3=0$ ----- ( 3 )

On solving equations ( 2 ) and ( 3 ), we get

$x=1$ and $y=2$

$\therefore$ Point of intersection of lines ( 2 ) and ( 3 ) is $(1,2).$

Let $M$ be the mid-point of $P{P}^{\prime }$ then $AM$ is perpendicular bisector of $P{P}^{\prime }$

$\therefore$ $AP=A{P}^{\prime }$

$\Rightarrow$ $\sqrt{(2-1{)}^2+(3-2{)}^2}=\sqrt{(h-1{)}^2+(k-2{)}^2}$

$\Rightarrow$ $\sqrt{2}=\sqrt{h^2+k^2-2h-4k+1+4}$

$\Rightarrow$ $\sqrt{2}=\sqrt{h^2+k^2-2h-4k+5}$

$\Rightarrow$ $h^2+k^2-2h-4k+5=2$

$\Rightarrow$ $h^2+k^2-2h-4k+3=0$

Thus, the required locus is $x^2+y^2-2x-4y+3=0$

Which is a equation of circle with

$\Rightarrow$ Radius $=\sqrt{1+4-3}=\sqrt{2}$