Locus of the image of the point (2,3) in the line (2x-3y+4)+k(x-2y+3)=0, k R, is a:

Question

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

Accepted Answer

Intersection point of $2x - 3y + 4 = 0$

and $x-2y+3=0$ is $(1, 2)$

Since, $P$ is the fixed point for given family of lines

So, $PB=PA$

$${\left( {\alpha  - 1} \right)^2} + {\left( {\beta  - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$

$${\left( {\alpha  - 1} \right)^2} + {\left( {\beta  - 2} \right)^2} = 1 + 1 = 2$$

$${\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2}$$

$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$

Therefore, given locus is a circle with center $(1, 2)$ and radius $$\sqrt 2 .$$

Solutions · 1