Question

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

Answer 1

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

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

this equation cAntains two lines $2x-3y+4=0$ and $x-2y+3=0$

solve thse lines together and we Will get point

of intersection i.e (1,2)

so given family of lines always passes through (1,2)

Let image of $P\equiv (2,3)$ be $Q\equiv (h,L)$

so line joining $P$ and $Q$ ulill be perpendicular to the line given about which image is taken

Let line joining $P$ and $Q$ intersect given line at $B$

$\therefore PB=QBB$ and $PQ$ perpendicular to given line

In $△ABP$ and $△ABQ$

$PB=QB$

$\angle PBA=\angle QBA$

$AB=AB$

$\therefore △ABP\cong △ABQ$

so $AP=AQ$

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

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

squaring both sides

$2=h^2+l^2+5-2h-4l$

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

Equation of locus of $(h,l)$ is

$x^2+y^2-2x-4y+3=0$

it is a equation of circle of radius $=\sqrt{2}$

and centre $\equiv (1,2)$.

Ansuer: option (c).