Question

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

• A. $x^2+y^2-3x-4y-4=0$
• B. $2x^2+3y^2+3x+4y-7=0$
• C. $x^2+y^2-2x-4y+3=0$
• D. None of these

Answer 1

Answer: (C)

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

Given,

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

image of point $(2,3)$

we have,

$x-2y+3=0$

$2x-3y+4=0$

solving the above equations, we get,

$x=1,y=2$

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

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

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

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

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

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