Browse Questions

# The image of the point (4,-3) with respect to the line $y=x$ is :

$(a)\;(-4,3)\qquad(b)\;(4,3)\qquad(c)\;(2,3)\qquad(d)\;(5,3)$

The line $y=x$ passes through (0,0) and $y=x$ line has the same distance from the point (4,-3) and the mirror image point hence (0,0) is the mid-point of the points (4,-3) and mirror image points.(Mirror image point will lie on perpedicular (i.e) $y=-x$)
So,
$0=\large\frac{x_1+4}{2}$,$0=\large\frac{y_1-3}{2}$
$\therefore x_1=-4,y_1=3$
That is (-4,3)
Hence (a) is the correct answer.