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

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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.

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