Question

Image of the point $ (4,-3) $ with respect to the line $ y=x, $ is

• A. $ (-4,-3) $
• B. $ (3,-4) $
• C. $ (-3,4) $
• D. $ (-3,-4) $

Answer 1

Answer: (C)

Let $ P({{x}_{1}},{{y}_{1}}) $ is the image of the point $ Q(4,-3) $ .
Midpoint of $ PQ\left( \frac{{{x}_{1}}+4}{2},\frac{{{y}_{1}}-3}{2} \right) $ .
This pointlies on $ y=x $ . $ \therefore $ $ \frac{{{x}_{1}}+4}{2}=\frac{{{y}_{1}}-3}{2} $
$ \Rightarrow $ $ {{x}_{1}}-{{y}_{1}}=-7 $ ..(i)
Slope of $ PQ=\frac{-3-{{y}_{1}}}{4-{{x}_{1}}} $ and slope of $ y=x $
is 1 Since, PQ is perpendicular to
$ y=x $
$ \therefore $ $ \left( \frac{-3-{{y}_{1}}}{4-{{x}_{1}}} \right).1=-1 $
$ \Rightarrow $ $ -3-{{y}_{1}}=-4+{{x}_{1}} $
$ \Rightarrow $ $ {{x}_{1}}+{{y}_{1}}=1 $ ...(ii)
On solving Eqs. (i) and (ii), $ {{x}_{1}}=-3,{{y}_{1}}=4 $
Hence, required point is $ (-3,4) $ .