Question

. The position of reflection of the point (4, 1) about the line $ y=x-1 $ is:

• A. (1, 2)
• B. (3, 4)
• C. $ (-\text{ }1,0) $
• D. $ (-2,-1) $
• E. (2, 3)

Answer 1

Answer: (E)

Let $ Q(x,\text{ }y) $ be the image of the point $ P(4,\text{ 1}) $ to the line $ y-x+1=0 $ . Then, PQ is $ \bot $ to $ y-x+1=0 $ $ \therefore $ $ \frac{y-1}{x-4}\times 1=-1 $ $ \Rightarrow $ $ y-1=-x+4 $ $ \Rightarrow $ $ y+1=4+1=5 $ ...(i) And also midpoint of PQ i. e., $ \left( \frac{4+x}{2},\frac{y+1}{2} \right) $ lies on $ y-x+1=0 $ $ \therefore $ $ \frac{y+1}{2}-\frac{(4+x)}{2}+1=0 $ $ \Rightarrow $ $ y+1-4-x+2=0 $ $ \Rightarrow $ $ y-x-1=0 $ $ \Rightarrow $ $ y-x=1 $ ...(ii) On solving Eqs. (i) and (ii), we get $ y=3 $ $ x=2 $ $ \therefore $ Required point is (2, 3).