Question

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

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

Answer 1

Answer: (E)

Let $Q(x,y)$ be the image of the point $P(4,\text{ 1})$ to the line $y-x+1=0$ . Then, PQ is $\perp$ 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).