The image of the point P(3, 5) with respect to the line y = x is the point Q and the image of Q with respect to the line y = 0 is the point R(a, b), then (a, b) =

Q. Find the image of the point ( 3, 2) with respect to the line x + 2y + 1 = 0

Join BYJU'S Learning Program

The image of the point P(4, 1) with respect to the line y=x-1 is .

The correct option is A (2, 3)Let Q(x', y') be the image of the point P w.r.t. the line x-y-1 = 0. ⇒x'−41=y'−1−1=−2(4−1−1)[12+(−1)2]⇒x'−4=1−y'=−2⇒x'=2 and y'=3

The correct option is A (2, 3)
Let Q(x', y') be the image of the point P w.r.t. the line x-y-1 = 0.
$\Rightarrow \frac{x' - 4}{1} = \frac{y' - 1}{- 1} = \frac{- 2 (4 - 1 - 1)}{[1^{2} + (- 1)^{2}]} \Rightarrow x' - 4 = 1 - y' = - 2 \Rightarrow x' = 2 and y' = 3$