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Class-12-science»Maths

What is the image of the point (1, 3) in the line x+y-6=0? Thanks!

$Given point (1,3) Line is x + y - 6 = 0 - - - (i) Let (h,k) be the image of A w.r.t (i) We know that P(h,k) be the image of A(x_{1}, y_{1}) w.r.t ax+by+c=0 then \frac{h - x_{1}}{a} = \frac{k - y_{1}}{b} = \frac{- 2 (a x_{1} + b y_{1} + c_{1})}{a^{2} + b^{2}} \frac{h - 1}{1} = \frac{k - 3}{1} = \frac{- 2 (1 + 3 - 6)}{1 + 1} \frac{h - 1}{1} = \frac{k - 3}{1} = \frac{- 2 (- 2)}{2} = 2 h - 1 = 2 k - 3 = 2 h = 3 k = 5 Image of (1,3) w.r.t to line x+y-6=0 is (3,5)$

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