Find the reflection of the point $(4, - 13)$ about the line $5 x + y + 6 = 0$ .

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Solution

$R e f l e c t i o n o f p o i n t (4, - 13) a b o u t 5 x + y + 6 = 0 A B i s l i n e o f e q^{n} 5 x + y + 6 5 x + y + 6 = 0 \Rightarrow y = - 5 x + 6 \to (i) \Rightarrow s l o p e o f A B = - 5 \Rightarrow P N ⊥ A B a n d P N = N Q \Rightarrow L e t t h e s l o p e o f P Q i s (m) T h e n m \times - 5 = - 1 ∴ m = \frac{1}{5} [∵ A B ⊥ P Q] e q^{n} o f l i n e P Q i s \Rightarrow \frac{y + 13}{x - 4} = \frac{1}{5} \Rightarrow 5 y + 65 = x - 4 \Rightarrow x - 5 y - 69 = 0 \to (i i) S o l v i n g e q^{n} (i) a n d (i i), w e g e t \Rightarrow \frac{x}{- 69 + 30} = \frac{y}{(6 + 5 \times 69)} = \frac{1}{(- 25 - 1)} \Rightarrow \frac{x}{- 39} = \frac{y}{351} = \frac{1}{- 26} ∴ x = \frac{- 39}{- 26} = \frac{- 27}{2} A B a n d P Q i n t e r s e c t a t p o i n t N (\frac{3}{2}, \frac{- 27}{2}) ∴ \frac{4 + x_{1}}{2} = \frac{3}{2}, x_{1} = - 1 a n d \frac{- 13 + y_{1}}{2} = \frac{- 27}{2}, y_{1} = - 14$

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