Question

what is the reflection of a point (4,-13) about the line 5x+y+6=0

Answer 1

Reflection of  point (4,-13) about 5x + y + 6 = 0

AB is line of eqⁿ  5x + y + 6

5x + y + 6 = 0 ⇒ y = - 5x + 6  .....(i)

⇒ slop of AB = -5 

⇒ PN ⊥ AB and PN = NQ

⇒ Let the slop of PQ is (m)

Then m x - 5 = -1

∴ m = \(\frac{1}{5}\)      [∵ AB ⊥ PQ]

eqⁿ of line PQ is 

⇒ \(\frac{y+13}{x-4}=\frac{1}{5}\)

⇒  5y + 65 = x - 4

⇒  x - 5y - 69 = 0 .....(ii)

Solving eqⁿ (i) and (ii), we get

⇒  \(\frac{x}{-69+30}=\frac{y}{6+5\times69}\) = \(\frac{1}{(-25-1)}\)

⇒ \(\frac{x}{-39}=\frac{y}{351}=\frac{1}{(-26)}\)

∴ x = \(\frac{-39}{-26}\) = \(\frac{-27}{2}\)

AB and PQ intersect at point N \(\Big(\frac{3}{2},\frac{-27}{2}\Big)\)

∴ \(\frac{4+x_1}{2}=\frac{3}{2},\) x₁ = -1

and \(\frac{-13+y_1}{2}=\frac{-27}{2},\)

y₁ = -14