Let the point_1 = p₁ = (1,4)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
![\sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28-2-1%29%5E2%2B%281-4%29%5E2%7D%20%3D%20%5Csqrt%7B%28x-1%29%5E2%2B%28y-4%29%5E2%7D%20)
⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
![\pm \sqrt{9} = \pm 3](https://tex.z-dn.net/?f=%5Cpm%20%5Csqrt%7B9%7D%20%3D%20%5Cpm%203)
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)