Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt2 is the same as the distance between pt1 and pt3 *
Accepted Solution
A:
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 = [tex] \sqrt{Δx^2+Δy^2} [/tex] d₁ = [tex] \sqrt{(-2-1)^2+(1-4)^2} [/tex] d₂ = [tex] \sqrt{(x-1)^2+(y-4)^2} [/tex] d₁ = d₂ ∴ [tex] \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} [/tex] ⇒⇒ 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 = [tex]\pm \sqrt{9} = \pm 3[/tex] ∴ x = 4 or x = -2 ∴ y = 1 or y = 7