Question

if A(5,3),B(11,-5) and (P(12,y) are the vertices of a right triangle angled at P ,then y =

Answer 1

On solving this equation
2y²-4y-16=0
y²-2y-8=0
(y²+2y)-(4y-8)=0
y(y+2)-4(y+2)=0
(y-4)(y+2)=0
y= 4 or y= - 2

Answer 2

Given:

In right-angled triangle ABP,

A = (5, 3)

B = (11, -5)

P = (12, y)

To Find:

Value of y.

Solution:

As given,

ΔABP is right-angled at P,

so,

AP ⊥ BP

∴ m₁ × m₂ = -1

where m₁ is the slope of AP and m₂ is the slope of BP.

⇒ the slope is given by (y₂ - y₁)/(x₂ - x₁)

∴ (y - 3)/(12 - 5) × ( y + 5)/(12 - 11) = -1

⇒  (y - 3)/ 7 × ( y + 5)/ 1 = -1

⇒ (y - 3) × ( y + 5) = -7

⇒ y² + 2y -15 = -7

⇒ y² + 2y - 8 = 0

⇒ y² + 4y - 2y - 8 = 0

⇒ y(y + 4) -2(y + 4) = 0

⇒ (y + 4).(y - 2) = 0

⇒ y = -4, y =2

Hence, the value of y can be -4 and 2.