Question

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

Options

• −2, 4

•  −2, −4

• 2, −4

•  2, 4

Answer 1

Disclaimer: option (b) and (c) are given to be same in the book. So, we are considering option (b) as −2, −4
instead of −2, 4.

We have a right angled triangle ΔAPC  whose co-ordinates are A (5, 3); B (11,−5);

P(12 , y)  . So clearly the triangle is, right angled at A. So,

`AP^2 = (12 - 5)^2 + (y -3)^2`

`BP^2 = (12-11)^2 + (y + 5)^2`

`AB^2 = ( 11 - 5)^2 + (-5 -3)^2`

Now apply Pythagoras theorem to get,

`AB^2 = AP^2 +BP^2`

So,

`100 = 50 + 2y^2 + 34 + 4y`

On further simplification we get the quadratic equation as,

`2y^2 + 4y - 16 = 0`

    `y^2 + 2y - 8 = 0`

Now solve this equation using factorization method to get,

 (y + 4 )(y-2 ) = 0

Therefore, y = 2 , - 4