If $A (5, 3), B (11, - 5) & P (12, y)$ are the vertices of a right triangle, right angled at $P$ , then the value of $y$ is equal to

-2, 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-2, -4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2, -4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2, 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 2, -4
$Given, APB is right angled triangle. ∠ P = 90^{\circ} A P^{2} = (12 - 5)^{2} + (y - 3)^{2} = 49 + y^{2} - 6 y + 9 = y^{2} - 6 y + 58 units P B^{2} = (12 - 11)^{2} + (y + 5)^{2} = 1 + y^{2} + 10 y + 25 = y^{2} + 10 y + 26 units A B^{2} = (11 - 5)^{2} + (- 5 - 3)^{2} = 36 + 64 = 100 units ∴ By pythagoras theorem, A P^{2} + P B^{2} = A B^{2} y^{2} - 6 y + 58 + y^{2} + 10 y + 26 = 100 \Rightarrow 2 y^{2} + 4 y - 16 = 0 \Rightarrow y^{2} + 2 y - 8 = 0 \Rightarrow (y + 4) (y - 2) = 0 \Rightarrow y = - 4 (or) y = 2$

Suggest Corrections

Similar questions

A (5, 3), B (11, - 5) & P (12, y)

P

y

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

(x_{1} - 4)^{2} + (y_{1} - 5)^{2} = (x_{2} - 4)^{2} + (y_{2} - 5)^{2} = (x_{3} - 4)^{2} + (y_{3} - 5)^{2},

(y_{1} + y_{2} + y_{3}) - (x_{1} + x_{2} + x_{3})

(- 4, - 4), (- 1, - 2)

(x, - 8)

(- 1, - 2)

x

Join BYJU'S Learning Program