Question

Let ΔABC is a right-angled triangle right angled at A such that A(1, 2), C(3, 1) and area of ΔABC = 55 then abscissa of B can be :

Answer 1

use section formula and distance formula if in class 10

Answer 2

Let ∆ABC is a right angled triangle at A such that A(1, 2) , C(3, 1) and area of ∆ABC = 55 sq unit

To find : abscissa of B can be ...

solution : B(x, y) , A(1, 2) and C(3, 1)

using Pythagoras theorem,

BC² = AB² + AC²

⇒(x - 3)² + (y - 1)² = (x - 1)² + (y - 2)² + (3 - 1)² + (1 - 2)²

⇒(x² - 6x + 9 - x² + 2x - 1) + (y² - 2y + 1 - y² + 4y - 4) = 4 + 1

⇒- 4x + 8 + 2y - 3 = 5

⇒-4x + 2y = 16

⇒-2x + y = 8 ........(1)

again, area of triangle = 1/2 |x₁(y₂ -y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

⇒55 = 1/2 |1(y - 1) + x(1 - 2) + 3(2 - y)|

⇒110 = |y - 1 - x + 6 - 3y|

⇒110 = |-2y - x + 5|

⇒±110 = -2y - x + 5

⇒-2y - x = 105, -115

x + 2y = 115 .......(2)

x + 2y = -105 ......(3)

from equations (1) and (2) we get ,

x = 99/5, y = 238/5

and solving equations (1) and (3) we get,

x = -101/5, y = -202/5

Therefore the abscissa of B can be 99/5, -101/5