NTA Abhyas JEE Main2020MathematicsStraight LinesPractice
The equation of the external bisector of ∠ B A C of Δ A B C with vertices A 5,2 ,   B 2,3 and C 6,5 is
Options
- A2 x + y + 12 = 0
- Bx + 2 y - 12 = 0
- C2 x + y - 12 = 0
- Dx - 2 y - 1 = 0
Correct answer
D. x - 2 y - 1 = 0
Step-by-step solution
A B = 9 + 1 = 10 A C = 1 + 9 = 10 ⇒ A B = A C ⇒ Δ A B C is an isosceles triangle ⇒ Median through A is also an internal angle bisector. Now, mid-point of B & C is 4,4 = D The slope of A D is 4 - 2 4 - 5 = - 2 So, the slope of external angle bisector of ∠ B A C is 1 2 ⇒ Required equation is y - 2 x - 5 = 1 2 ⇒ x - 2 y - 1 = 0