In a triangle ABC, if A(2, –1) and 7x – 10y + 1 = 0 and 3x... | MATH - PROBLEMS RELATED TO TRIANGLE | SolveeIt | Solveeit

Today, August 24

Question

In a triangle ABC, if A(2, –1) and 7x – 10y + 1 = 0 and

3x – 2y + 5 = 0 are equations of an altitude and an angle bisector respectively drawn from B, then equation of BC is-

A

x + y + 1 = 0

B

4x + 9y + 30 = 0

C

5x + y + 17 = 0

D

x – 5y – 7 = 0

Answer

5x + y + 17 = 0

Explanation

Solution

BE ŗ 7x – 10y + 1 = 0

BN ŗ 3x – 2y + 5 = 0

So B ŗ (–3, –2), m_AB = $\frac{1}{5}$

DABN

tan q = $\left| \frac{\frac{1}{5} - \frac{3}{2}}{1 + \frac{1}{5}.\frac{3}{2}} \right|$ = 1 (Q q < p/2)

DABN 1 = $\left| \frac{m - \frac{3}{2}}{1 + m - \frac{3}{2}} \right|$

Ž 5m² + 24 m –5 = 0 Ž m = $\frac{1}{5}$ or – 5

BC ŗ y + 2 = –5 (x + 3) Ž 5x + y + 17 = 0