Question

In a triangle ABC with $\angle A={90}^{\circ }$, P is a point on BC such that PA : PB = 3:4. If AB $\sqrt{7}$ and AC = $\sqrt{5}$, then BP:PC is

• A. 2:1
• B. 4:3
• C. 4:5
• D. 8:7

Answer 1

The correct option is A

2:1

Equation of line AB is

$\frac{x}{\sqrt{7}}+\frac{y}{\sqrt{5}}=1$

Let $P[a,\sqrt{5}(1-\frac{a}{\sqrt{7}}\left)\right]$

On solving $16(PA{)}^2=9(PB{)}^2$

$p[\frac{\sqrt{7}}{3},\frac{2\sqrt{5}}{3}]$

$LetBP:PC=\lambda :1then\lambda =2BP:PC=2:1$