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

• Correct answer
  A

  $2: 1$
• B

  $4: 3$
• C

  $4: 5$
• D

  $8: 7$

Solution

Equation of line $A B$ is $\frac{x}{\sqrt{7}}+\frac{y}{\sqrt{5}}=1$ Let $\mathrm{P}\left[\alpha, \sqrt{5}\left(1-\frac{\alpha}{\sqrt{7}}\right)\right]$ on solving $16(\mathrm{PA}) 2=9(\mathrm{~PB}) 2$ $P\left[\frac{\sqrt{7}}{3}, \frac{2 \sqrt{5}}{3}\right]$ Let $\mathrm{BP}: \mathrm{PC}=\lambda: 1$ then $\lambda=2$ $\mathrm{BP}: \mathrm{PC}=2: 1$

Hence, the correct answer is option: 'A'.