Let $\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$ be the vertices of a $\triangle \mathrm{ABC}$. If the median through $\mathrm{A}$ is equally inclined to the coordinate axes, then:

• A

  $5 \lambda-8 \mu=0$
• B

  $8 \lambda-5 \mu=0$
• Correct answer
  C

  $10 \lambda-7 \mu=0$
• D

  $7 \lambda-10 \mu=0$

Solution

If D be the mid-point of $B C$, then $$ D=\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right) $$
Direction ratios of $\mathrm{AD}$ are $\frac{\lambda-5}{2}, 1, \frac{\mu-8}{2}$ Since median AD is equally inclined with coordinate axes, therefore direction ratios of $\mathrm{AD}$ will be equal, i.e, $$ \begin{aligned} &\frac{\left(\frac{\lambda-5}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\ &=\frac{1}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\ &=\frac{\left(\frac{\mu-8}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\ &\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2 \end{aligned} $$ $$ \begin{aligned} &\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2 \\ &\Rightarrow \lambda=7,3 \text { and } \mu=10,6 \end{aligned} $$ If $\lambda=7$ and $\mu=10$ Then $\frac{\lambda}{\mu}=\frac{7}{10} \Rightarrow 10 \lambda-7 \mu=0$

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