$\mathrm{ABC}$ is a triangle in a plane with vertices $\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$ and…
$\mathrm{ABC}$ is a triangle in a plane with vertices $\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$. If median through $\mathrm{A}$ is equally inclined to the co-ordinate axes, then value of $\lambda+\mu$ is
17
10
7
3
Solution
Let $\mathrm{AD}$ be the median
$\therefore \quad$ Co-ordinates of
$\begin{aligned}
& \mathrm{D} \equiv\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{\mathrm{z}_1+\mathrm{z}_2}{2}\right) \\
& \mathrm{D} \equiv\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right) \\
\therefore \quad & \overline{\mathrm{AD}}=\left(\frac{\lambda-1}{2}-2\right) \hat{\mathrm{i}}+(4-3) \hat{\mathrm{j}}+\left(\frac{\mu+2}{2}-5\right) \hat{\mathrm{k}} \\
\therefore \quad & \overline{\mathrm{AD}}=\left(\frac{\lambda-5}{2}\right) \hat{\mathrm{i}}+\hat{\mathrm{j}}+\left(\frac{\mu-8}{2}\right) \hat{\mathrm{k}}
\end{aligned}$
Since $\mathrm{AD}$ makes equal angle with co-ordinate axes, the direction ratios are equal.
$\therefore \quad \frac{\lambda-5}{2}=1=\frac{\mu-8}{2}$
Consider,
$\begin{aligned}
& \frac{\lambda-5}{2}=1 \\
& \Rightarrow \lambda-5=2
\end{aligned}$
$\begin{aligned} & \Rightarrow \lambda=7 \\ & \text { and } \frac{\mu-8}{2}=1 \\ & \Rightarrow \mu-8=2 \\ & \Rightarrow \mu=10 \\ & \therefore \quad \lambda+\mu=7+10=17 \\ & \end{aligned}$