$\mathrm{A}(2,3,5), \mathrm{B}(\alpha, 3,3)$ and $\mathrm{C}(7,5, \beta)$ are the vertices of a triangle. If…
$\mathrm{A}(2,3,5), \mathrm{B}(\alpha, 3,3)$ and $\mathrm{C}(7,5, \beta)$ are the vertices of a triangle. If the median through $\mathrm{A}$ is equally inclined with the coordinate axes, then $\frac{\beta}{\alpha}=$
-9
$\frac{-1}{9}$
$\frac{-2}{9}$
$\frac{9}{2}$
Solution
Given, points $\mathrm{A}(2,3,5), \mathrm{B}(\alpha, 3,3)$ and $\mathrm{C}(7,5, \beta)$
$\therefore$ Mid-point of $\mathrm{BC}$ is $\mathrm{D}\left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$
$\because$ Direction ratio of line joining points $\mathrm{A}(2,3,5)$ and $\mathrm{D}$
$\left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$ is $\left(\frac{\alpha+3}{2}, 1, \frac{\beta-7}{2}\right)$.
$\because$ The line segment $\mathrm{AD}$ is equally inclined with co-ordinate axes. So, $\frac{\alpha+3}{2}=1=\frac{\beta-7}{2}$
$\Rightarrow \alpha=-1$ and $\beta=9 \Rightarrow \frac{\beta}{\alpha}=-9$