A(2,3,5), B(alpha, 3,3) and C(7,5, beta) are | vedclass

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(A) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left(\frac{\alpha+7}{2}, \frac{3+5}{2}, \frac{3+\beta}{2}\right) = \left(\frac{\alph

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-$9$

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(A) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left(\frac{\alpha+7}{2}, \frac{3+5}{2}, \frac{3+\beta}{2}\right) = \left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$.The direction ratios of the median $AD$ are given by the difference of coordinates of $D$ and $A$:$\left(\frac{\alpha+7}{2} - 2, 4 - 3, \frac{3+\beta}{2} - 5\right) = \left(\frac{\alpha+3}{2}, 1, \frac{\beta-7}{2}\right)$.Since the median $AD$ is equally inclined to the coordinate axes,its direction cosines are equal,which implies that its direction ratios must be proportional to $(1, 1, 1)$.Thus,$\frac{\alpha+3}{2} = 1$ and $\frac{\beta-7}{2} = 1$.Solving for $\alpha$: $\alpha+3 = 2 \Rightarrow \alpha = -1$.Solving for $\beta$: $\beta-7 = 2 \Rightarrow \beta = 9$.Therefore,$\frac{\beta}{\alpha} = \frac{9}{-1} = -9$.

$A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$ are the vertices of a triangle. If the median through $A$ is equally inclined with the coordinate axes,then $\frac{\beta}{\alpha}=$

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