Let A (2, 3, 5),B (-1, 3, 2) and C (lambda, 5 | vedclass

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(C) Let $D$ be the mid-point of $BC$. The coordinates of $D$ are given by:$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right

Vedclass · Accepted Answer

$10\lambda - 7\mu = 0$

Vedclass · Answer

(C) Let $D$ be the mid-point of $BC$. The coordinates of $D$ are given by:$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} ight) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} ight)$The direction ratios of the median $AD$ are:$AD = \left( \frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5 ight) = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} ight)$Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal in magnitude:$\left| \frac{\lambda - 5}{2} ight| = |1| = \left| \frac{\mu - 8}{2} ight|$This implies:$\frac{\lambda - 5}{2} = \pm 1 \Rightarrow \lambda - 5 = \pm 2 \Rightarrow \lambda = 7 	ext{ or } 3$$\frac{\mu - 8}{2} = \pm 1 \Rightarrow \mu - 8 = \pm 2 \Rightarrow \mu = 10 	ext{ or } 6$For the median to be equally inclined,the direction ratios must be equal,i.e.,$\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.Thus,$\lambda = 7$ and $\mu = 10$.Checking the options,for $\lambda = 7$ and $\mu = 10$:$10\lambda - 7\mu = 10(7) - 7(10) = 70 - 70 = 0$.Therefore,the correct relation is $10\lambda - 7\mu = 0$.

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

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