ABC is a triangle in a plane with vertices A( | vedclass

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(A) Let $AD$ be the median through vertex $A$ to side $BC$.Since $D$ is the midpoint of $BC$,its coordinates are $D = \left(\frac{\lambda - 1}{2}, \fr

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) Let $AD$ be the median through vertex $A$ to side $BC$.Since $D$ is the midpoint of $BC$,its coordinates are $D = \left(\frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2}\right) = \left(\frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2}\right)$.The vector $\vec{AD}$ is given by $D - A = \left(\frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5\right) = \left(\frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2}\right)$.Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal.Therefore,$\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.Solving for $\lambda$: $\frac{\lambda - 5}{2} = 1 \implies \lambda - 5 = 2 \implies \lambda = 7$.Solving for $\mu$: $\frac{\mu - 8}{2} = 1 \implies \mu - 8 = 2 \implies \mu = 10$.Thus,$\lambda + \mu = 7 + 10 = 17$.

$ABC$ is a triangle in a plane with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $\lambda + \mu$ is:

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