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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If $(1,0,3), (2,1,5), (-2,3,6)$ are the mid-points of the sides of a triangle, then the centroid of the triangle is

If $G(3, -5, r)$ is the centroid of triangle $ABC$ where $A(7, -8, 1)$,$B(p, q, 5)$,and $C(q+1, 5p, 0)$ are the vertices of the triangle,then the values of $p, q, r$ are respectively . . . . . . .

Find the lengths of the medians of the triangle with vertices $A(0,0,6)$,$B(0,4,0)$,and $C(6,0,0)$.

If $D(2, 1, 0)$, $E(2, 0, 0)$, and $F(0, 1, 0)$ are the mid-points of the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, respectively, then the centroid of $\triangle ABC$ is:

If the points $P = \hat{i} + 2 \hat{j}$,$Q = 4 \hat{i} + 6 \hat{j}$,$R = 5 \hat{i} + 7 \hat{j}$,and $S = a \hat{i} + b \hat{j}$ are the consecutive vertices of a parallelogram $PQRS$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module