Question

$AB$ and $CD$ are $2$ line segments ; where $A(2,3,0),B(6,9,0),C(-6,-9,0).P$ and $Q$ are midpoint of $AB$ and $CD$, respectively and $L$ is the midpoint of $PQ$. Find the distance of $L$ from the plane $3x+4z+25=0$

• A. 25
• B. 15
• C. 5
• D. 40

Answer 1

Answer: (C)

Let co-ordinate of $D$ is $(x,y,z)$
Using mid-point formula
$Q=\left(\frac{x-6}{2},\frac{y-9}{2},\frac{z+0}{2}\right)$
Also, $P=\left(\frac{2+6}{2},\frac{3+9}{2},\frac{0+0}{2}\right)=\left(4,6,0\right)$
Since, $AC||PQ$
$\therefore$ D.r's of line $AC$ = D.r's of line $PQ$
$\Rightarrow \left(-8,12,0\right)=\left(\frac{x-14}{2},\frac{y-21}{2},\frac{z}{2}\right)$
$\Rightarrow x=-2,y=-3,z=0$
$\Rightarrow D(-2,-3,0)\Rightarrow Q(-4,-6,0)$
If $L$ is midpoint of $PQ$ then
$L\left(\frac{4-4}{2},\frac{6-6}{2},0\right)=\left(0,0,0\right)$
$\therefore$ $\perp$ distance of $L(0,0,0)$ from the plane $3x+4z+25=0$ is
$=\left|\frac{3\left(0\right)+4\left(0\right)+25}{\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}}\right|=\left|\frac{25}{\sqrt{25}}\right|=\sqrt{25}=5$