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)

5

Answer 2

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
$=\left|\frac{3\left(0\right)+4\left(0\right)+25}{\sqrt{\left(3\right)^{2}+\left(4\right)^{2}}}\right|= \left|\frac{25}{\sqrt{25}}\right| =\sqrt{25} =5$