Question

Find the value of $x$, if line AB is perpendicular to a line having slope of $-\frac{1}{3}.$

• A. $x=\frac{3}{8}$
• B. $x=\frac{3}{8}$
• C. $x=\frac{8}{3}$
• D. $x=\frac{8}{3}$

Answer 1

The correct option is D $x=\frac{8}{3}$

Given, The line is perpendicular to the line with slope be $m_2=-\frac{1}{3}$.
Let. the slope of the required line would be $m_1$

Concept: If two line are perpendicular to each other, then product of the slope of both line is $-1.$
$m_1\times m_2=-1$
$m_1\times \frac{-1}{3}=-1$
$m_1=3$

Now, find the value of $x$
We know that, slope of the line segment $=\frac{y_2-y_1}{x_2-x_1}$

Here, $(x_1,y_1)\equiv (4,7)$
$(x_2,y_2)\equiv (x,3)$
$\therefore$ Slope of line segment $AB=m_1=\frac{3-7}{x-4}$

We know, $m_1=3$
$\therefore \frac{3-7}{x-4}=3$

$\frac{-4}{x-4}=3$

$-4=3\times (x-4)$

$-4=3x-12$

$3x=8$

$x=\frac{8}{3}$
Hence, value of $x$ is $\frac{8}{3}.$
Option (d) is correct.