Question
The line through the points (– 2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
✓
Answer
Find the slope of the line joining the point $(-2,6)$ and $(4,8)$
Slope of line $\left( m _1\right)=\frac{y_2-y_1}{x_2-x_1}$
$
=\frac{8-6}{4+2}=\frac{2}{6}=\frac{1}{3}
$
Find the slope of the line joining the points $(8,12)$ and $(x, 24)$
Slope of a line $\left( m _2\right)=\frac{24-12}{x-8}=\frac{12}{x-8}$
Since the two lines are perpendicular.
$
\begin{aligned}
& m _1 \times m _2=-1 \\
& \frac{1}{3} \times \frac{12}{x-8}=-1 \\
& \Rightarrow \frac{12}{3(x-8)}=-1 \\
& -1 \times 3(x-8)=12 \\
& -3 x+24=12
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow-3 x=12-24 \\
& -3 x=-12 \\
& \Rightarrow x=\frac{12}{3}=4
\end{aligned}
$
$\therefore$ The value of $x =4$
Slope of line $\left( m _1\right)=\frac{y_2-y_1}{x_2-x_1}$
$
=\frac{8-6}{4+2}=\frac{2}{6}=\frac{1}{3}
$
Find the slope of the line joining the points $(8,12)$ and $(x, 24)$
Slope of a line $\left( m _2\right)=\frac{24-12}{x-8}=\frac{12}{x-8}$
Since the two lines are perpendicular.
$
\begin{aligned}
& m _1 \times m _2=-1 \\
& \frac{1}{3} \times \frac{12}{x-8}=-1 \\
& \Rightarrow \frac{12}{3(x-8)}=-1 \\
& -1 \times 3(x-8)=12 \\
& -3 x+24=12
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow-3 x=12-24 \\
& -3 x=-12 \\
& \Rightarrow x=\frac{12}{3}=4
\end{aligned}
$
$\therefore$ The value of $x =4$
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