Question

A line passing through the points $(1,0)$ and $(4,3)$ is perpendicular to the line joining $(-2,-1)$ and $(m,0)$. Find the value of m.

Answer 1

Let $A(1,0),$ $B(4,3),$ $C(-2,-1)$ and $D(m,0)$.

We know that the slope of the line joining two points $(x_1,y_1)$ and $(x_2,y_2)$ is:

$m=\frac{y_2-y_1}{x_2-x_1}$

Let us first find the slope of the line $AB$ with points $A(1,0)$ and $B(4,3)$ as shown below:

$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{4-1}=\frac{3}{3}=1$

Now, we find the slope of the line $CD$ with points $C(-2,-1)$ and $D(m,0)$ as shown below:

$m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-1)}{m-(-2)}=\frac{1}{m+2}$

We also know that if the slope of two lines have the relation $m_1\times m_2=-1$, then the lines are perpendicular or vice versa.

Here, it is given that the lines $AB$ and $CD$ are perpendicular, therefore,$m_1\times m_2=-1$ that is:

$m_1\times m_2=-1\Rightarrow 1\times \frac{1}{m+2}=-1\Rightarrow \frac{1}{m+2}=-1\Rightarrow -(m+2)=1\Rightarrow -m-2=1\Rightarrow -m=1+2\Rightarrow -m=3\Rightarrow m=-3$

Hence, $m=-3$.