Question

How do you find the equation of the line through a point $(-6,-1)$ perpendicular to the line $5x-3y=2?$

Answer 1

Step- 1: Finding the slope of the line:

The slope-intercept form of the line is $y=mx+c$.

Where $m$is the slope of the line and $c$ is the y-intercept.

Rearranging the given equation:

$\begin{array}{rcl}5x-3y& =& 2\\ \Rightarrow y& =& \frac{5}{3}x-\frac{2}{3}\end{array}$

Comparing the equation with standard form:

Therefore,$m=\frac{5}{3}$

Also,

$$\begin{gathered} m_{\mathrm{perpendicular}}=-\frac{1}{m} \\ \Rightarrow m_{\mathrm{perpendicular}}=-\frac{1}{{\displaystyle \frac{5}{3}}} \\ \Rightarrow m_{\mathrm{perpendicular}}=-\frac{3}{5} \end{gathered}$$

Step- 2: Finding the value of $c$:

$y=-\frac{3}{5}x+c\left[m=-\frac{3}{5}\right]$

To find the value of $c$ substitute $(-6,-1)$ in the above equation

$\begin{array}{rcl}-1& =& -\frac{3}{5}(-6)+c\\ & \Rightarrow & -1=\frac{18}{5}+c\\ & \Rightarrow & c=-\frac{23}{5}\end{array}$

Step-3: Finding the equation of the line:

Substituting the value of $c$ and $m$ in the standard form of the equation:

The equation of line is $y=-\frac{3}{5}x-\frac{23}{5}$.

Hence, the equation of the line passing through point $(-6,-1)$ and perpendicular to line $5x-3y=2$ is $y=-\frac{3}{5}x-\frac{23}{5}$.