Question

What is an equation of the line that passes through the point $(-5,2)$ and is parallel to the line $4\mathrm{x}-5\mathrm{y}=5$?

Answer 1

Find the equation of the line that passes through the given point and parallel to the given line:

Given that the points are $(-5,2)$ which parallel to the line $4\mathrm{x}-5\mathrm{y}=5$.

Step- 1: Find the slope:

Rewrite the equation $4\mathrm{x}-5\mathrm{y}=5$ in slope-intercept form.

The slope-intercept form is $\mathrm{y}=\mathrm{mx}+\mathrm{b}$, where $\mathrm{m}$ is the slope and $\mathrm{b}$ is the $\mathrm{y}-$intercept.

Subtract $4\mathrm{x}$ from both sides of the equation:

$\begin{array}{rcl}& \Rightarrow & -5\mathrm{y}=5-4\mathrm{x}\end{array}$

Divide each term by ${\displaystyle -5}$ and simplify:

$\begin{array}{rcl}& \Rightarrow & \mathrm{y}=-1+\frac{4\mathrm{x}}{5}\end{array}$

Since the slope form is $\mathrm{y}=\mathrm{mx}+\mathrm{b}$, the slope is $\frac{4}{5}$.

Step-2: Find an equation that is parallel to the line.

To find an equation that is parallel, the slopes must be equal.

So, find the parallel line using the point-slope formula.

Use the slope ${\displaystyle \frac{4}{5}}$ and a given point ${\displaystyle (-5,2)}$ to substitute for ${\mathrm{x}}_1$ and ${\mathrm{y}}_1$ in the point-slope form $\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1)$.

$\begin{array}{rcl}& \Rightarrow & \mathrm{y}-2=\frac{4}{5}(\mathrm{x}-\left(-5\right))\\ \mathrm{y}-2& =& \frac{4}{5}(\mathrm{x}+5)\\ \mathrm{y}-2& =& \frac{4\mathrm{x}}{5}+4\\ \mathrm{y}& =& \frac{4\mathrm{x}}{5}+6\end{array}$

Hence, the equation of the line that passes through the point $\mathbf{(}\mathbf{-}\mathbf{5}\mathbf{,}\mathbf{2}\mathbf{)}$ and is parallel to the line $\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{5}$ is $\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \frac{\mathbf{4}\mathbf{x}}{\mathbf{5}}\mathbf{+}\mathbf{6}.\end{array}$