Chapter B, Problem 23E

To determine

To find: The equation of the line for the given conditions.

Answer to Problem 23E

The equation of the line is $y=\frac{5}{2}x+\frac{1}{2}$.

Explanation of Solution

Given:

The required line is passing through the points $\left(-1,-2\right)$ and parallel to $2x+5y+8=0$.

Property used:

Two lines with slopes $m_1\text{ and }{m}_2$ are perpendicular if and only if $m_1{m}_2=-1$ their slopes are negative reciprocal $m_2=-\frac{1}{m_1}$.

The equation of the line passes through the point $\left(x_1,y_1\right)$ and having slope m is, $y-y_1=m\left(x-x_1\right)$.

Calculation:

To find the slope, rewrite the equation,

$\begin{array}{l}2x+5y+8=0\\ 5y=-2x-8\\ y=-\frac{2}{5}x-\frac{8}{5}\\ y=mx+b\end{array}$

Therefore, the slope becomes,

$\begin{array}{l}{m}_1=-\frac{2}{5}\\ m_2=-\frac{1}{m_1}\\ m_2=-\frac{1}{\left(-\frac{2}{5}\right)}\\ m_2=\frac{5}{2}\end{array}$.

Substitute the values in the property mentioned above.

$\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\left(-2\right)=\left(\frac{5}{2}\right)\left(x-\left(-1\right)\right)\\ y+2=\frac{5}{2}\left(x+1\right)\\ y+2=\frac{5}{2}x+\frac{5}{2}\\ y=\frac{5}{2}x+\frac{5}{2}-2\end{array}$

On further simplification the equation becomes,

$\begin{array}{l}y=\frac{5}{2}x+\frac{5}{2}-\frac{2\times 2}{2}\\ y=\frac{5}{2}x+\frac{5}{2}-\frac{4}{2}\\ y=\frac{5}{2}x+\frac{1}{2}\end{array}$

Thus, the equation of the line is $y=\frac{5}{2}x+\frac{1}{2}$.