Chapter B, Problem 24E

To determine

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

Answer to Problem 24E

The equation of the line is $y=-2x+\frac{1}{3}$.

Explanation of Solution

Given:

The required line is passing through the points $\left(\frac{1}{2},-\frac{2}{3}\right)$ and parallel to $4x-8y=1$.

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,

$\begin{array}{l}4x-8y=1\\ -8y=-4x+1\\ 8y=4x-1\\ y=\frac{4}{8}x-\frac{1}{8}\\ y=\frac{1}{2}x-\frac{1}{8}\\ y=mx+b\end{array}$

Therefore, the slope becomes,

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

Substitute the values in the property mentioned above.

$\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\left(-\frac{2}{3}\right)=\left(-2\right)\left(x-\frac{1}{2}\right)\\ y+\frac{2}{3}=-2x+\frac{2}{2}\\ y+\frac{2}{3}=-2x+1\end{array}$

On further simplification the equation becomes,

$\begin{array}{l}y=-2x+1-\frac{2}{3}\\ y=-2x+\frac{1\times 3}{3}-\frac{2}{3}\\ y=-2x+\frac{1}{3}\end{array}$

Thus, the equation of the line is $y=-2x+\frac{1}{3}$.