Question

Find the equation of the line that satisfies the given conditions.(Lesson 1.10)

Through$\left(\frac{1}{2},-\frac{2}{3}\right)$; perpendicular to the line$4x-8y=1$

Short answer

Equation of the line is $6x+3y-1=0$.

Step-by-step solution

Step 1. Definition point-slope form of a line.

Point-slope form is the equation of a straight line in the form$y-y_1=m\left(x-x_1\right)$, where m is the slope of the line and$\left(x_1,y_1\right)$are the coordinates of the given point on the line.

Step 2. Find the slope for the perpendicular line.

Identify the slope from the equation of the given line and take negative reciprocal of the resulting slope for the line perpendicular to the given value.

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

Compare with slope-intercept form $y=mx+b$, we get

$\begin{array}{l}{m}_{\text{perpendicular}}=-\left(\frac{2}{1}\right)\\ =2\end{array}$

Step 3. Find the equation of the line.

Use$x_1=\frac{1}{2},y_1=-\frac{2}{3}$ and $m=2$

Plug into the point-slope form:

$\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\left(-\frac{2}{3}\right)=-2\left(x-\frac{1}{2}\right)\\ y+\frac{2}{3}=-\left(2x-1\right)\\ \frac{3y+2}{3}=-2x+1\\ 3y-2=3\left(-2x+1\right)\\ 3y-2=-6x+3\\ -6x-3-3y+2=0\\ 6x+3y-1=0\end{array}$