Question

Find the equation of the line through the point (1,-4) that is parallel to the line with equation -13x-3y=-66

Answer 1

Given:

A line is passing through the point (1, -4) and parallel to the line

$-13x-3y=-66$

To find:

The equation of the line.

Explanation:

Let us write the given equation in the slope-intercept form,

$\begin{gathered} -3y=13x-66 \\ y=-\frac{13}{3}x+22 \end{gathered}$

So, the slope of the line is,

$m=-\frac{13}{3}$

Since the lines are parallel. So, the slopes are equal.

Using the point and slope formula,

$\begin{gathered} (y-y_1)=m(x-x_1) \\ (y-(-4))=-\frac{13}{3}(x-1) \\ y+4=-\frac{13}{3}x+\frac{13}{3} \\ y=-\frac{13}{3}x+\frac{13}{3}-4 \\ y=-\frac{13}{3}x+\frac{1}{3} \\ y=\frac{-13x+1}{3} \\ 3y=-13x+1 \\ 13x+3y-1=0 \end{gathered}$

Therefore, the equation of the line is,

$13x+3y-1=0$

Final answer:

The equation of the line is,

$13x+3y-1=0$