Question

3) write the equation of a line that passes through the point (8.3) and is parallel to the line that passes through the points (-5, 1) and (1, -4)

Answer 1

First we need to find the line that passes through the points (-5, 1) and (1, -4).

To do so, we can use the linear equation below:

$y=mx+b$

Where m is the slope and b is the y-intercept.

Using the points given, that is, applying the x and y values, we have that:

$\begin{gathered} (-5,1)\colon \\ 1=m\cdot(-5)+b \\ -5m+b=1 \\ \\ (1,-4) \\ -4=m\cdot1+b \\ m+b=-4 \end{gathered}$

If we subtract the first equation by the second one, we have:

$\begin{gathered} -5m+b-(m+b)=1-(-4) \\ -5m+b-m-b=1+4 \\ -6m=5 \\ m=-\frac{5}{6} \end{gathered}$

Parallel lines have the same slope, so the slope of the first line is also m = -5/6.

Now, using this slope and the point (8, 3), we can find the equation of the first line:

$\begin{gathered} (8,3)\colon \\ 3=m\cdot8+b \\ 3=-\frac{5}{6}\cdot8+b \\ 3=-\frac{40}{6}+b \\ 3=-\frac{20}{3}+b \\ 9=-20+3b \\ 3b=9+20 \\ 3b=29 \\ b=\frac{29}{3} \end{gathered}$

So the equation of the line we want is:

$y=-\frac{5}{6}x+\frac{29}{3}$