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=-(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=-(5)/(6)\cdot8+b \\ 3=-(40)/(6)+b \\ 3=-(20)/(3)+b \\ 9=-20+3b \\ 3b=9+20 \\ 3b=29 \\ b=(29)/(3) \end{gathered}`

So the equation of the line we want is:

`y=-(5)/(6)x+(29)/(3)`