Question

Question #43c33

Answer 1

First, we need the gradient of the original line (the line that it is parallel to).

`m=(y_2-y_1)/(x_2-x_1)=(-5-(-3))/(5-(-2))=(-5+3)/(5+2)=-2/7`

The equation of a line is `y=mx+c`, we know `m` since it is parallel, and we know `x` and `y` from a set of coordinates.

`-5=-2/7(3)+c`

`c=-5+2/7(3)=-5+6/7=6/7-5=6/7-35/7=(6-35)/7=-29/7`

`y=-(2x)/7-29/7`