Step 1

To find the equation of a line which is parallel to the line whose equation is \(y=-3x+4\). Since we know that the parallel lines have the same slope so the value of the slope of the required line is -3 that is

\(m=-3\)

Also the given point is

\(\displaystyle{\left({x}_{{{1}}},{y}_{{{1}}}\right)}={\left(-{2},-{8}\right)}\)

Step 2

Using the formula of the equation of the line in

Slope intercept form

y=mx+c

point slope form

\(\displaystyle{\left({y}-{y}_{{{1}}}\right)}={m}{\left({x}-{x}_{{{1}}}\right)}\)

Step 3

The equation of the line in slope-intercept form is given by

\(y=-3x+c\)

Since line is passing through \((-2,-8)\) so

\(-8=-3(-2)+c\)

\(-8=6+c\)

\(\displaystyle\Rightarrow{c}=-{14}\)

So equation becomes

\(y=-3x-14\)

Step 4

Also, the equation of the line in point-slope form is given by

\((y-(-8))=-3(x-(-2))\)

\(y+8=-3(x+2)\)