Answer :

y=-5x-26

Explanation

the equation of a line can be written as:

[tex]\begin{gathered} y=mx+b \\ where\text{ m is the slope} \\ and\text{ b is the y-intercept} \end{gathered}[/tex]

now, when we have the slope and a passing point, we need to use the slope-point formula , it says

[tex]\begin{gathered} y-y_1=m(x-x_1) \\ where\text{ m is the slope and \lparen x}_1,y_1)\text{ is a point from the line} \end{gathered}[/tex]

so

Step 1

a)let

[tex]\begin{gathered} slope=-5 \\ (x_1,y_1)=(-7,9) \end{gathered}[/tex]

b) now replace in the slope-point formula and solve for y

[tex]\begin{gathered} y-y_{1}=m(x-x_{1}) \\ replace \\ y-9=-5(x-(-7)) \\ y-9=-5(x+7) \\ y-9=-5x-35 \\ add\text{ 9 in both sides} \\ y-9+9=-5x-35+9 \\ y=-5x-26 \end{gathered}[/tex]

therefore, the equaton of the line is

y=-5x-26

I hope this helps you

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