We can get the equation of the line by using the Point-Slope Formula.
The Point-Slope Formula is used when we have one point on the line that isn't the y-intercept and the slope. It is given as:
[tex]y-y_1=m(x-x_1)[/tex]where:
[tex]\begin{gathered} m=\text{ slope} \\ (x_1,y_1)=\text{ point on the line} \end{gathered}[/tex]Parameters:
The point is given in the question to be:
[tex](x_1,y_1)=(-4,15)[/tex]The slope: We are given that the line is parallel to the line
[tex]y=-2x+5[/tex]Comparing with the Slope-Intercept form of a straight line given as:
[tex]y=mx+b[/tex]where m is the slope, we have the slope of the line to be:
[tex]m=-2[/tex]Formula:
We can now input these into the formula and we get the equation below:
[tex]\begin{gathered} y-15=-2(x-\lbrack-4\rbrack) \\ y-15=-2(x+4) \end{gathered}[/tex]Hence, we can rewrite the equation to be:
[tex]\begin{gathered} y-15=-2x-8 \\ y=-2x-8+15 \\ y=-2x+7 \end{gathered}[/tex]The answer is:
[tex]y=-2x+7[/tex]