Question

what is an equation of the line that is perpendicular to y-3=4(x+2) and passes through the point (-5, 7)

Answer 1

Answer:  $y-7=-\frac{1}{4}(x+5)$

Step-by-step explanation:

The equation of the line in Point-Slope form is:

$y-y_1=m(x-x_1)$

Where "m" is the slope and ($x_1,y_1$) is a point on the line.

You can identify that in the equation of the line $y-3=4(x+2)$, the slope is:

$m=4$

By definition, the slopes of perpendicular lines are negative reciprocals. Then, the slope of the other line is:

$m=-\frac{1}{4}$

 Finally, knowing that this line passes through the point (-5, 7),you can substitute this point and the slope into the equation  $y-y_1=m(x-x_1)$ to get the equation of this line:

 $y-7=-\frac{1}{4}(x-(-5))$

 $y-7=-\frac{1}{4}(x+5)$