Question

What is the equation of line perpendicular to y-3=-4(x+2) and passee through the point (-5,7

Answer 1

Answer: $y=\frac{1}{4}x+\frac{33}{4}$

Step-by-step explanation:

1. By definition, two slopes are perpendicular if their slopes are negative reciprocals of each other. So, let's find the slope of the other line.

2. The equation given in the problem is written in Point-slope form:

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

Where m is the slope.

3. Therefore, the slope of its perpendicular line must be:

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

4. You have the point (-5,7), so you can substitute it into the point-slope formula to find the equation of the new line:

$y-7=\frac{1}{4}(x+5)$

5. In slope intercept form is:

$y=\frac{1}{4}x+\frac{5}{4}+7\\y=\frac{1}{4}x+\frac{33}{4}$