Question

What is an equation of the line that is perpendicular to y+1=-3(x-5) and passes through the point 4,-6

Answer 1

Answer:

y + 6 = 1/3 (x - 4)

Step-by-step explanation:

Answer 2

Answer:

$y=\frac{1}{3}x-\frac{16}{3}$

Step-by-step explanation:

Arranging the given equation in slope-intercept form ( y = mx + b ):

$y+1=-3(x-5)\\y+1=-3x+15\\y=-3x+15-1\\y=-3x+14$

We know that the perpendicular line would have a slope (m) that is negative reciprocal of this ($\frac{1}{3}$). Thus we can write the perpendicular line's equation as  $y=\frac{1}{3}x+b$.

Now putting x = 4 and y = -6 into the equation, we can solve for b:

$y=\frac{1}{3}x+b\\-6=\frac{1}{3}(4)+b\\-6=\frac{4}{3}+b\\b=-6-\frac{4}{3}\\b=-\frac{16}{3}$

Thus, the equation of the perpendicular line is  $y=\frac{1}{3}x-\frac{16}{3}$