Question

What is the equation of the line through (3,7) that is perpendicular to the line through points (-1,-2) and (5,3)?

Please help :')

Answer 1

the line through points (-1,-2) and (5,3)
slope=(3-(-2))/(5-(-1))=5/6
for a perpendicular line slope =-6/5=-1.2
y=(-6/5)x+b
7=(-6/5)*3+b
7=-18/5+b
7=-36/10+b
7=-3.6+b
b=10.6
y=-1.2x+10.6

Answer 2

The equation of the line would by y = -6/5x + 53/5

Explanation:

In order to find this, we must first find the slope of the original line. We can do this with the slope formula.

m (slope) = (y2 - y1)/(x2 - x1)

m = (3 - -2)/(5 - -1)

m = (3 + 2)/(5 + 1)

m = 5/6

Now that we have the slope of this line, we can find the slope to the other line by using the reciprocal and opposite of it. So we flip 5/6 to get 6/5 and then we make it negative to get -6/5. Now that we have this, we can use the slope and a point in point-slope form to find the line.

y - y1 = m(x - x1)

y - 7 = -6/5(x - 3)

y - 7 = -6/5x + 18/5

y = -6/5x + 53/5