Question

A line passes through the point (–2, 4), and its y-intercept is (0, –6). What is the equation of the line that is perpendicular to the first line and passes through the point (5, –4)?
a.) y=1/5x-3
b.) y=5x-3
c.) y=1/5x-5
d.) y=5/3x-2

Answer 1

The product of the slopes of 2 perpendicular lines is always -1. The slope of the first line can be found in many ways. One way is from the formula below:

S1 = Ya - Yb / Xa - Xb where a and b are any 2 points on the line (the y intercept can be used as one of the points).

S1 = (4 - (-6)) / (-2 - 0)
= 10 / -2 = -5

Or you can find the slope by substituting the y-intercept and the point (-2,4) coordinates into the general equation of a straight line and then solving for the slope:

y = ax + b
4 = -2a - 6
So a = S1 = 10 / -2 = -5 as above

Then S1 x S2 = -1
S2 = -1 / S1 = -1 / -5
= 1/5

Knowing the slope of the second line and a point on the line is enough to find the equation. Substitute the slope and point coordinates into the general equation of the line to find the y-intercept b:

y = ax + b where a is the slope of the line (S2) and x and y are the coordinates of any point on the line.

-4 = (1/5)(5) + b
b = -4 - 1 = -5

So the equation of the line is y = (1/5)x - 5

Hope this helps

Answer 2

So what I should do first is find the slope of the second line, and then apply the last formula
Thank you very much :)