Write an equation of the line passing through point P(-8,0) that is perpendicular to the line 3x-5y=6.

Answer 1

Let’s do a similar problem. Instead of line 3x - 5y = 6, let’s say the original line is

x - 6y = 5, and let’s note this given line as L1.

Let’s re-write this L1 in a slope-intercept form as:

L1: y = 1/6 * x - 5/6, slope k1 is 1/6

Let’s note the line in question as L2, and its slope as k2.

For L2 to be perpendicular to L1, their slopes k1 and k2 satisfy a nice property, that is,

k1 * k2 = -1

We know k1= 1/6, hence we have k2 = -6.

Now, we additionally know that L2 is passing a point, (-8, 0). Hence, we can apply the slope-point method to get L2, as

L2: (y - 0) = k2 * (x + 8) = (-6) * (x + 8) // note: x - (-8) = x + 8

Simplify it, we have

L2: y = -6 * (x + 8)

If you master this method, you now can try it on the original problem.