Question

Find the equation of a line perpendicular to 3x−4y=6 that contains the point (2,−4).

Answer 1

Explanation:

y=mx+b

3x-4y=6

4y=3x-6

y=(3x-6)/4

So the slope, m, is 3/4

For lines to be perpendicular m1(m2)=-1, in this case

3m/4=-1

m=-4/3 so our line so far is

y=-4x/3+b, using point (2,-4) we can solve for the y intercept or b

-4=-4(2)/3+b

b=-4/3 so

y=(-4x-4)/3

Answer 2

Answer: y= -4x/3 - 4/3

Explanation:

The equation of the line in the slope-intercept form is y=3x4−32.

The slope of the perpendicular line is negative inverse: m=−43.

So, the equation of the perpendicular line is y=−4x3+a.

To find a, we use the fact that the line should pass through the given point: −4=(−43)⋅(2)+a.

Thus, a=−43.

Therefore, the equation of the line is y=−4x3−43.