Consider the line y=-4/3x-6

Question

Find the equation of the line that is perpendicular to this line and passes through the point
, −8− 4
.

Find the equation of the line that is parallel to this line and passes through the point
, −8− 4

Answer 1

y - (-4) = (3/4)(x-(-8)) ⇔ y=3/4 x + 2 perpendicular to the original;

 

y- (-4) = (-4/3)(x-(-8)) ⇔ y=-4/3 x -44/3 parallel to the original.

Answer 2

The line y = -4/3x - 6 has slope m = -4/3

 

A line perpendicular to this line has slope m_⊥ = 3/4

When it goes through (-8, -4), (y + 4) = (3/4)(x + 8) = 3/4x + 6

y = 3/4x + 2

 

A line parallel to the first line has the same slope

When it goes through (-8, -4), (y + 4) = (-4/3)(x + 8) = -4/3x - 32/3

y = -4/3x - 44/3

Answer 3

Hi Aaron

If you have a line y = m*x + b , the perpendicular line is given by y' = (-1/m)*x + b'

The slope of the line y = -(4/3)x + 6 is -(4/3). Then, the slope of the line perpendicular is:

-(1/m) = -(1/(-4/3)) = 3/4

Then the perpendicular line is: y' = (3/4)x + b'

For getting the value of b' you use the fact that this perpendicular line pass through the point x=-8 and y=-4.

-4 = (3/4)*(-8) + b'

-4 = -6 + b'

b' = 2

The perpendicular line is : y' = (3/4)*x + 2

 

Is clear the procedure for you? I hope it helps.