Equation that is perpendicular to the equation y=4x-2 and through points (8,3)

Question

I need the correct answer to this question please

 

Answer 1

Hey Mikiah,

 

slope of a line: m = (y₂ - y₁) / (x₂ - x₁)

(slope of line 1) x (slope of line 2) = -1
which it verifies the definition of perpendicular (between 2 lines)

 

m of y = 4x-2 is 4/1 therefore you need a line that has (-1/4) that goes through (8,3).  To get that, use point-slope form:

 

m (x - x₁) = (y - y₁)

 

With new slope being (-1/4) and point is (8,3), here is the equation:

 

(-1/4) (x - 8) = (y - 3) 

 

If you distribute left side, you will get (-1/4 x + 2) = y - 3

 

After that, you add 3 to both sides to get y = mx + b (slope y-intercept form).  For your equation, it will be:

 

y = (-1/4) x + 5

 

If your teacher also wants in standard form, multiply entire equation by 4 and move the x to the left side to make

Ax + By = C.  And it will look like this:

 

x +  4y = 20 (for standard form)

 

Good luck to you.

 

Answer 2

The key fact is that the slopes of perpendicular lines are negative reciprocals of each other.

 

The first equation is in slope-intercept form: y = mx + b, where m is the slope.

 

 

The slope of the perpendicular is negative (1/4).

 

Use the slope-intercept equation to set up the equation you seek.

 

y = mx + b, where m = - (1/4)

 

y = - (1/4)x + b

 

 

Use the given point (8, 3) to find the y-intercept, b of the line you seek.

 

3 = - (1/4)(8) + b

 

3 = - 2 + b

 

5 = b

 

y = - (1/4)x + 5.