I’m assuming you are finding the equation of the line of Line J

Slope-intercept form: y = mx + b

[m is the slope, b is the y-intercept or the y value when x = 0 –> (0, y) or the point where the line crosses through the y-axis]

For lines to be perpendicular, the slopes have to be negative reciprocals of each other. (basically changing the sign (+/-) and flipping the fraction/switching the numerator and the denominator)

For example:

slope = 2 or

Perpendicular line’s slope = [changed sign from + to -, and flipped the fraction]

slope =

Perpendicular line’s slope = [changed sign from – to +, and flipped the fraction]

Since you know the slope is 3, the perpendicular line’s slope is . Plug this into the equation

y = mx + b

To find b, plug in the point (3, 8)

8 = -1 + b Add 1 on both sides to get “b” by itself

## Answers ( )

I’m assuming you are finding the equation of the line of Line J

Slope-intercept form: y = mx + b

[m is the slope, b is the y-intercept or the y value when x = 0 –> (0, y) or the point where the line crosses through the y-axis]

For lines to be perpendicular, the slopes have to be negative reciprocals of each other. (basically changing the sign (+/-) and flipping the fraction/switching the numerator and the denominator)

For example:

slope = 2 or

Perpendicular line’s slope = [changed sign from + to -, and flipped the fraction]

slope =

Perpendicular line’s slope = [changed sign from – to +, and flipped the fraction]

Since you know the slope is 3, the perpendicular line’s slope is . Plug this into the equation

y = mx + b

To find b, plug in the point (3, 8)

8 = -1 + b Add 1 on both sides to get “b” by itself

8 + 1 = -1 + 1 + b

9 = b