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 [tex]\frac{2}{1}[/tex]
Perpendicular line's slope = [tex]-\frac{1}{2}[/tex] [changed sign from + to -, and flipped the fraction]
slope = [tex]-\frac{2}{5}[/tex]
Perpendicular line's slope = [tex]\frac{5}{2}[/tex] [changed sign from - to +, and flipped the fraction]
Since you know the slope is 3, the perpendicular line's slope is [tex]-\frac{1}{3}[/tex]. Plug this into the equation
y = mx + b
[tex]y=-\frac{1}{3} x+b[/tex] To find b, plug in the point (3, 8)
[tex]8=-\frac{1}{3}(3)+b[/tex]
8 = -1 + b Add 1 on both sides to get "b" by itself
8 + 1 = -1 + 1 + b
9 = b
[tex]y=-\frac{1}{3} x+9[/tex]
