Question

Line K has a slope of 3. Line J is perpendicular to like K and passes through the point (3,8). Type the equation of the line in y=Mx+b

Answer 1

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  $\frac{2}{1}$

Perpendicular line's slope = $-\frac{1}{2}$    [changed sign from + to -, and flipped the fraction]

slope = $-\frac{2}{5}$

Perpendicular line's slope = $\frac{5}{2}$   [changed sign from - to +, and flipped the fraction]

Since you know the slope is 3, the perpendicular line's slope is $-\frac{1}{3}$.  Plug this into the equation

y = mx + b

$y=-\frac{1}{3} x+b$    To find b, plug in the point (3, 8)

$8=-\frac{1}{3}(3)+b$

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

8 + 1 = -1 + 1 + b

9 = b

$y=-\frac{1}{3} x+9$