Question

find the equation of the line passing through the given point and perpendicular to the given equation write your answer in slope intercept form

Answer 1

Slope-intercept form:

y = mx + b      "m" is the slope, "b" is the y-intercept

For lines to be perpendicular, their slopes have to be the opposite/negative reciprocals (flipped sign and number)

For example:

slope is 2

perpendicular line's slope is -1/2

slope is -2/3

perpendicular line's slope is 3/2

3.) y = 2x - 2

The given line's slope is 2, so the perpendicular line's slope is -1/2

$y=-\frac{1}{2}x+b$ To find "b", plug in the point (-5 , 5) into the equation

$5=-\frac{1}{2}(-5)+b$

$5=\frac{5}{2}+b$     Subtract 5/2 on both sides

$5-\frac{5}{2}=b$   Make the denominators the same

$\frac{10}{2}-\frac{5}{2}=b$

$\frac{5}{2}=b$

$y=-\frac{1}{2}x+\frac{5}{2}$

4.) -6x + 5y = -10     Get "y" by itself, add 6x on both sides

5y = -10 + 6x          Divide 5 on both sides

$y=-2+\frac{6}{5}x$

The given line's slope is 6/5, so the perpendicular line's slope is -5/6.

$y=-\frac{5}{6}x+b$       Plug in (-2, 5)

$5 = -\frac{5}{6}(-2)+b$

$5=\frac{10}{6}+b\\ 5=\frac{5}{3}+b$    Subtract 5/3 on both sides

$5-\frac{5}{3} =b$    Make the denominators the same

$\frac{15}{3}-\frac{5}{3}=b\\\frac{10}{3} =b$

$y = -\frac{5}{6}x+\frac{10}{3}$

7.) Perpendicular line's slope is -2

y = -2x + b      Plug in (1,4)

4 = -2(1) + b

4 = -2 + b

6 = b

y = -2x + 6

8.) Perpendicular line's slope is -1/4

$y = -\frac{1}{4}x+b$     Plug in (-5 , 2)

$2=-\frac{1}{4}(-5)+b$

$2 = \frac{5}{4}+b$    Subtract 5/4 on both sides

$2-\frac{5}{4}=b$     Make the denominators the same

$\frac{8}{4}-\frac{5}{4}=b$

$\frac{3}{4}=b$

$y=-\frac{1}{4}x+\frac{3}{4}$