Question

Help please

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

1) (1 , 0) and 3x - 2y = 2


2) ( 3 , -1) and 5x + 6y = 12


3) ( -5 , 5) and y = 2x - 2

Answer 1

Slope-intercept form is:

y = mx + b "m" is the slope, "b" is the y-intercept (the y value when x = 0)

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

1.) Isolate the "y" in the given equation.

3x - 2y = 2 Subtract 3x on both sides

-2y = 2 - 3x Divide -2 on both sides

y = -1 + 3/2x

The given line's slope is 3/2, so the perpendicular line's slope is -2/3.

y = -2/3x + b

To find "b", plug in the point (1, 0) into the equation

y = -2/3x + b

0 = -2/3(1) + b

0 = -2/3 + b Add 2/3 on both sides

2/3 = b

`y=-(2)/(3)x +(2)/(3)`

2.) Isolate "y"

5x + 6y = 12 Subtract 5x on both sides

6y = 12 - 5x Divide 6 on both sides

y = 2 - 5/6x

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

y = 6/5x + b Plug in the point (3, -1)

-1 = 6/5(3) + b

-1 = 18/5 + b Subtract 18/5 on both sides

-1 - 18/5 = b Make the denominators the same

-5/5 - 18/5 = b

-23/5 = b

`y=(6)/(5)x-(23)/(5)`

3. y = 2x - 2

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

y = -1/2x + b Plug in (-5,5)

5 = -1/2(-5) + b

5 = 5/2 + b subtract 5/2 on both sides

5 - 5/2 = b Make the denominators the same

10/2 - 5/2 = b

5/2 = b

`y=-(1)/(2) x+(5)/(2)`