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

Respuesta :

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


[tex]y=-\frac{2}{3}x +\frac{2}{3}[/tex]


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


[tex]y=\frac{6}{5}x-\frac{23}{5}[/tex]


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


[tex]y=-\frac{1}{2} x+\frac{5}{2}[/tex]

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