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Write the equation of a line that is perpendicular to the given line and passes through the given point.

 

y-3=8/3(x+20;(-2,3)

 Dec 28, 2015

Best Answer 

 #3
avatar+129852 
+10

Good answer, Solveit.......you are exactly correct....!!!!

 

A perpendicular line will have the negative reciprocal slope to 8/3   =  -3/8

 

So.....we have

 

y - 3 = -(3/8)(x - [-2] )

 

y = (-3/8)x -6/8 + 3

 

y = (-3/8)x + 18/8

 

y = (-3/8)x + 9/4

 

Here is the graph of both lines......https://www.desmos.com/calculator/btb92kqc5y

 

 

cool cool cool

 Dec 28, 2015
 #1
avatar+2498 
+5

y-3=8/3(x+20)

y=8/3x+160/3+3   y=mx+b

our b is 169/3

mx is 8/3x slope is 8/3

formula of perpendicular is -1/m 

-1/8/3=-3/8 is the slope of perpendicular line 

we have:

m=-3/8

x=-2

y=3

let s put it into y=mx+b

3=-3/8(-2)+b   3=3/4+b  b is 9/4

equation:y=-3/8x+9/4

 Dec 28, 2015
 #2
avatar+2498 
0

i am not sure :)

 Dec 28, 2015
 #3
avatar+129852 
+10
Best Answer

Good answer, Solveit.......you are exactly correct....!!!!

 

A perpendicular line will have the negative reciprocal slope to 8/3   =  -3/8

 

So.....we have

 

y - 3 = -(3/8)(x - [-2] )

 

y = (-3/8)x -6/8 + 3

 

y = (-3/8)x + 18/8

 

y = (-3/8)x + 9/4

 

Here is the graph of both lines......https://www.desmos.com/calculator/btb92kqc5y

 

 

cool cool cool

CPhill Dec 28, 2015

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