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What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y = −2/3x + 5?

 Dec 11, 2017

Best Answer 

 #1
avatar+8394 
+1

The slope of the line    y  =  -\(\frac23\)x + 5    is   -\(\frac23\)

 

And the slope of a perpendicular line is the negative reciprocal of  - \(\frac23\) , which is  \(\frac32\)  .

 

Using the point  (8, 1)  and the slope  \(\frac32\) , the equation of our line in point-slope form is...

 

y - 1  =  \(\frac32\)(x - 8)    This equation is correct as is, but if you need it in slope-intercept form,

                              distribute the  \(\frac32\)  to both terms in parenthesees.

y - 1  =  \(\frac32\)x - 12

                              Add  1  to both sides of the equation.

y  =  \(\frac32\)x - 11

 Dec 11, 2017
 #1
avatar+8394 
+1
Best Answer

The slope of the line    y  =  -\(\frac23\)x + 5    is   -\(\frac23\)

 

And the slope of a perpendicular line is the negative reciprocal of  - \(\frac23\) , which is  \(\frac32\)  .

 

Using the point  (8, 1)  and the slope  \(\frac32\) , the equation of our line in point-slope form is...

 

y - 1  =  \(\frac32\)(x - 8)    This equation is correct as is, but if you need it in slope-intercept form,

                              distribute the  \(\frac32\)  to both terms in parenthesees.

y - 1  =  \(\frac32\)x - 12

                              Add  1  to both sides of the equation.

y  =  \(\frac32\)x - 11

hectictar Dec 11, 2017

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