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# algebra 2

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fine the equation of a line that passes through (-7, -6) and (3, -6)

Guest Aug 24, 2017

#1
+4480
+3

First let's find the slope.

slope  =  $$\frac{\text{difference of y values}}{\text{difference of x values}}=\frac{-6--6}{-7-3}=\frac{0}{-10}=0$$

So, in point-slope form, using the point  (-7, -6) ,  the equation is....

y - -6  =  0(x - -7)

y + 6  =  0

y  =  -6

And..we could have noticed from the beginning that the  y  value of both points is  -6  . That means it must be a horizontal line at  y = -6  .

hectictar  Aug 25, 2017
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#1
+4480
+3

First let's find the slope.

slope  =  $$\frac{\text{difference of y values}}{\text{difference of x values}}=\frac{-6--6}{-7-3}=\frac{0}{-10}=0$$

So, in point-slope form, using the point  (-7, -6) ,  the equation is....

y - -6  =  0(x - -7)

y + 6  =  0

y  =  -6

And..we could have noticed from the beginning that the  y  value of both points is  -6  . That means it must be a horizontal line at  y = -6  .

hectictar  Aug 25, 2017

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