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Find the point with x-coordinate -92 on the line through (26, 1052) and (-53,-1950).

Guest Apr 17, 2017

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 #1
avatar+4777 
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The slope of a line that passes through (26 , 1052) and (-53 , -1950) = \(\frac{\text{change in y's}}{\text{change in x's}}=\frac{1052-(-1950)}{26-(-53)}=\frac{3002}{79}=38\)

 

The equation of this line is

(y - 1052) = 38(x - 26)

y = 38x - 988 + 1052

y = 38x + 64

 

Now, just plug in -92 for x and solve for y.

y = 38(-92) + 64 = -3496 + 64 = -3432

 

So, the point with an x-coordinate of -92 is : (-92 , -3432)

hectictar  Apr 18, 2017
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1+0 Answers

 #1
avatar+4777 
+2
Best Answer

The slope of a line that passes through (26 , 1052) and (-53 , -1950) = \(\frac{\text{change in y's}}{\text{change in x's}}=\frac{1052-(-1950)}{26-(-53)}=\frac{3002}{79}=38\)

 

The equation of this line is

(y - 1052) = 38(x - 26)

y = 38x - 988 + 1052

y = 38x + 64

 

Now, just plug in -92 for x and solve for y.

y = 38(-92) + 64 = -3496 + 64 = -3432

 

So, the point with an x-coordinate of -92 is : (-92 , -3432)

hectictar  Apr 18, 2017

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