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

Guest Apr 17, 2017

#1**+2 **

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

#1**+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