Find the point with x-coordinate -92 on the line through (26, 1052) and (-53,-1950).
+9466 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)
+9466 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)
