Answer :
[tex]\sf~m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
(-10, -3), (-12, -10)
x1 y1 x2 y2
Plug in what we know:
[tex]\sf~m=\dfrac{-10-(-3)}{-12-(-10)}[/tex]
Subtract:
[tex]\sf~m=\dfrac{-7}{-2}[/tex]
Divide:
[tex]\sf~m=\boxed{\sf3.5}[/tex]
(-10, -3), (-12, -10)
x1 y1 x2 y2
Plug in what we know:
[tex]\sf~m=\dfrac{-10-(-3)}{-12-(-10)}[/tex]
Subtract:
[tex]\sf~m=\dfrac{-7}{-2}[/tex]
Divide:
[tex]\sf~m=\boxed{\sf3.5}[/tex]
[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-10 - (-3)}{-12 - (-10)} = \frac{-10 + 3}{-12 + 10} = \frac{-7}{-2} = 3\frac{1}{2}[/tex]
The slope of the line is equal to 3¹/₂.
[tex]y - y_1 = m(x - x_1) \\y - (-3) = 3\frac{1}{2}(x - (-10) \\y + 3 = 3\frac{1}{2}(x + 10) \\y + 3 = 3\frac{1}{2}(x) + 3\frac{1}{2}(10) \\y + 3 = 3\frac{1}{2}x + 35 \\y = 3\frac{1}{2}x + 32[/tex]
The slope of the line is equal to 3¹/₂.
[tex]y - y_1 = m(x - x_1) \\y - (-3) = 3\frac{1}{2}(x - (-10) \\y + 3 = 3\frac{1}{2}(x + 10) \\y + 3 = 3\frac{1}{2}(x) + 3\frac{1}{2}(10) \\y + 3 = 3\frac{1}{2}x + 35 \\y = 3\frac{1}{2}x + 32[/tex]
