Question

Find the slope of each side of the triangle: A(-3,5) B(2,6) C(-5,-2)​

Answer 1

Answer:

${\sf AB}$: $(1/5)$.

${\sf BC}$: $(8/7)$.

${\sf AC}$: $(7/2)$.

Step-by-step explanation:

Let $(x_{0},\, y_{0})$ and $(x_{1},\, y_{1})$ denote two points in the plane. As long as $x_{0} \ne y_{0}$ (that is, these two points are not on the same vertical line,) the slope of the line between these two points would be:

$\displaystyle \frac{y_{1} - y_{0}}{x_{1} - x_{0}}$.

For example, for side ${\sf AB}$, $x_{0} = (-3)$ and $y_{0} = 5$ (for point ${\sf A}$ at $(-3,\, 5)$) while $x_{1} = 2$ and $y_{1} = 6$ (for point ${\sf B}$ at $(2,\, 6)$.)

Since $x_{0} \ne x_{1}$, the slope of the line between ${\sf A}$ and ${\sf B}$ would be:

$\begin{aligned}m({\sf AB}) &=\frac{6 - 5}{2 - (-3)} \\ &= \frac{6 - 5}{2 + 3} \\ &= \frac{1}{5}\end{aligned}$,

Likewise:

$\begin{aligned}m({\sf BC}) &=\frac{(-2) - 6}{(-5) - 2} \\ &= \frac{(-8)}{(-7)} \\ &= \frac{8}{7}\end{aligned}$.

$\begin{aligned}m({\sf AC}) &=\frac{(-2) - 5}{(-5) - (-3)} \\ &= \frac{(-7)}{(-2)} \\ &= \frac{7}{2}\end{aligned}$.