Slope: Additional Practice (Set 2)

Both points share $x = -3$, so the run is zero.

$m = \frac{2 - (-7)}{-3 - (-3)} = \frac{9}{0}$

Dividing by zero has no value, so the slope is undefined (a vertical line).

Set the slope formula equal to $4$ and solve for $k$.

$\frac{10 - 2}{7 - k} = 4, \qquad \frac{8}{7 - k} = 4, \qquad 7 - k = 2$

So $k = 5$.

Read $\tfrac{2}{3}$ as a run of $3$ and a rise of $2$. Step twice from $(-2, 1)$.

$(-2, 1) \to (1, 3) \to (4, 5)$

So $(4, 5)$ is on the line.

Both points share $x = 3$, so the run is zero.

$m = \frac{-5 - 3}{3 - 3} = \frac{-8}{0}$

Dividing by zero has no value, so the slope is undefined (a vertical line).

Apply the slope formula and simplify.

$m = \frac{-2 - 4}{-2 - (-6)} = \frac{-6}{4} = -\frac{3}{2}$

So the slope is $-\tfrac{3}{2}$.

Both points share the height $y = 2$, so the rise is zero.

$m = \frac{2 - 2}{5 - (-5)} = \frac{0}{10} = 0$

So the slope is $0$ (a horizontal line).

Slope $0$ needs a rise of $0$, so the two points must share a $y$-coordinate.

$m = \frac{3 - 3}{5 - 2} = \frac{0}{3} = 0$

The pair $(2, 3)$ and $(5, 3)$ works. The pair sharing $x = 2$ is undefined, not $0$.

Apply the slope formula and simplify.

$m = \frac{7 - (-1)}{3 - (-1)} = \frac{8}{4} = 2$

So the slope is $2$.

The rise is the slope times the run.

$\text{rise} = -\tfrac{3}{4} \times 8 = -6$

So the rise is $-6$ (the line drops $6$ units).

Find each slope and compare sizes.

$m_1 = \frac{11 - 2}{4 - 1} = 3, \qquad m_2 = \frac{4 - 0}{2 - 0} = 2$

Since $3 > 2$, the first line is steeper.

Set the slope formula equal to $\tfrac{1}{2}$ and solve for $y$.

$\frac{3 - y}{2 - (-4)} = \frac{1}{2}, \qquad \frac{3 - y}{6} = \frac{1}{2}, \qquad 3 - y = 3$

So $y = 0$.

Take $(7, 2)$ as point one and simplify.

$m = \frac{5 - 2}{1 - 7} = \frac{3}{-6} = -\frac{1}{2}$

So the slope is $-\tfrac{1}{2}$.