Parallel and Perpendicular Lines: Practice

Parallel lines have equal slopes, so read the slope straight off the equation.

$m_{\parallel} = -\frac{2}{3}$

The flipped value $\tfrac{3}{2}$ belongs to a perpendicular line (after a sign change), not a parallel one.

The given slope is $4$. Take the negative reciprocal: flip and change the sign.

$m_{\perp} = -\frac{1}{4}$

As a check, $4 \cdot \left(-\tfrac{1}{4}\right) = -1$.

First find the slope of the given line by solving for $y$.

$x + 3y = 6 \;\Rightarrow\; y = -\tfrac{1}{3}x + 2$

Its slope is $-\tfrac{1}{3}$, so the perpendicular slope is the negative reciprocal: flip to $-3$ and change the sign, giving $3$.

Compare the two slopes directly from slope-intercept form.

$m_1 = m_2 = \tfrac{1}{4}, \quad b_1 = 2 \neq -5 = b_2$

Equal slopes with different intercepts means the lines are parallel.

Convert the second line to slope-intercept form.

$x + 5y = 10 \;\Rightarrow\; y = -\tfrac{1}{5}x + 2$

The slopes are $5$ and $-\tfrac{1}{5}$, and $5 \cdot \left(-\tfrac{1}{5}\right) = -1$, so the lines are perpendicular.

A parallel line has the same slope, $2$. The point $(0, -3)$ is the $y$-intercept, so $b = -3$.

$y = 2x - 3$

The line $y = 2x + 1$ is parallel but passes through $(0, 1)$, not $(0, -3)$, so it is not the answer.

A parallel line has slope $-1$, and passing through the origin makes $b = 0$.

$y = -x + 0 = -x$

The line $y = -x + 8$ is parallel but crosses the $y$-axis at $(0, 8)$, not the origin.

Solve the second line for $y$.

$2x + 3y = 9 \;\Rightarrow\; y = -\tfrac{2}{3}x + 3$

The slopes are $\tfrac{3}{2}$ and $-\tfrac{2}{3}$, and $\tfrac{3}{2} \cdot \left(-\tfrac{2}{3}\right) = -1$, so the lines are perpendicular.

A horizontal line has slope $0$, and $0$ has no negative reciprocal, so the product rule does not apply here.

$\text{horizontal} \perp \text{vertical}$

The partner that meets a horizontal line at a right angle is a vertical line, of the form $x = c$.

Read the steepness of each line off the grid. Each one rises $1$ unit for every $2$ units it runs, so both have slope $\tfrac{1}{2}$.

$m_1 = m_2 = \tfrac{1}{2}$

Equal slopes with different $y$-intercepts, and the lines never meet, so they are parallel.

Solve the second equation for $y$.

$x - 2y = -8 \;\Rightarrow\; y = \tfrac{1}{2}x + 4$

Both the slope and the intercept match the first line, $\tfrac{1}{2}$ and $4$, so the two equations describe the same line.

Solve the given line for $y$.

$4x + 5y = 20 \;\Rightarrow\; y = -\tfrac{4}{5}x + 4$

Its slope is $-\tfrac{4}{5}$, and a parallel line has the same slope, $-\tfrac{4}{5}$.