Parallel and Perpendicular Lines: Additional Practice (Set 2)

Solve the given line for $y$: $2y = -3x + 4$, so $y = -\tfrac{3}{2}x + 2$, with slope $-\tfrac{3}{2}$. A parallel line has the same slope. Use point-slope form with $(2, 5)$.

$y - 5 = -\tfrac{3}{2}(x - 2) \;\Rightarrow\; y = -\tfrac{3}{2}x + 8$

The line $y = -\tfrac{3}{2}x + 2$ is parallel but passes through $(2, -1)$, not $(2, 5)$.

Solve the given line for $y$: $-2y = -5x + 9$, so $y = \tfrac{5}{2}x - \tfrac{9}{2}$, with slope $\tfrac{5}{2}$.

$m_{\perp} = -\frac{2}{5}$

The perpendicular slope is the negative reciprocal of $\tfrac{5}{2}$, which is $-\tfrac{2}{5}$.

The line $x = 3$ is vertical (undefined slope) and $y = -4$ is horizontal (slope $0$).

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

A vertical line and a horizontal line meet at a right angle, so they are perpendicular. The slope product rule cannot be used, since a vertical line has no slope number.

The line $y = 8$ is horizontal, so a perpendicular line is vertical, of the form $x = c$.

$x = -3$

To pass through $(-3, 5)$ the vertical line must be $x = -3$.

Read each slope off the grid. The steep line rises $2$ for every $1$ across, slope $2$; the shallow line falls $1$ for every $2$ across, slope $-\tfrac{1}{2}$.

$2 \cdot \left(-\tfrac{1}{2}\right) = -1$

The product is $-1$, and the small square confirms the right angle, so the lines are perpendicular.

Solve the second line for $y$: $6y = -4x + 5$, so $y = -\tfrac{2}{3}x + \tfrac{5}{6}$, with slope $-\tfrac{2}{3}$. For perpendicular lines, $k$ is the negative reciprocal of $-\tfrac{2}{3}$.

$k = \frac{3}{2}$

As a check, $\tfrac{3}{2} \cdot \left(-\tfrac{2}{3}\right) = -1$.

For perpendicular lines, $k$ is the negative reciprocal of $\tfrac{2}{9}$: flip to $\tfrac{9}{2}$ and change the sign.

$k = -\frac{9}{2}$

As a check, $\tfrac{2}{9} \cdot \left(-\tfrac{9}{2}\right) = -1$.

Solve the given line for $y$: $-2y = -x + 6$, so $y = \tfrac{1}{2}x - 3$, with slope $\tfrac{1}{2}$. The perpendicular slope is $-2$. Use point-slope form with $(-1, -5)$.

$y + 5 = -2(x + 1) \;\Rightarrow\; y = -2x - 7$

At $x = -1$ this gives $y = 2 - 7 = -5$, so the line passes through $(-1, -5)$.

The slope $m$ is the negative reciprocal of $-\tfrac{3}{7}$: flip to $-\tfrac{7}{3}$ and change the sign.

$m = \frac{7}{3}$

As a check, $\tfrac{7}{3} \cdot \left(-\tfrac{3}{7}\right) = -1$.

Flip $\tfrac{11}{4}$ to $\tfrac{4}{11}$ and change the sign.

$m_{\perp} = -\frac{4}{11}$

As a check, $\tfrac{11}{4} \cdot \left(-\tfrac{4}{11}\right) = -1$.

Both lines run straight up and down, side by side, and never meet.

$x = 2 \parallel x = -3$

All vertical lines are parallel to one another, so these two are parallel. They have no slope number, so the slope rules are not needed here.

First find the slope of $\ell$ with the slope formula.

$m = \frac{6 - 0}{0 - 2} = \frac{6}{-2} = -3$

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