Parallel and Perpendicular Lines: Practice

The perpendicular slope is the negative reciprocal of $\tfrac{3}{5}$, which is $-\tfrac{5}{3}$. Use point-slope form with $(3, -2)$.

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

At $x = 3$ this gives $y = -5 + 3 = -2$, so the line passes through $(3, -2)$.

Solve each line for $y$.

$2x - 5y = 10 \;\Rightarrow\; y = \tfrac{2}{5}x - 2, \qquad 5x + 2y = 8 \;\Rightarrow\; y = -\tfrac{5}{2}x + 4$

The slopes are $\tfrac{2}{5}$ and $-\tfrac{5}{2}$, and $\tfrac{2}{5} \cdot \left(-\tfrac{5}{2}\right) = -1$, so the lines are perpendicular.

First find the slope of the given line with the slope formula.

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

The perpendicular slope is the negative reciprocal of $-\tfrac{1}{2}$: flip to $-2$ and change the sign, giving $2$.

The line $y = 6$ is horizontal (slope $0$), so any parallel line is also horizontal, of the form $y = c$.

$y = -5$

To pass through $(2, -5)$ the constant height must be $-5$, giving $y = -5$.

Solve the second line for $y$: $-2y = -3x + 8$, so $y = \tfrac{3}{2}x - 4$, with slope $\tfrac{3}{2}$.

$\tfrac{2}{3} \cdot \tfrac{3}{2} = 1 \neq -1$

The slopes $\tfrac{2}{3}$ and $\tfrac{3}{2}$ are reciprocals but not negative reciprocals, and they are not equal, so the lines are neither parallel nor perpendicular.

Solve the given line for $y$: $y = 2x - 7$, with slope $2$. The perpendicular slope is $-\tfrac{1}{2}$. Use point-slope form with $(4, -3)$.

$y + 3 = -\tfrac{1}{2}(x - 4) \;\Rightarrow\; y = -\tfrac{1}{2}x - 1$

At $x = 4$ this gives $y = -2 - 1 = -3$, so the line passes through $(4, -3)$.

The line $x = -2$ is vertical, so a perpendicular line is horizontal, of the form $y = c$.

$y = 1$

To pass through $(5, 1)$ the height must be $1$, giving $y = 1$.

Parallel lines have equal slopes, so $k$ must match the slope of the given line.

$k = -\frac{7}{2}$

The negative reciprocal $\tfrac{2}{7}$ would make the lines perpendicular instead.

The perpendicular slope is $-3$, and the point $(0, 4)$ is the $y$-intercept, so $\ell$ is $y = -3x + 4$. Set $y = 0$ to find the $x$-intercept.

$0 = -3x + 4 \;\Rightarrow\; x = \tfrac{4}{3}$

So $\ell$ crosses the $x$-axis at $\left(\tfrac{4}{3}, 0\right)$.

Solve the given line for $y$: $3y = -4x + 12$, so $y = -\tfrac{4}{3}x + 4$, with slope $-\tfrac{4}{3}$. The perpendicular slope is $\tfrac{3}{4}$, and through the origin $b = 0$.

$y = \tfrac{3}{4}x$

The slope $\tfrac{3}{4}$ times $-\tfrac{4}{3}$ gives $-1$, confirming the right angle.

Solve the first line for $y$: $4y = -6x + 1$, so $y = -\tfrac{3}{2}x + \tfrac{1}{4}$, with slope $-\tfrac{3}{2}$.

$m_1 = m_2 = -\tfrac{3}{2}, \quad \tfrac{1}{4} \neq 10$

Equal slopes with different intercepts, so the lines are parallel.

Solve the given line for $y$: $-4y = -x + 8$, so $y = \tfrac{1}{4}x - 2$, with slope $\tfrac{1}{4}$. A parallel line has the same slope. Use point-slope form with $(-6, 1)$.

$y - 1 = \tfrac{1}{4}(x + 6) \;\Rightarrow\; y = \tfrac{1}{4}x + \tfrac{5}{2}$

At $x = -6$ this gives $y = -\tfrac{3}{2} + \tfrac{5}{2} = 1$, so the line passes through $(-6, 1)$.