Parallel and Perpendicular Lines: Additional Practice (Set 1)

A parallel line has the same slope, $\tfrac{3}{4}$. Use point-slope form with $(-4, 3)$.

$y - 3 = \tfrac{3}{4}(x + 4) \;\Rightarrow\; y = \tfrac{3}{4}x + 6$

At $x = -4$ this gives $y = -3 + 6 = 3$, so the line passes through $(-4, 3)$.

Solve the second line for $y$.

$x + 7y = 14 \;\Rightarrow\; y = -\tfrac{1}{7}x + 2$

The slopes are $7$ and $-\tfrac{1}{7}$, and $7 \cdot \left(-\tfrac{1}{7}\right) = -1$, so the lines are perpendicular. They cross at their shared intercept $(0, 2)$, and they cross there at a right angle.

Solve the given line for $y$: $4y = -3x + 8$, so $y = -\tfrac{3}{4}x + 2$, with slope $-\tfrac{3}{4}$. The perpendicular slope is $\tfrac{4}{3}$. Use point-slope form with $(6, -1)$.

$y + 1 = \tfrac{4}{3}(x - 6) \;\Rightarrow\; y = \tfrac{4}{3}x - 9$

At $x = 6$ this gives $y = 8 - 9 = -1$, so the line passes through $(6, -1)$.

Multiply the two slopes and compare with $-1$.

$-\frac{9}{4} \cdot \frac{4}{9} = -\frac{36}{36} = -1$

The product is $-1$, so the lines are perpendicular. The two slopes are negative reciprocals of each other.

Solve the given line for $y$: $y = 3x$, with slope $3$. A parallel line has slope $3$, and $(0, -7)$ is its $y$-intercept.

$y = 3x - 7$

The line $y = 3x$ is parallel but passes through the origin, not $(0, -7)$.

Solve the second line for $y$: $-5y = -6x + 20$, so $y = \tfrac{6}{5}x - 4$, with slope $\tfrac{6}{5}$.

$-\frac{5}{6} \cdot \frac{6}{5} = -1$

The product of the slopes is $-1$, so the lines are perpendicular.

Solve the first line for $y$: $3y = -kx + 1$, so $y = -\tfrac{k}{3}x + \tfrac{1}{3}$, with slope $-\tfrac{k}{3}$. Parallel lines have equal slopes, so set this equal to $2$.

$-\frac{k}{3} = 2 \;\Rightarrow\; k = -6$

So $k = -6$ makes the two lines parallel.

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

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

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

A line parallel to $\ell$ has the same slope as $\ell$, so $\ell$ has slope $\tfrac{5}{9}$. A line perpendicular to $\ell$ has the negative reciprocal slope.

$m_{\perp} = -\frac{9}{5}$

Flip $\tfrac{5}{9}$ to $\tfrac{9}{5}$ and change the sign to get $-\tfrac{9}{5}$.

Find the slopes of the two sides meeting at $A$ with the slope formula.

$m_{AB} = \frac{2 - 0}{4 - 0} = \tfrac{1}{2}, \qquad m_{AC} = \frac{-4 - 0}{2 - 0} = -2$

Their product is $\tfrac{1}{2} \cdot (-2) = -1$, so $AB$ and $AC$ are perpendicular and the angle at $A$ is a right angle.

The perpendicular slope is the negative reciprocal of $\tfrac{6}{5}$, which is $-\tfrac{5}{6}$. The $y$-intercept $(0, 4)$ makes $b = 4$.

$y = -\tfrac{5}{6}x + 4$

The slope $-\tfrac{5}{6}$ times $\tfrac{6}{5}$ gives $-1$, confirming the right angle.

Perpendicular lines have slopes whose product is $-1$. Check each pair's slopes.

$\tfrac{1}{4} \cdot (-4) = -1$

Only $y = \tfrac{1}{4}x$ and $y = -4x + 2$ give a product of $-1$. The first pair is parallel (slopes $2$ and $2$), and the other two pairs give products $-9$ and $-\tfrac{4}{9}$, neither equal to $-1$.