Question

practice writing equations of parallel and perpendicular lines. which lines are perpendicular to the line $y - 1=\frac{1}{3}(x + 2)$? check all that apply. $y + 2=-3(x - 4)$ $y - 5 = 3(x + 11)$ $y=-3x-\frac{5}{3}$ $y=\frac{1}{3}x - 2$ $3x + y = 7$

Explanation:

Step1: Identify the slope of the given line

The given line $y - 1=\frac{1}{3}(x + 2)$ is in point - slope form $y - y_1=m(x - x_1)$. Its slope $m_1=\frac{1}{3}$.

Step2: Recall the slope - relationship for perpendicular lines

For two perpendicular lines with slopes $m_1$ and $m_2$, $m_1\times m_2=-1$. So, the slope of a line perpendicular to the given line is $m_2=-3$.

Step3: Rewrite each option in slope - intercept form $y = mx + b$ and find the slope

• For $y + 2=-3(x - 4)$, expand to get $y+2=-3x + 12$, then $y=-3x+10$. Slope $m=-3$.
• For $y - 5 = 3(x + 11)$, expand to get $y-5=3x+33$, then $y = 3x+38$. Slope $m = 3$.
• For $y=-3x-\frac{5}{3}$, slope $m=-3$.
• For $y=\frac{1}{3}x - 2$, slope $m=\frac{1}{3}$.
• For $3x + y=7$, rewrite as $y=-3x + 7$. Slope $m=-3$.

Answer:

1. $y + 2=-3(x - 4)$
2. $y=-3x-\frac{5}{3}$
3. $3x + y = 7$