Question

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

Explanation:

Step1: Find slope of given line

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

Step2: Determine perpendicular slope

Perpendicular lines have slopes that are negative reciprocals, so $m_2 = -3$ (since $\frac{1}{3} \times (-3) = -1$).

Step3: Check each option's slope

1. $y + 2 = -3(x - 4)$: Slope is $-3$ (matches).
2. $y - 5 = 3(x + 11)$: Slope is $3$ (does not match).
3. $y = -3x - \frac{5}{3}$: Slope is $-3$ (matches).
4. $y = \frac{1}{3}x - 2$: Slope is $\frac{1}{3}$ (does not match).
5. $3x + y = 7$: Rewrite as $y = -3x + 7$, slope is $-3$ (matches).

Answer:

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