Question

1. select all the equations that represent lines that are perpendicular to the graph of 6x + 18y = 5. circle the letter of the correct answer. a. y = 3x - 10 b. x = 3 c. y + 6 = 3(x - 15) d. 3x + 9y = 8 e. 2x - 3y = 5

Explanation:

Step1: Rewrite given line in slope - intercept form

Rewrite $6x + 18y=5$ as $y=mx + b$. Solve for $y$:
$18y=-6x + 5$, so $y=-\frac{6}{18}x+\frac{5}{18}=-\frac{1}{3}x+\frac{5}{18}$. The slope of this line is $m_1 =-\frac{1}{3}$.

Step2: Determine the slope of perpendicular lines

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the perpendicular line be $m_2$. Then $m_1\times m_2=-1$. Substituting $m_1 =-\frac{1}{3}$, we get $-\frac{1}{3}m_2=-1$, so $m_2 = 3$.

Step3: Check the slopes of each option

Option A

The equation $y = 3x-10$ is in slope - intercept form $y=mx + b$ with $m = 3$.

Option B

The equation $x = 3$ is a vertical line. The slope of a vertical line is undefined, not $3$.

Option C

Rewrite $y + 6=3(x - 15)$ in slope - intercept form: $y+6 = 3x-45$, so $y=3x-51$. The slope is $3$.

Option D

Rewrite $3x + 9y=8$ as $y=mx + b$. Solve for $y$: $9y=-3x + 8$, so $y=-\frac{3}{9}x+\frac{8}{9}=-\frac{1}{3}x+\frac{8}{9}$. The slope is $-\frac{1}{3}$, not $3$.

Option E

Rewrite $2x-3y=5$ as $y=mx + b$. Solve for $y$: $-3y=-2x + 5$, so $y=\frac{2}{3}x-\frac{5}{3}$. The slope is $\frac{2}{3}$, not $3$.

Answer:

A. $y = 3x-10$, C. $y + 6=3(x - 15)$