Question

question
which equation represents a line which is perpendicular to the line $y = \frac{3}{8}x - 1?$
answer
$3x + 8y = -40$
$3y - 8x = 3$
$3x - 8y = -56$
$8x + 3y = 9$

Explanation:

Step1: Identify given line slope

The given line is $y = \frac{3}{8}x - 1$, so its slope $m_1 = \frac{3}{8}$.

Step2: Find perpendicular slope

Perpendicular slopes multiply to $-1$. Let $m_2$ be the perpendicular slope:
$$m_2 = -\frac{1}{m_1} = -\frac{8}{3}$$

Step3: Convert options to slope-intercept form

Option1: $3x + 8y = -40$

Rearrange to $y = -\frac{3}{8}x - 5$, slope $= -\frac{3}{8}$

Option2: $3x - 8y = -56$

Rearrange to $y = \frac{3}{8}x + 7$, slope $= \frac{3}{8}$

Option3: $3y - 8x = 3$

Rearrange to $y = \frac{8}{3}x + 1$, slope $= \frac{8}{3}$

Option4: $8x + 3y = 9$

Rearrange to $y = -\frac{8}{3}x + 3$, slope $= -\frac{8}{3}$

Step4: Match to perpendicular slope

The slope $-\frac{8}{3}$ matches the required perpendicular slope.

Answer:

$8x + 3y = 9$