Question

line fg is represented by the equation 8x + 3y = 10. which value is the slope of a line perpendicular to line fg? -3/8 -1/8 3/8 8/3

Explanation:

Step1: Rewrite the line equation in slope - intercept form

The slope - intercept form is $y = mx + b$, where $m$ is the slope. Given $8x + 3y=10$, we solve for $y$:
$3y=-8x + 10$, so $y=-\frac{8}{3}x+\frac{10}{3}$. The slope of line FG, $m_1 =-\frac{8}{3}$.

Step2: Use the perpendicular - slope relationship

If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1\times m_2=- 1$. Let the slope of the perpendicular line be $m_2$. We know $m_1 =-\frac{8}{3}$, so $-\frac{8}{3}\times m_2=-1$. Solving for $m_2$ gives $m_2=\frac{3}{8}$.

Answer:

C. $\frac{3}{8}$