Question

what is the equation of the line that is perpendicular to the given line and has an x - intercept of 6?
o y = -\frac{3}{4}x + 8
o y = -\frac{3}{4}x + 6
o y = \frac{4}{3}x - 8
o y = \frac{4}{3}x - 6

Explanation:

Step1: Find the slope of the given line

The slope $m_1$ of the line passing through points $(-4,4)$ and $(4, - 2)$ is $m_1=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-2 - 4}{4+4}=\frac{-6}{8}=-\frac{3}{4}$.

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is - 1. Let the slope of the perpendicular line be $m_2$. Then $m_1m_2=-1$. Substituting $m_1 = -\frac{3}{4}$, we get $-\frac{3}{4}m_2=-1$, so $m_2=\frac{4}{3}$.

Step3: Use the x - intercept to find the equation

The x - intercept is 6, which means the point $(6,0)$ lies on the line. Using the point - slope form $y - y_1=m(x - x_1)$ with $m=\frac{4}{3}$ and $(x_1,y_1)=(6,0)$, we have $y-0=\frac{4}{3}(x - 6)$. Simplifying gives $y=\frac{4}{3}x-8$.

Answer:

$y=\frac{4}{3}x - 8$