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find the equation of the line perpendicular to $y = -\frac{3}{4}x + 8$,…

Question

find the equation of the line perpendicular to $y = -\frac{3}{4}x + 8$, that passes through $(-9, -2)$. $y = \frac{?}{?}x + \square$

Explanation:

Step1: Find perpendicular slope

The slope of the given line is $m_1 = -\frac{3}{4}$. The slope of a perpendicular line is the negative reciprocal: $m_2 = \frac{4}{3}$.

Step2: Solve for intercept b

Substitute $x=-9$, $y=-2$, and $m_2=\frac{4}{3}$ into $y=mx+b$:
$$-2 = \frac{4}{3}(-9) + b$$
$$-2 = -12 + b$$
$$b = -2 +12 = 10$$

Answer:

$y = \frac{4}{3}x + 10$