Question

1. find the equation of a line perpendicular to y = 3x - 4 that passes through (6, -7)

Explanation:

Step1: Identify slope of given line

The given line $y=3x-4$ is in slope-intercept form $y=mx+b$, so its slope $m_1=3$.

Step2: Find perpendicular slope

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

Step3: Use point-slope form

Point-slope formula: $y - y_1 = m(x - x_1)$, where $(x_1,y_1)=(6,-7)$:
$$y - (-7) = -\frac{1}{3}(x - 6)$$

Step4: Simplify to slope-intercept form

Simplify the equation:
$$y + 7 = -\frac{1}{3}x + 2$$
$$y = -\frac{1}{3}x + 2 - 7$$
$$y = -\frac{1}{3}x - 5$$

Answer:

$y = -\frac{1}{3}x - 5$