Question

what is the equation of the line that is parallel to the line $y = -\frac{1}{3}x + 4$ and passes through the point $(6, 5)$?
$y = -\frac{1}{3}x + 7$
$y = 3x + 5$
$y = 3x - 13$
$y = -\frac{1}{3}x + 3$

Explanation:

Step1: Determine the slope

Parallel lines have the same slope. The given line is \( y = -\frac{1}{3}x + 4 \), so the slope \( m = -\frac{1}{3} \).

Step2: Use point - slope form

The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(6,5) \) and \( m = -\frac{1}{3} \).
Substitute the values: \( y - 5=-\frac{1}{3}(x - 6) \)

Step3: Simplify the equation

Expand the right - hand side: \( y - 5=-\frac{1}{3}x+2 \)
Add 5 to both sides: \( y=-\frac{1}{3}x + 2 + 5 \)
\( y=-\frac{1}{3}x+7 \)

Answer:

\( y = -\frac{1}{3}x + 7 \)