Question

which equation represents the line that passes through (-6, 7) and (-3, 6)?
○ $y = -\frac{1}{3}x + 9$
○ $y = -\frac{1}{3}x + 5$
○ $y = -3x - 11y$
○ $y = -3x + 25$

Explanation:

Step1: Calculate the slope (m)

The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Let \((x_1, y_1) = (-6, 7)\) and \((x_2, y_2) = (-3, 6)\).
\( m = \frac{6 - 7}{-3 - (-6)} = \frac{-1}{3} = -\frac{1}{3} \)

Step2: Use point - slope form to find the equation

The point - slope form is \( y - y_1 = m(x - x_1) \). Using the point \((-6, 7)\) and \( m = -\frac{1}{3} \):
\( y - 7 = -\frac{1}{3}(x + 6) \)
Expand the right - hand side: \( y - 7 = -\frac{1}{3}x - 2 \)
Add 7 to both sides: \( y = -\frac{1}{3}x + 5 \) (We can also check by plugging in the points into the equations. For the point \((-6,7)\):

• For \( y = -\frac{1}{3}x + 9 \), when \( x=-6 \), \( y = -\frac{1}{3}(-6)+9 = 2 + 9 = 11

eq7 \)

• For \( y = -\frac{1}{3}x + 5 \), when \( x = - 6 \), \( y=-\frac{1}{3}(-6)+5=2 + 5 = 7 \); when \( x=-3 \), \( y = -\frac{1}{3}(-3)+5 = 1 + 5 = 6 \)
• For \( y=-3x - 11y \), this is not a linear equation in the form \( y=mx + b \) (it has a \( y \) on both sides and is not in slope - intercept form)
• For \( y=-3x + 25 \), when \( x=-6 \), \( y=-3(-6)+25=18 + 25 = 43

eq7 \)

Answer:

B. \( y = -\frac{1}{3}x + 5 \)