Question

find an equation for the line that passes through the points (2, -4) and (-4, -2).

Explanation:

Step1: Calculate the slope

The slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
For the points \( (2, -4) \) and \( (-4, -2) \), we have \( x_1 = 2 \), \( y_1 = -4 \), \( x_2 = -4 \), \( y_2 = -2 \).
So, \( m = \frac{-2 - (-4)}{-4 - 2} = \frac{-2 + 4}{-6} = \frac{2}{-6} = -\frac{1}{3} \).

Step2: Use point - slope form

The point - slope form of a line is \( y - y_1 = m(x - x_1) \). Let's use the point \( (2, -4) \) and \( m = -\frac{1}{3} \).
Substitute into the formula: \( y - (-4)=-\frac{1}{3}(x - 2) \).
Simplify: \( y + 4 = -\frac{1}{3}x+\frac{2}{3} \).
Subtract 4 from both sides: \( y=-\frac{1}{3}x+\frac{2}{3}-4 \).
Since \( 4=\frac{12}{3} \), we have \( y = -\frac{1}{3}x+\frac{2 - 12}{3}=-\frac{1}{3}x-\frac{10}{3} \).
We can also write it in standard form \( x + 3y=- 10 \) (by multiplying through by 3: \( 3y=-x - 10\Rightarrow x + 3y=-10 \)) or in slope - intercept form \( y=-\frac{1}{3}x-\frac{10}{3} \).

Answer:

\( y = -\frac{1}{3}x-\frac{10}{3} \) (or \( x + 3y=-10 \))