Question

find an equation for the line that passes through the points (-4, -5) and (2, -3).

Explanation:

Step1: Calculate the slope

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Here, \((x_1,y_1)=(-4,-5)\) and \((x_2,y_2)=(2,-3)\). So, \( m=\frac{-3 - (-5)}{2 - (-4)}=\frac{-3 + 5}{2 + 4}=\frac{2}{6}=\frac{1}{3} \).

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 \((2,-3)\) (we could also use \((-4,-5)\)). Substituting \( m = \frac{1}{3} \), \( x_1 = 2 \) and \( y_1=-3 \) into the formula, we get \( y - (-3)=\frac{1}{3}(x - 2) \).
Simplify this equation: \( y + 3=\frac{1}{3}x-\frac{2}{3} \).
Subtract 3 from both sides: \( y=\frac{1}{3}x-\frac{2}{3}-3 \).
Since \( 3=\frac{9}{3} \), then \( y=\frac{1}{3}x-\frac{2 + 9}{3}=\frac{1}{3}x-\frac{11}{3} \). We can also write it in standard form \( x-3y = 11 \) (by multiplying both sides by 3: \( 3y=x - 11 \), then \( x-3y=11 \)) or in slope - intercept form \( y=\frac{1}{3}x-\frac{11}{3} \).

Answer:

The equation of the line is \( y=\frac{1}{3}x-\frac{11}{3} \) (or \( x - 3y=11 \))