Question

which linear equation includes the points \\((6, 0)\\) and \\((0, -4)\\)?

hint: use the formula sheet to determine the formula(s) needed to solve the problem.

a. \\(2x + 3y = -12\\)
b. \\(-2x + 3y = 12\\)
c. \\(-2x - 3y = -12\\)
d. \\(2x - 3y = 12\\)

Explanation:

⚡ Using what you learned: slope of a line · definition and standard forms

Step 1: Find the slope

$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 0}{0 - 6} = \frac{-4}{-6} = \frac{2}{3} $$

Step 2: Write the equation in slope-intercept form

Using the \(y\)-intercept \((0, -4)\):

$$ y = \frac{2}{3}x - 4 $$

Step 3: Convert to standard form

Multiply the entire equation by \(3\):

$$ 3y = 2x - 12 $$

Rearrange to match the given options:

$$ -2x + 3y = -12 $$

Multiply by \(-1\) to check alternative signs:

$$ 2x - 3y = 12 $$

This matches option D.

Answer:

D (2x - 3y = 12)