Question

what is an equation of the line that passes through the points (-8,6) and (-6,3)?

Explanation:

Step1: Find the slope (m)

The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. Let $(x_1, y_1) = (-8, 6)$ and $(x_2, y_2) = (-6, 3)$. Then $m = \frac{3 - 6}{-6 - (-8)} = \frac{-3}{2} = -\frac{3}{2}$.

Step2: Use point - slope form

The point - slope form of a line is $y - y_1 = m(x - x_1)$. We can use the point $(-8, 6)$ and $m = -\frac{3}{2}$. So $y - 6 = -\frac{3}{2}(x - (-8))$, which simplifies to $y - 6 = -\frac{3}{2}(x + 8)$.

Step3: Convert to slope - intercept form

Expand the right - hand side: $y - 6 = -\frac{3}{2}x - 12$. Then add 6 to both sides: $y = -\frac{3}{2}x - 12 + 6$, so $y = -\frac{3}{2}x - 6$. We can also convert it to standard form $Ax+By = C$. Multiply through by 2: $2y=-3x - 12$, then $3x+2y=-12$.

Answer:

The equation of the line can be $y = -\frac{3}{2}x - 6$ (slope - intercept form) or $3x + 2y=-12$ (standard form) or $y - 6 = -\frac{3}{2}(x + 8)$ (point - slope form)