Question

What is an equation of the line that passes through the points (-6, -2) and (6,8)?

Answer 1

Answer:

$y-8=\dfrac{2}{3}(x-6)\ -\ \text{point-slope form}\\\\y=\dfrac{2}{3}x+4\ -\ \text{slope-intercept form}\\\\2x-3y=-12\ -\ \text{standard form}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

=============================================

We have two points (-6, -2) and (6, 8).

Calculate the slope:

$m=\dfrac{8-(-2)}{6-(-6)}=\dfrac{8}{12}=\dfrac{8:4}{12:4}=\dfrac{2}{3}$

Put it and the coordinates of the point (6, 8) to the equation of a line:

$y-8=\dfrac{2}{3}(x-6)$

Convert to the slope-intercept form y = mx + b:

$y-8=\dfrac{2}{3}(x-6)$     use the distributive property a(b + c) = ab + ac

$y-8=\dfrac{2}{3}x-\left(\dfrac{2}{3\!\!\!\!\diagup_1}\right)(6\!\!\!\!\diagup^2)$

$y-8=\dfrac{2}{3}x-4$            add 8 to both sides

$y=\dfrac{2}{3}x+4$

Convert to the standard form Ax + By = C:

$y=\dfrac{2}{3}x+4$            multiply both sides by 3

$3y=3\!\!\!\!\diagup^1\cdot\dfrac{2}{3\!\!\!\!\diagup_1}x+(3)(4)$

$3y=2x+12$        subtract 2x from both sides

$-2x+3y=12$          change the signs

$2x-3y=-12$