Question

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

NEED ASAP!!

Answer 1

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

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=(y_2-y_1)/(x_2-x_1)`

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

Calculate the slope:

`m=(8-(-2))/(6-(-6))=(8)/(12)=(8:4)/(12:4)=(2)/(3)`

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

`y-8=(2)/(3)(x-6)`

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

`y-8=(2)/(3)(x-6)` use the distributive property a(b + c) = ab + ac

`y-8=(2)/(3)x-\left((2)/(3\!\!\!\!\diagup_1)\right)(6\!\!\!\!\diagup^2)`

`y-8=(2)/(3)x-4` add 8 to both sides

`y=(2)/(3)x+4`

Convert to the standard form Ax + By = C:

`y=(2)/(3)x+4` multiply both sides by 3

`3y=3\!\!\!\!\diagup^1\cdot(2)/(3\!\!\!\!\diagup_1)x+(3)(4)`

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

`-2x+3y=12` change the signs

`2x-3y=-12`