Question

The point-slope form of the equation of the line that passes through (-4,-3) and (12, 1) is y-1= 164–12). What is the standard form of the equation for this line?​

Answer 1

`y = (1)/(4)x -2`

Explanation:

Step 1: Find the standard form of the equation

The equation that was given made no sense so I will recreate the entire equation using the point slope formula.

Use the point slope formula

`y - y_(1) = m(x - x_(1))`

`y - (-3) = m(x - (-4))`

`y +3 = m(x + 4)`

Find the slope

`m = (y_(2)-y_(1))/(x_(2)-x_(1))`

`m = (1-(-3))/(12-(-4))`

`m = (1+3)/(12+4)`

`m = (4)/(16)`

`m=(1)/(4)`

Combine them together

`y +3 = (1)/(4)(x + 4)`

Convert to standard form

`y +3 = (1)/(4)x + 1`

`y +3 - 3 = (1)/(4)x + 1 - 3`

`y = (1)/(4)x -2`

Answer:
`y = (1)/(4)x -2`