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

Answer:

$y = \frac{1}{4}x -2$

Step-by-step 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

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m = \frac{1-(-3)}{12-(-4)}$

$m = \frac{1+3}{12+4}$

$m = \frac{4}{16}$

$m=\frac{1}{4}$

Combine them together

$y +3 = \frac{1}{4}(x + 4)$

Convert to standard form

$y +3 = \frac{1}{4}x + 1$

$y +3 - 3 = \frac{1}{4}x + 1 - 3$

$y = \frac{1}{4}x -2$

Answer:  $y = \frac{1}{4}x -2$