Question

The point-slope form of the equation of the line that passes through (-5,-1) and (10,-7) 157Wt is the ), What is the standard form of the equation for this line?

Answer 1

Answer:

$y + 1 = -\frac{2}{5}(x+5)$

and

2x + 15y = -15

Step-by-step explanation:

To write the point slope form, a point and a slope is required. Find the slope using the two points given and the slope formula.

$m = \frac{y_2-y_1}{x_2-x_1}= \frac{-1 --7}{-5-10}= \frac{6}{-15} = -\frac{2}{5}$

Substitute -2/5 and the point (-5,-1) into the form. Then convert to make the standard form of the equation.

$y - y_1 = m(x-x_1)\\y --1=-\frac{2}{5}(x--5)\\y + 1 = -\frac{2}{5}(x+5)$

Now convert to standard form by applying the distributive property and moving terms.

$y + 1 = -\frac{2}{5}(x+5)\\y + 1 = -\frac{2}{5}x - 2\\y + 3 = -\frac{2}{5}x\\\frac{2}{5}x + y +3=0\\2x + 5y + 15 = 0\\2x + 15y = -15$