Question

Find an equation for the line that passes through the points (5.-5) and (-4,-2)

Answer 1

Answer:

$\large\boxed{y=-\dfrac{1}{3}x-\dfrac{10}{3}\to x+3y=-10}$

Step-by-step explanation:

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

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

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

We have the points (5, -5) and (-4, -2).

Substiute:

$m=\dfrac{-2-(-5)}{-4-5}=\dfrac{-2+5}{-9}=\dfrac{3}{-9}=-\dfrac{1}{3}$

Put the value of the slope and the coordinates of the point (5, -5) to the equation of a line:

$-5=-\dfrac{1}{3}(5)+b$

$-5=-\dfrac{5}{3}+b$               add 5/3 to both sides

$-\dfrac{15}{3}+\dfrac{5}{3}=b\\\\-\dfrac{10}{3}=b\to b=-\dfrac{10}{3}$

Finally we have the equation of a line in the slope-intercept form:

$y=-\dfrac{1}{3}x-\dfrac{10}{3}$

Convert to the standard form (Ax + By = C):

$y=-\dfrac{1}{3}x-\dfrac{10}{3}$          multiply both sides by 3

$3y=-x-10$              add x to both sides

$x+3y=-10$