Question

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

Answer 1

`\large\boxed{y=-(1)/(3)x-(10)/(3)\to x+3y=-10}`

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

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

Substiute:

`m=(-2-(-5))/(-4-5)=(-2+5)/(-9)=(3)/(-9)=-(1)/(3)`

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

`-5=-(1)/(3)(5)+b`

`-5=-(5)/(3)+b` add 5/3 to both sides

`-(15)/(3)+(5)/(3)=b\\\\-(10)/(3)=b\to b=-(10)/(3)`

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

`y=-(1)/(3)x-(10)/(3)`

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

`y=-(1)/(3)x-(10)/(3)` multiply both sides by 3

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

`x+3y=-10`