Question

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

Answer 1

Point-Slope form:
`y+1=-(6)/(15)(x+5)\\`

standard form:
`2x+5y=-15`

Explanation:

The standard form is given by:
`Ax+By=C`

The point-slope form is given by:
`y-y_1=m(x-x_1)`

• Let's call
  `(-5,-1)` as point
  `(x_1,y_1)`, and

point
`(10,-7)` as point
`(x_2,y_2)`

• We also know that the slope
  `m` is given by
  `m=(y_2-y_1)/(x_2-x_1)`

Let's find the slope:

`m=(-7-(-1))/(10-(-5))=-(6)/(15)`

1. So, point slope form can be written as:

`y-(-1)=-(6)/(15)(x-(-5))\\y+1=-(6)/(15)(x+5)\\`

2. Rearranging this equation and bringing all x's and y's to one side and the number to another side will give us the standard form. So:

`y+1=-(6)/(15)x-2\\y+(6)/(15)x=-2-1\\y+(6)/(15)x=-3`

But A, B, and C (standard form) needs to be integers, so to get rid of the denominator , we multiply the whole equation by 15. So we have:

`y+(6)/(15)x=-3\\15y+6x=-45`

We can factor out a 3, so it becomes:

`15y+6x=-45\\5y+2x=-15`

So standard form is
`2x+5y=-15`