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

Answer:

Point-Slope form:  $y+1=-\frac{6}{15}(x+5)$

standard form:  $2x+5y=-15$

Step-by-step 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 $m=\frac{y_2-y_1}{x_2-x_1}$

Let's find the slope:

$m=\frac{-7-(-1)}{10-(-5)}=-\frac{6}{15}$

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

$y-(-1)=-\frac{6}{15}(x-(-5))\\y+1=-\frac{6}{15}(x+5)$

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:

$y+1=-\frac{6}{15}x-2\\y+\frac{6}{15}x=-2-1\\y+\frac{6}{15}x=-3$

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:

$y+\frac{6}{15}x=-3\\15y+6x=-45$

We can factor out a 3, so it becomes:

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

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