Question

How do you write an equation in standard form given point (-1,0) and slope -2/3?

Answer 1

`2x+3y+2=0`

Explanation:

The equation of a line in `color(blue)"standard form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(ax+by+c=0)color(white)(2/2)|)))`
where a is a positive integer and b, c are integers.

To begin, express the equation in `color(blue)"point-slope form"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`
where m represents the slope and `(x_1,y_1)" a point on the line"`

`"here "m=-2/3" and "(x_1,y_1)=(-1,0)`

substitute values into point-slope equation.

`y-0=-2/3(x-(-1))`

`rArry=-2/3(x+1)larrcolor(red)" point-slope form"`

distributing.

`y=-2/3x-2/3`

add `2/3x" to both sides"`

`2/3x+y=cancel(-2/3x)cancel(+2/3x)-2/3`

`rArr2/3x+y=-2/3`

`"add "2/3" to both sides"`

`2/3x+y+2/3=cancel(-2/3)cancel(+2/3)`

`rArr2/3x+y+2/3=0`

multiply ALL terms by 3

`rArr2x+3y+2=0larrcolor(red)" standard form"`