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

##### 1 Answer

#### 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"#