The point-slope form of the equation of the line that passes through #(-5,-1)# and #(10, -7)# is #y + 7 =-2/5(x-10).# What is the standard form of the equation for this line?

2 Answers
Jan 23, 2018

#2/5x + y = -3#

Explanation:

The format of standard form for an equation of a line is #Ax+By = C#.

The equation that we have, #y+7 = -2/5(x-10)# is currently in point-slope form.

The first thing to do is to distribute the #-2/5(x-10)#:
#y + 7 = -2/5(x-10)#

#y + 7 = -2/5x+ 4#

Now let's subtract #4# from both sides of the equation:
#y + 3 = -2/5x#

Since the equation needs to be #Ax+By = C#, let's move #3# to the other side of the equation and #-2/5x# to the other side of the equation:
#2/5x + y = -3#

This equation is now in standard form.

Jan 23, 2018

#2x-5y=15#

Explanation:

#"the equation of a line in standard form is."#

#color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))#

#"where A is a positive integer and B, C are integers"#

#"rearrange "y+7=-2/5(x-10)" into this form"#

#y+7=2/5x+4larrcolor(blue)"distributing"#

#rArry=2/5x-3larrcolor(blue)"collecting like terms"#

#"multiply through by 5"#

#rArr5y=2x-15#

#rArr2x-5y=15larrcolor(red)"in standard form"#