How do you write an equation in standard form for a line that passes through (-5, -2) and (1, -6)?

1 Answer
May 10, 2018

#2x+3y=-16#

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)color(white)(2/2)|)))#

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

#"obtain the equation in "color(blue)"point-slope form"#
#"and rearrange into standard form"#

#•color(white)(x)y-y_1=m(x-x_1)larrcolor(blue)"point-slope form"#

#"where m is the slope and "(x_1,y_1)" a point on the line"#

#"to calculate m use the "color(blue)"gradient formula"#

#•color(white)(x)m=(y_2-y_1)/(x_2-x_1)#

#"let "(x_1,y_1)=(-5,-2)" and "(x_2,y_2)=(1,-6)#

#rArrm=(-6-(-2))/(1-(-5))=(-4)/6=-2/3#

#"using "m=-2/3" and "(x_1,y_1)=(1,-6)" then"#

#y-(-6)=-2/3(x-1)#

#rArry+6=-2/3(x-1)larrcolor(magenta)"in point-slope form"#

#rArry+6=-2/3x+2/3#

#rArry=-2/3x+2/3-6#

#rArry=-2/3x-16/3#

#"multiply all terms by 3"#

#rArr3y=-2x-16#

#rArr2x+3y=-16larrcolor(magenta)"in standard form"#