How do you determine the standard form of the equation of the line that passes through (-7, 8) and (0, -8)?

1 Answer
Apr 25, 2018

#6x+7y=-56#

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"#

#•color(white)(x)y-y_1=m(x-x_1)#

#"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)=(-7,8)" and "(x_2,y_2)=(0,-8)#

#rArrm=(-8-8)/(0-(-7))=(-16)/7=-16/7#

#"using "m=-16/7" and "(x_1,y_1)=(-7,8)" then"#

#y-8=-16/7(x-(-7))#

#rArry-8=-16/7(x+7)larrcolor(red)"in point-slope form"#

#"distribute and rearrange"#

#y-8=-16/7x-16#

#"multiply all terms by 7"#

#7y-56=-16x-112#

#rArr16x+7y=-56larrcolor(red)"in standard form"#