How do you write an equation in standard form for a line passing through (3, 4) and (–3, –8)?

The equation of line passing through the points #(x_1, y_1)\equiv(3, 4)# & #(x_2, y_2)\equiv(-3, -8)# is given by following general formula

#y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)#

#y-4=\frac{-8-4}{-3-3}(x-3)#

#y-4=2(x-3)#

#y-4=2x-6#

#2x-y-2=0#

#"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)"slope-intercept form"#

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"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)=(3,4)" and "(x_2,y_2)=(-3,-8)#

#m=(-8-4)/(-3-3)=(-12)/(-6)=2#

#y=2x+blarrcolor(blue)"is the partial equation"#

#"to find b substitute either of the 2 given points into"#
#"the partial equation"#

#"using "(3,4)" then"#

#4=6+brArrb=4-6=-2#

#y=2x-2larrcolor(red)"in slope-intercept form"#

#2x-y=2larrcolor(red)"in standard form"#

If you are given the co-ordinates of two points on a line, here is a good formula to use to get the equation of the line:

#(y-y_1)/(x-x_1) = (y_2-y_1)/(x_2-x_1)#

Use #(3,4)# as #(x_2,y_2)#

#(y-(-8))/(x-(-3)) = (4-(-8))/(3-(-3))#

#(y+8)/(x+3) = (4+8)/(3+3) = 12/6 =2/1" "larr# this is the slope.

#(y+8)/(x+3) = 2/1" "larr# cross multiply

#2x+6 = y+8" "larr#re-arrange into standard form.

#2x-y = 2#