How do you write the standard form of a line given (4,3) and (7, -2)?

The slope can be computed as
#m=(y_2-y_1)/(x_2-x_1)=(-2-3)/(7-4)=-5/3#
so we get
#y=-5/3x+n#
substituting
#x=4,y=3#
we get
#3+20/3=n# so #n=29/3#

#"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"#

#"to begin 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)=(4,3)" and "(x_2,y_2)=(7,-2)#

#m=(-2-3)/(7-4)=(-5)/3=-5/3#

#y=-5/3x+blarrcolor(blue)"is the partial equation"#

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

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

#3=-20/3+brArrb=9/3+20/3=29/3#

#y=-5/3x+29/3larrcolor(red)"in slope-intercept form"#

#"multiply all terms by 3"#

#3y=-5x+29#

#"add "5x" to both sides"#

#5x+3y=29larrcolor(red)"in standard form"#

#color(crimson)("Standard form of linear equation is " ax + by = c#

#"Given points are " (x_1, y_1) = 4,3), (x_2,y_2) = (7,-2)#

Knowing two points on a line, we can form the equation using the formula,

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

#(y - 3) / (-2 -3) = (x - 4) / (7 - 4)#

#(y - 3) / -5 = (x - 4) / 3#

#3y - 9 = -5x + 20, " cross-multiplying"#

#color(green)(5x + 3y = 29 " is the standard form"#

graph{-(5/3)x + (29/3) [-10, 10, -5, 5]}