Consider the line with #x#-intercept of 2 and #y#-intercept of -2. What is the equation of this line in standard form?

An x-intercept of two means we have the point

#(2,0)#

A y-intercept of negative two means we have the point

#(0,-2)#

Recall the formula of the slope of a line:

#m=(y_2-y_1)/(x_2-x_1)#

Since we have two points, we can calculate the slope using ( #(x_1,y_1)=(2,0), (x_2,y_2)=(0,-2)#

#m=(-2-0)/(0-2)#

#m=1#

The point-slope form of a line is:

#y-y_1=m(x-x_1)#

Where #(x_1,y_1)# is a point through which the line passes and #m# is the slope.

We already have m; we can use either of our points for #(x_1,y_1)#, let's use #(2,0)#.

#y-0=1(x-2)#
#y=x-2#

#"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"#

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•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)=(2,0)" and "(x_2,y_2)=(0,-2)#

#rArrm=(-2-0)/(0-2)=1#

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

#"y-intercept "=(0,-2)rArrb=-2#

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

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