How do you the equation of the line that passes through point (4,2) and (6,6)?

The slope of the line through #(4,2)# and #(6,6)#
is #color(green)(m)=(Deltay)/(Deltax)=(6-2)/(6-4)=4/2=color(green)(2)#

The slope-point form for a line
with slope #color(green)(m)#
through the point #(color(red)(a),color(blue)(b))#
is #y-color(blue)b=color(green)(m)(x-color(red)(a))#

Using #(color(red)4,color(blue)2)# as our point (we could have used either of the given points)
and the previously determined #color(green)(m=2)#

The slope-point form for the required line is
#color(white)("XXX")y-color(blue)(2)=color(green)(2)(x-color(red)4)#

This could easily be converted into slope-point form as
#color(white)("XXX")y=color(green)(2)x-6#

or into standard form as
#color(white)("XXX")x-2y=6#

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

#color(red)(bar(ul(|color(white)(a/a)color(black)(y=mx+b)color(white)(a/a)|)))#
where m represents the slope and b, the y-intercept.

To calcuate m, use the #color(blue)"gradient formula"#

#color(red)(bar(ul(|color(white)(a/a)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(a/a)|)))#
where # (x_1,y_1)" and " (x_2,y_2)" are 2 coordinate points"#

The 2 points here are (4 ,2) and (6 ,6)

let # (x_1,y_1)=(4,2)" and " (x_2,y_2)=(6,6)#

#rArrm=(6-2)/(6-4)=4/2=2#

Thus partial equation is : #y=2x+b#

To find b, substitute either of the 2 given points into the partial equation and solve for b.

Using (4 ,2) : #(2xx4)+b=2rArrb=2-8=-6#

#rArry=2x-6" is the equation of the line"#