Question 176441
Note: a y-intercept of 2 means that the point (0,2) is on the line



So let's find the equation of the line through (7,-3) and (0,2)





First let's find the slope of the line through the points *[Tex \LARGE \left(7,-3\right)] and *[Tex \LARGE \left(0,2\right)]



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(2--3)/(0-7)}}} Plug in {{{y[2]=2}}}, {{{y[1]=-3}}}, {{{x[2]=0}}}, and {{{x[1]=7}}}



{{{m=(5)/(0-7)}}} Subtract {{{-3}}} from {{{2}}} to get {{{5}}}



{{{m=(5)/(-7)}}} Subtract {{{7}}} from {{{0}}} to get {{{-7}}}



{{{m=-5/7}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(7,-3\right)] and *[Tex \LARGE \left(0,2\right)] is {{{m=-5/7}}}



Now let's use the point slope formula:



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y--3=(-5/7)(x-7)}}} Plug in {{{m=-5/7}}}, {{{x[1]=7}}}, and {{{y[1]=-3}}}



{{{y+3=(-5/7)(x-7)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=(-5/7)x+(-5/7)(-7)}}} Distribute



{{{y+3=(-5/7)x+5}}} Multiply



{{{y=(-5/7)x+5-3}}} Subtract 3 from both sides. 



{{{y=(-5/7)x+2}}} Combine like terms. 



{{{y=(-5/7)x+2}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(7,-3\right)] and *[Tex \LARGE \left(0,2\right)] is {{{y=(-5/7)x+2}}}



 Notice how the graph of {{{y=(-5/7)x+2}}} goes through the points *[Tex \LARGE \left(7,-3\right)] and *[Tex \LARGE \left(0,2\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(-5/7)x+2),
 circle(7,-3,0.08),
 circle(7,-3,0.10),
 circle(7,-3,0.12),
 circle(0,2,0.08),
 circle(0,2,0.10),
 circle(0,2,0.12)
 )}}} Graph of {{{y=(-5/7)x+2}}} through the points *[Tex \LARGE \left(7,-3\right)] and *[Tex \LARGE \left(0,2\right)]