Question 189488
First let's find the slope of the line through the points *[Tex \LARGE \left(2,0\right)] and *[Tex \LARGE \left(0,5\right)]



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



{{{m=(5-0)/(0-2)}}} Plug in {{{y[2]=5}}}, {{{y[1]=0}}}, {{{x[2]=0}}}, and {{{x[1]=2}}}



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



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



{{{m=-5/2}}} Reduce



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



Now let's use the point slope formula:



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



{{{y-0=(-5/2)(x-2)}}} Plug in {{{m=-5/2}}}, {{{x[1]=2}}}, and {{{y[1]=0}}}



{{{y-0=(-5/2)x+(-5/2)(-2)}}} Distribute



{{{y-0=(-5/2)x+5}}} Multiply



{{{y=(-5/2)x+5+0}}} Add 0 to both sides. 



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



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


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

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