Question 994012
If you have two points, {{{A(x[A],y[A])}}} and {{{B(x[B],y[B])}}} ,
you can find the slope, {{{m}}} , of the line connecting them as
{{{m=(y[A]-y[B])/(x[A]-x[B])}}} .


If you have the slope, {{{m}}} , of a line and a point, {{{C(x[C],y[C])}}} , on that line
(point {{{C(x[C],y[C])}}} can be the same as {{{A(x[A],y[A])}}} or {{{B(x[B],y[B])}}} ),
you can write the equation of that line in {{{highlight(point-slope)}}} form as
{{{y-y[C]=m(x-x[C])}}} .


If you have the equation of that line in {{{highlight(point-slope)}}} form,
you can use algebra to transform that equation into an equivalent equation
for that line in {{{highlight(slope-intercept)}}} form by solving for {{{y}}} .


IN THIS CASE:
{{{A(x[A],y[A])=A(-2,-1)}}} and {{{B(x[B],y[B])=B(-9,-4)}}}, so
{{{system(x[A]=-2,y[A]=-1,x[B]=-9,y[B]=-4)}}} , and {{{m=(-1-(-4))/(-2-(-9))=(-1+4)/(-2+9)=3/7}}} .
An equation for the line in {{{highlight(point-slope)}}} form,
using point {{{C(x[C],y[C])=C(-2,-1)}}} , is
{{{y-(-1)=(3/7)(x-(-2))}}}<-->{{{y+1=(3/7)(x+2)}}} .
Solving for {{{y}}} ,
{{{y+1=(3/7)(x+2)}}}<-->{{{y+1=(3/7)x+6/7}}}<-->{{{y=(3/7)x+6/7-1}}}<-->{{{highlight(y=(3/7)x-1/7)}}} ,
gives you the one and only equation in {{{highlight(slope-intercept)}}} form for that line.