Question 167433
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{{{y+9=(-2/5)(x-3)}}}, EQN 1
It follows Point Slope Form Eqn: {{{y=mx+b}}}
Continuing,
{{{y=(-2/5)x+(6/5)-9}}}
{{{y=(-2/5)x+(6-45)/5}}}
{{{y=(-2/5)x+(-39/5)}}} ---> {{{highlight(slope=(-2/5))}}}
In standard form:{{{ax+by=c}}}, then EQN 1 becomes:
{{{y+9=-2/5(x-3)}}} ---> {{{y+(2/5)x=(6/5)-9}}}
{{{(2/5)x+y=(6-45)/5=-39/5}}}, multiply eqn by "5"
{{{highlight(2x+5y=-39)}}} --------> Standard Equation
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Using Point-Slope Form for points (-5,6)(3,10):
{{{m=(y[2]-y[1])/(x[2]-x[1])}}}
{{{m=(10-6)/(3-(-5))=4/(3+5)=4/8}}}
{{{highlight(m=1/2)}}}, Slope
Thru points (-5,6), standard line eqn:
Via {{{y=mx+b}}}
{{{6=(1/2)(-5)+b}}}
{{{6+(5/2)=b}}} --> {{{b=(12+5)/2}}} ---> {{{b=17/2}}}, y-intercept
Therefore it follows,
{{{y=(1/2)x+(17/2)}}} ---> {{{(-1/2)x+y=17/2}}}, Multiply by "2" the eqn:
{{{highlight(-x+2y=17)}}}, STANDARD FORM
We'll see the graph:
{{{drawing(300,300,-20,8,-15,15,grid(1),graph(300,300,-20,8,-15,15,(-2/5)x-(39/5),(1/2)x+(17/2)),circle(-5,6,.20),circle(3,10,.20))}}}--- See Green Line {{{-x+2y=17}}} passes thru points (-5,6)(3,10)
Thank you,
Jojo</pre>