Question 173829
Graph the line with slope -2/3 passong through the point (-5,5). 
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Okay, good you have the {{{slope=m=-2/3}}}
Thru point (-5,5), via Slope-Intercept form, {{{y=mx+b}}}, solving for y-inyercept {{{b}}}:
{{{5=(-2/3)(-5)+b}}}
{{{5=10/3+b}}}---->{{{b=5-(10/3)=(15-10)/3=5/3}}}, Y-intercept
Then, it follows our eqn-----> {{{y=(-2/3)x+(5/3)}}}
As we see in the graph:
{{{drawing(300,300,-8,5,-5,8,grid(1),graph(300,300,-8,5,-5,8,(-2/3)x+(5/3)),blue(circle(-5,5,.20)))}}}
Thank you,
Jojo</pre>