Question 838152
Slope-Intercept form for a line, y=mx+b.
{{{m=(0-7)/(1-(-2))=-7/3}}};
b=y-mx, using either point of the line:
{{{b=7-(-7/3)(-2)=7-14/3=(21-14)/3=7/3}}}.
Line equation is {{{highlight_green(y=-(7/3)x+7/3)}}}


Find distance from P to R, using Distance Formula.
{{{PR=sqrt((-2-1)^2+(7-0)^2)}}}
{{{sqrt((-3)^2+(7)^2)}}}

{{{sqrt(9+49)}}}
{{{highlight_green(PR=sqrt(58))}}}


Any point (x,y) on the line for PR will be in the form, (x,-(7/3)x+7/3), the expression for y being that of the equation's expression for the line.


You want the point Q(x, -(7/3)x+7/3) so that 
{{{highlight(PQ=(2/3)*sqrt(58)=sqrt((x-(-2))^2+((-(7/3)x+7/3)-7)^2))}}}
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Simplify that and solve for x.  Realize, that one value will be good and the other value will need to be rejected.