Question 1178012

P,Q,R are ({{{-1}}} ,{{{11}}} ),({{{2}}} ,{{{5}}} ) and ({{{t}}} ,{{{3}}} ) respectively


first find the slope of the line passes through 
P({{{-1}}} ,{{{11}}} ) and Q({{{2}}} ,{{{5}}} )

{{{m=(5-11)/(2-(-1))=-6/3=-2}}}

{{{y=-2x+b}}}....use one point to calculate {{{b}}}

{{{5=-2*2+b}}}

{{{b=5+4}}}

{{{b=9}}}

{{{y=-2x+9}}}.....eq.1


since given that ∠{{{PQR = 90}}} °, find a line that is perpendicular  to line with a slope {{{-2}}} and passes through Q and R

{{{y=mx+b}}}

the slope of the perpendicular line will be negative reciprocal of the slope {{{-2}}} and it is

{{{m=-1/-2=1/2}}}

{{{y=(1/2)x+b}}}...use one point to calculate {{{b}}}
{{{5=(1/2)2+b}}}
{{{5=1+b}}}
{{{b=4}}}

{{{y=(1/2)x+4}}}.....eq.2

since given ({{{t}}} ,{{{3}}} ), we use line{{{ y=3}}} and the intersection point of the {{{y=(1/2)x+4}}} and {{{ y=3}}} will have {{{x=t}}}

{{{y=(1/2)x+4}}}...substitute {{{ y}}} 
{{{3=(1/2)x+4}}}
{{{3-4=(1/2)x}}}
{{{-1*2=x}}}
{{{x=-2}}}=>{{{t=-2}}} and the point {{{R}}}({{{-2}}} ,{{{3}}} ) 


given also: the line{{{ PQ}}} is produced to {{{S}}} so that {{{QS = PQ}}}. Calculate the coordinates of {{{S}}}.

{{{S}}} will be on a line {{{y=(1/2)x+4}}}, from  {{{Q}}} same distance as {{{ PQ}}}


{{{d=sqrt((2-(-1))^2+(5-11)^2)}}}
{{{d=sqrt(3^2+(-6)^2)}}}

distance is  {{{d=sqrt(45)=6.7}}} which is radius {{{r}}} of the circle that passes through P, Q, and S

use the radius , draw a circle and read from the graph what are the  coordinates of the point S



{{{drawing ( 600, 600, -10, 15, -10, 15,
circle(-1,11,.12), locate(-1,11,P(-1,11)),
circle(2,5,.12), locate(2,5,Q(2,5)),
circle(-2,3,.12), locate(-2,3,R(-2,3)),
circle(2,5,sqrt(45)),blue(line(2,5,-1,11)),
green(line(-2,3,-1,11)),circle(-4,2,.12),locate(-4.3,2,S(-4,2)),
graph( 600, 600, -10, 15, -10, 15,3 ,-2x+9, x/2+4)) }}} 


{{{S}}}({{{-4}}} ,{{{2}}} )