Question 1127181
given:

{{{P}}} amd {{{Q}}} are ({{{-2}}},{{{6}}}) and ({{{9}}},{{{3}}}) , 

find:


a) cooridantes of point {{{R }}}that lies on {{{y}}} axis such that {{{PR =QR}}}

if point {{{R}}} that lies on {{{y }}}axis, cooridantes will be ({{{0}}},{{{y}}})

use distance formula to find {{{y}}}

({{{0}}},{{{y}}}) and ({{{-2}}},{{{6}}}) 

{{{d=sqrt((x-x[1])^2+(y-y[1])^2)}}}

{{{d=sqrt((0+2)^2+(y-6)^2)}}}

{{{d=sqrt(4+(y-6)^2)}}}



({{{0}}},{{{y}}}) and ({{{9}}},{{{3}}})

{{{d=sqrt((x-x1)^2+(y-y1)^2)}}}

{{{d=sqrt((0-9)^2+(y-3)^2)}}}

{{{d=sqrt(81+(y-3)^2)}}}


=>

{{{sqrt(4+(y-6)^2)=sqrt(81+(y-3)^2)}}}

{{{4+y^2-12y+36=81+y^2-6y+9}}}

{{{-12y+40=-6y+90}}}

{{{-90+40=-6y+12y}}}

{{{6y=-50}}}

{{{y=-8.33}}}

cooridantes of point {{{R}}}: ({{{0}}},{{{-8.33}}})


proof that {{{PR =QR}}}

 *[invoke formula_distance -2, 6, 0, -8.33]

{{{PR =14.47}}}

 *[invoke formula_distance 9, 3, 0, -8.33]

{{{QR=14.47}}}


b) 

coordinates of point {{{S}}} that lies on {{{x}}} axis such as {{{PS=QS}}}

 {{{S}}} that lies on {{{x}}} axis :({{{x}}},{{{0}}})


distance formula again

({{{x}}},{{{0}}}) and ({{{-2}}},{{{6}}} )


{{{d=sqrt((x-x1)^2+(y-y1)^2)}}}

{{{d=sqrt((x+2)^2+(0-6)^2)}}}

{{{d=sqrt((x+2)^2+36)}}}


({{{x}}},{{{0}}}) and ({{{9}}},{{{3}}} )

{{{d=sqrt((x-x1)^2+(y-y1)^2)}}}

{{{d=sqrt((x-9)^2+(0-3)^2)}}}

{{{d=sqrt((x-9)^2+9)}}}

=>
{{{sqrt((x+2)^2+36)=sqrt((x-9)^2+9)}}}

{{{(x+2)^2+36=(x-9)^2+9}}}

{{{x^2+4x+4+36=x^2-18x+81+9}}}


{{{4x+40=-18x+90}}}
{{{4x+18x=90-40}}}
{{{22x=50}}}
{{{x=2.272727272727273}}}
{{{x=2.27}}}


so, the point {{{S}}} is at ({{{2.27}}},{{{0}}})

proof: PS=QS

*[invoke formula_distance -2, 6, 2.27, 0]

{{{PS=7.36}}}

*[invoke formula_distance 9, 3, 2.27, 0]

{{{QS=7.36}}}



{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-2,6,.12),circle(9,3,.12),circle(0,-8.33,.13),circle(2.27,0,.13),
locate(-2,6,P(-2,6)),locate(8.8,3,Q(9,3)),locate(0.3,-8.33,R(0,-8.33)),
locate(2.27,0.5,S(2.27,0)),
 graph( 600, 600, -10, 10, -10, 10, 0)) }}}