The point P lies on the line #x-3y=2# and at equal distance from the two points (2,3) and (6,-5). Find the co-ordinates of P?

1 Answer
Feb 16, 2018

#P-=(x,y)=P-=(-10,-8)#

Explanation:

Given:
#P-=(x,y)# is at equal distance from #A-=(2,3,) and B-=(6,-5)#

Thus,

#(x-2)^2+(y-3)^2=(x-6)^2+(y+5)^2#

#x^2-4x+4+y^2-6y+9=x^2-12x+36+y^2+10y+25#

#-4x+12x-6y-10y+4+9-36-25=0#
#8x-16y-48=0#

Simplifying

#x-2y-6=0#

#x-2y=6#

Also given that:
#x-3y=2#

The system of equations are:
#x-2y=6#-------(1)
#x-3y=2#-------(2)

Subtracting (2) from (1)

#y=-8#

(1) becomes

#x-2(-8)=6#

#x+16=6#

#x=6-16#

#x=-10#

Check::

Distance of P from A is represented by

#(x-2)^2+(y-3)^2#
#=(-10-2)^2+(-8-3)^2#
#=(-12)^2+(-11)^2#
#=144+121=2465#

Distance from P to B is represented by

#(x-6)^2+(y+5)^2#
#=(-10-6)^2+(-8+5)^2#
#=(-16)^2+(-3)^2#
#=256+9=265#

Checked and found correct

Hence,

#P-=(x,y)=P-=(-10,-8)#