If the distance of point c(x,y)from the paint a(3,6,)and breequal,then prove that that 3x+y=5
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Iby distance formula
PA= √(3-x)^2+(6-y)^2
PB= √(-3-x)^2+(4-y)^2
it is given that pa = pb
then equating values of pa and pb
√(3-x)^2+(6-y)^2=√(-3-x)^2+(4-y)^2
(3-x)^2+(6-y)^2=(-3-x)^2+(4-y)^2
9+x^2- 6x +36+y^2 -12y=
9+x^2+6x+16+y^2-8y
-6x+36-12y=6x+16-8y
12x+4y-20=0
3x+y-5=0
hence proved
PA= √(3-x)^2+(6-y)^2
PB= √(-3-x)^2+(4-y)^2
it is given that pa = pb
then equating values of pa and pb
√(3-x)^2+(6-y)^2=√(-3-x)^2+(4-y)^2
(3-x)^2+(6-y)^2=(-3-x)^2+(4-y)^2
9+x^2- 6x +36+y^2 -12y=
9+x^2+6x+16+y^2-8y
-6x+36-12y=6x+16-8y
12x+4y-20=0
3x+y-5=0
hence proved
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