1.

If the point `P(3,4)` is equidistant from the points `A(a+b,b-a` and `B(a-b,a+b)`, then prove that `3b-4a=0`

Answer» `AP^2=(a+b-3)^2+(b-a-4)^2`
`=(a+b)^2-6(a+b)+a+(b-a)^2-8(b-a)+16`
`BP^2=(a-b-3)^2+(a+b-4)^2`
`=(a-b)^2-6(a-b)+a+(a+b)^2-8(a+b)+16`
`AP=BP`
`AP^2=BP^2`
`-6(a+b)-8(b-a)=-6(a-b)-8(a+b)`
`8(a+b-b+a)+6(a-b-a-b)=0`
`16a-12b=0`
`3b-4a=0`.


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