A point \(P(x,\,y)\) is at the same distance from \(A(3,\,-2)\) and \(B(-1,\,4)\). Which equation must \(x\) and \(y\) satisfy?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2x-3y=-1\)
\(PA=PB\) is the same as \(PA^2=PB^2\); writing both with the distance formula and simplifying gives the relation.
\(PA^2=(x-3)^2+(y+2)^2=x^2+y^2-6x+4y+13\)
\(PB^2=(x+1)^2+(y-4)^2=x^2+y^2+2x-8y+17\)
\(PA^2=PB^2\): \(-6x+4y+13=2x-8y+17\)
\(-8x+12y=4\)
Divide by \(-4\): \(2x-3y=-1\)
Check: the mid-point of AB, \((1,\,1)\), must satisfy it: \(2-3=-1\)