The relation between A and B such that the point (A, B) is equidistant from the points (1, 3) and (2, 5) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2A+4B=19\)
A point is equidistant from two points when its distances to them are equal. Squaring both distances removes the square roots of the distance formula \(d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\).
Distance of \((A,B)\) from \((1,3)\) equals its distance from \((2,5)\):
\((A-1)^{2}+(B-3)^{2}=(A-2)^{2}+(B-5)^{2}\)
Expanded: \(A^{2}-2A+1+B^{2}-6B+9=A^{2}-4A+4+B^{2}-10B+25\)
The squares cancel: \(-2A-6B+10=-4A-10B+29\)
So the relation is \(2A+4B=19\)