Find a relation between x and y such that the point (x, y) is equidistant from points (7, 3) and (5, 2).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4x+2y=29\)
A point equidistant from (7, 3) and (5, 2) has equal distances to both points. By the distance formula, the squares of the two distances are then equal too, and squaring removes the square roots.
Distance to (7, 3) squared equals distance to (5, 2) squared: \((x-7)^{2}+(y-3)^{2}=(x-5)^{2}+(y-2)^{2}\)
Expanding: \(x^{2}-14x+49+y^{2}-6y+9=x^{2}-10x+25+y^{2}-4y+4\)
The \(x^{2}\) and \(y^{2}\) terms cancel: \(-14x-6y+58=-10x-4y+29\)
\(-4x-2y=-29\), that is, \(4x+2y=29\)