If a point (x, y) in a plane is equidistant from the points (-1, 1) and (4, 3), then:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(10x + 4y = 23\)
A point equidistant from two points has equal distances to both, so the two distance formulas are set equal. Squaring both sides removes the square roots, and the \(x^2\) and \(y^2\) terms cancel.
Squared distance to \((-1,1)\) equals squared distance to \((4,3)\): \((x+1)^2+(y-1)^2=(x-4)^2+(y-3)^2\)
\(x^2+2x+1+y^2-2y+1=x^2-8x+16+y^2-6y+9\)
\(2x-2y+2=-8x-6y+25\)
\(2x+8x-2y+6y=25-2\), so \(10x+4y=23\)
Tip: The answer is the perpendicular bisector of the segment joining the two points, so a quick check is that the midpoint \(\left(\frac{3}{2},2\right)\) satisfies it: \(10\times\frac{3}{2}+4\times2=15+8=23\).