Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (-7, 0)
Every point on the x-axis has y-coordinate 0, so the point is P(\(x\), 0). It is equidistant from A(2, −5) and B(−2, 9), so by the distance formula PA2 = PB2 (squares avoid the square roots).
\((x-2)^2+(0+5)^2=(x+2)^2+(0-9)^2\)
\(x^2-4x+4+25=x^2+4x+4+81\)
\(-8x=56\), so \(x=-7\)
The point is (−7, 0); both distances are \(\sqrt{106}\)
