Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Step-by-Step Solution
Key Concept: Use the distance formula to express the distances from a generic point on the x‑axis to the two given points, set the distances equal (since the point is equidistant), and solve the resulting equation for the x‑coordinate.
1. Let the required point on the x‑axis be \((x,0)\).
2. Distance from \((x,0)\) to \((2,-5)\):
$$d_1 = \sqrt{(x-2)^2 + (0+5)^2} = \sqrt{(x-2)^2 + 25}.$$
3. Distance from \((x,0)\) to \((-2,9)\):
$$d_2 = \sqrt{(x+2)^2 + (0-9)^2} = \sqrt{(x+2)^2 + 81}.$$
4. Since the point is equidistant, set \(d_1 = d_2\) and square both sides to remove the square roots:
$$(x-2)^2 + 25 = (x+2)^2 + 81.$$
5. Expand the squares:
\[x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81.\]
6. Cancel \(x^2\) and simplify:
\[-4x + 29 = 4x + 85.\]
7. Bring like terms together:
\[-4x - 4x = 85 - 29 \quad\Rightarrow\quad -8x = 56.\]
8. Solve for \(x\):
$$x = \frac{-56}{8} = -7.$$
9. Therefore the required point on the x‑axis is \((-7,0)\).
Correct Answer: (-7, 0)