NCERT Solutions
Class 10 - Mathematics - Chapter 7: COORDINATE GEOMETRY - Exercise 7.1
Question 7

Question. 7

Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).

Answer:

(−7, 0)

Video Explanation:

Detailed Answer with Explanation:

Let the required point on the x-axis be \((x, 0)\). A point on the x-axis has y–coordinate 0.

This point must be equidistant from \((2, -5)\) and \((-2, 9)\). So we set their distances equal.

Distance from \((x, 0)\) to \((2, -5)\): \( \sqrt{(x - 2)^2 + (0 + 5)^2} \).

Distance from \((x, 0)\) to \((-2, 9)\): \( \sqrt{(x + 2)^2 + (0 - 9)^2} \).

Equating the two distances (squaring both sides to remove the square roots):

\((x - 2)^2 + 25 = (x + 2)^2 + 81\).

Expand both sides: \( x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \).

Simplify: \( x^2 - 4x + 29 = x^2 + 4x + 85 \).

Cancel \(x^2\): \( -4x + 29 = 4x + 85 \).

Bring variables together: \( -4x - 4x = 85 - 29 \).

So, \( -8x = 56 \).

Thus, \( x = -7 \).

The required point on the x-axis is \((-7, 0)\).

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