NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 7

NCERT Class X Chapter 7: Coordinate Geometry

Question:

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

Given:

  • Point A: \( (2, -5) \)
  • Point B: \( (-2, 9) \)

To Find:

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

Formula:

Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Solution:

Step 1: Let the required point on the x-axis be \( (x, 0) \).

Step 2: Find the distance from \( (x, 0) \) to \( (2, -5) \) using the distance formula.

$$ \text{Distance}_1 = \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - 2)^2 + 25} $$

Step 3: Find the distance from \( (x, 0) \) to \( (-2, 9) \) using the distance formula.

$$ \text{Distance}_2 = \sqrt{(x + 2)^2 + (0 - 9)^2} = \sqrt{(x + 2)^2 + 81} $$

Step 4: Set the two distances equal since the point is equidistant from both points.

$$ \sqrt{(x - 2)^2 + 25} = \sqrt{(x + 2)^2 + 81} $$

Step 5: Square both sides to remove the square roots.

$$ (x - 2)^2 + 25 = (x + 2)^2 + 81 $$

Step 6: Expand both sides.

$$ (x^2 - 4x + 4) + 25 = (x^2 + 4x + 4) + 81 $$

Step 7: Simplify the equation.

$$ x^2 - 4x + 29 = x^2 + 4x + 85 $$

Step 8: Bring like terms together and solve for \( x \).

$$ x^2 - 4x + 29 - x^2 - 4x - 85 = 0 \\ -8x + 29 - 85 = 0 \\ -8x - 56 = 0 \\ -8x = 56 \\ x = -7 $$

Result:

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

© Kaliyuga Ekalavya. All rights reserved.

Comments

Post a Comment