NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.2 Question 7

NCERT Class X Chapter 7: Coordinate Geometry

Question:

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).

Given:

  • The centre of the circle is at \( O(2, -3) \).
  • One end of the diameter is \( B(1, 4) \).

To Find:

The coordinates of point \( A \), the other end of the diameter \( AB \).

Formula:

Midpoint formula: If the endpoints of a line segment are \( (x_1, y_1) \) and \( (x_2, y_2) \), then the midpoint is given by:

$$ \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) $$

Solution:

Step 1: Let the coordinates of point \( A \) be \( (x, y) \).

Step 2: Since \( O \) is the midpoint of \( AB \), using the midpoint formula:

$$ \left( \frac{x + 1}{2},\ \frac{y + 4}{2} \right) = (2,\ -3) $$

Step 3: Equate the \( x \)-coordinates and solve for \( x \):

$$ \frac{x + 1}{2} = 2 \\ x + 1 = 4 \\ x = 3 $$

Step 4: Equate the \( y \)-coordinates and solve for \( y \):

$$ \frac{y + 4}{2} = -3 \\ y + 4 = -6 \\ y = -10 $$

Step 5: Therefore, the coordinates of point \( A \) are \( (3, -10) \).

Result:

The coordinates of point \( A \) are \( (3, -10) \).

© Kaliyuga Ekalavya. All rights reserved.

Comments

Post a Comment