NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 10

NCERT Class X Chapter 7: Coordinate Geometry

Question:

Find a relation between \( x \) and \( y \) such that the point \( (x, y) \) is equidistant from the point \( (3, 6) \) and \( (–3, 4) \).

Given:

Let \( P(x, y) \) be a point equidistant from \( A(3, 6) \) and \( B(-3, 4) \).

To Find:

The relation between \( x \) and \( y \).

Formula:

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

Solution:

Step 1: Let the distances from \( P(x, y) \) to \( A(3, 6) \) and \( B(-3, 4) \) be equal.

$$ PA = PB $$ $$ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2} $$

Step 2: Square both sides to remove the square roots.

$$ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 $$

Step 3: Expand both sides.

$$ (x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16) $$

Step 4: Combine like terms and simplify.

$$ x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 $$ $$ -6x - 12y + 45 = 6x - 8y + 25 $$

Step 5: Bring all terms to one side to form the required relation.

$$ -6x - 12y + 45 - 6x + 8y - 25 = 0 $$ $$ -12x - 4y + 20 = 0 $$ $$ 12x + 4y - 20 = 0 $$ $$ 3x + y - 5 = 0 $$

Result:

The relation between \( x \) and \( y \) is:

$$ 3x + y - 5 = 0 $$

Result:

The distance between the points (2, 3) and (4, 1) is \(2\sqrt{2}\).

© Kaliyuga Ekalavya. All rights reserved.

Comments

Post a Comment