NCERT Class X Chapter 7: Coordinate Geometry Example 4

NCERT Class X Chapter 7: Coordinate Geometry Example 4

Question:

Find a relation between \( x \) and \( y \) such that the point \( (x, y) \) is equidistant from the points \( (7, 1) \) and \( (3, 5) \).

Given:

Let \( A = (7, 1) \), \( B = (3, 5) \), and \( P = (x, y) \) is a point equidistant from \( A \) and \( B \).

To Find:

A relation between \( x \) and \( y \) such that \( P \) is equidistant from \( A \) and \( B \).

Formula:

The 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 distances from \( P(x, y) \) to \( A(7, 1) \) and \( B(3, 5) \) be equal.

$$ PA = PB $$ $$ \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 3)^2 + (y - 5)^2} $$

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

$$ (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 $$

Step 3: Expand both sides.

$$ (x^2 - 14x + 49) + (y^2 - 2y + 1) = (x^2 - 6x + 9) + (y^2 - 10y + 25) $$

Step 4: Combine like terms.

$$ x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25 $$ $$ (x^2 - x^2) + (y^2 - y^2) - 14x + 49 - 2y + 1 = -6x + 9 - 10y + 25 $$ $$ -14x + 49 - 2y + 1 = -6x + 9 - 10y + 25 $$

Step 5: Bring all terms to one side.

$$ -14x + 49 - 2y + 1 - (-6x + 9 - 10y + 25) = 0 $$ $$ -14x + 49 - 2y + 1 + 6x - 9 + 10y - 25 = 0 $$ $$ (-14x + 6x) + (-2y + 10y) + (49 + 1 - 9 - 25) = 0 $$ $$ -8x + 8y + 16 = 0 $$

Step 6: Rearrange the equation.

$$ -8x + 8y + 16 = 0 $$ $$ 8x - 8y = 16 $$

Step 7: Divide both sides by 2 to simplify.

$$ 4x - 4y = 8 $$

Step 8: Bring all terms to one side to write the relation.

$$ 4x - 4y - 8 = 0 $$

Or, equivalently,

$$ 4x - 4y + 12 = 0 $$

Result:

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

$$ 4x - 4y + 12 = 0 $$
© Kaliyuga Ekalavya. All rights reserved.

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