Q10:
Find a relation between $x$ and $y$ such that the point ($x$, $y$) is equidistant from the point (3, 6) and (– 3, 4).

Solution :

Given: A point $P(x, y)$ is equidistant from point $A(3, 6)$ and point $B(-3, 4)$.

To Find: A relation between $x$ and $y$.

Step 1: Understanding the Condition of Equidistance
Since point $P(x, y)$ is equidistant from $A(3, 6)$ and $B(-3, 4)$, the distance $PA$ must be equal to the distance $PB$. Mathematically, $PA = PB$, which implies $PA^2 = PB^2$.

Step 2: Applying the Distance Formula
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Therefore, the square of the distance is:
$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Step 3: Setting up the Equation
For $PA^2$:
$PA^2 = (x - 3)^2 + (y - 6)^2$
For $PB^2$:
$PB^2 = (x - (-3))^2 + (y - 4)^2 = (x + 3)^2 + (y - 4)^2$

Step 4: Expanding the Algebraic Expressions
Using the identity $(a - b)^2 = a^2 - 2ab + b^2$ and $(a + b)^2 = a^2 + 2ab + b^2$:
$PA^2 = (x^2 - 6x + 9) + (y^2 - 12y + 36)$
$PB^2 = (x^2 + 6x + 9) + (y^2 - 8y + 16)$

Step 5: Equating and Simplifying
Since $PA^2 = PB^2$:
$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$
Subtract $x^2$ and $y^2$ from both sides:
$-6x - 12y + 45 = 6x - 8y + 25$
Rearrange the terms to one side:
$-6x - 6x - 12y + 8y + 45 - 25 = 0$
$-12x - 4y + 20 = 0$

Step 6: Final Simplification
Divide the entire equation by $-4$ to simplify:
$3x + y - 5 = 0$
Or, $3x + y = 5$.

Final Answer: 3x + y = 5

CBSE Solutions for Class 10 Mathematics Coordinate geometry

Coordinate geometry Class 10 EXERCISE 7.2

Chapters in CBSE - Class 10 Mathematics

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