We use the distance formula, which states that the distance between two points (x_1, y_1) and (x_2, y_2) is √((x_2 - x_1)² + (y_2 - y_1)²). Since PA = PB, we can square both sides to eliminate the square root, giving PA² = PB².

We use the distance formula, which states that the distance between two points (x_1, y_1) and (x_2, y_2) is √((x_2 - x_1)² + (y_2 - y_1)²). Since PA = PB, we can square both sides to eliminate the square root, giving PA² = PB².

Expand the squared terms on both sides of the equation. (a-b)² = a² - 2ab + b² is used for each term. Then, we combine the constant terms.

(a) Find the relation between x and y such that the point P(x, y) is equidistant from the points A(7, 1) and B(3, 5). OR (b) Find the coordinates of the points of trisection of the line segment joining the points A(2, -2) and B( -7, 4).

Answer: The relation between $x$ and $y$ is $x - y = 2$ .

Step-by-step solution

Step 1: Understand the problem

The problem asks for the relation between $x$ and $y$ such that point $P(x, y)$ is equidistant from points $A(7, 1)$ and $B(3, 5)$ . This means the distance from $P$ to $A$ is equal to the distance from $P$ to $B$ .

Step 2: Apply the distance formula

We use the distance formula, which states that the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . Since $PA = PB$ , we can square both sides to eliminate the square root, giving $PA^2 = PB^2$ .

Step 3: Expand and simplify the equation

Expand the squared terms on both sides of the equation. $(a-b)^2 = a^2 - 2ab + b^2$ is used for each term. Then, we combine the constant terms.

Step 4: Rearrange and solve for the relation

Cancel out $x^2$ and $y^2$ from both sides. Then, move all $x$ and $y$ terms to one side and constant terms to the other side. Simplify the equation by dividing by a common factor to get the final relation.

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