Find a relation between and such that the point is equidistant from the point and .

Understand the Question A point is equidistant from two points and if the distance equals the distance (). The distance formula between two points and is: Equating eliminates the square roots, which allows us to expand the terms, cancel the quadratic terms (), and simplify to find a linear relation between and . Step 1 · Set Up the Distance Equation Let the given points be , , and . Since point is equidistant from and , . Using the distance formula : Equating both distances: Step 2 · Square Both Sides and Expand Squaring both sides: Expanding each term using algebraic identities: Step 3 · Simplify and Find the Relation Combine constant terms on both sides: Cancelling and from both sides: Rearranging terms: Dividing the entire equation by : Answer > Common Mistakes Sign Errors in Expansion: Incorrectly expanding as . Since , the correct expansion is . Forgetting to Square Both Sides: Trying to simplify without eliminating the radical signs, leading to algebraic errors.

Coordinate Geometry | Exercise 7.1

Question 10

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

Solution

Understand the Question

A point $P(x, y)$ is equidistant from two points $A(3, 6)$ and $B(-3, 4)$ if the distance $PA$ equals the distance $PB$ ( $PA = PB$ ).
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Equating $PA^2 = PB^2$ eliminates the square roots, which allows us to expand the terms, cancel the quadratic terms ( $x^2, y^2$ ), and simplify to find a linear relation between $x$ and $y$ .

Step 1 · Set Up the Distance Equation

Let the given points be $P(x, y)$ , $A(3, 6)$ , and $B(-3, 4)$ . Diagram 1

Since point $P$ is equidistant from $A$ and $B$ , $PA = PB$ .

Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ :

\begin{aligned} PA &= \sqrt{(3 - x)^2 + (6 - y)^2} \\[0.6em] PB &= \sqrt{(-3 - x)^2 + (4 - y)^2} \end{aligned}

Equating both distances: $\sqrt{(3 - x)^2 + (6 - y)^2} = \sqrt{(-3 - x)^2 + (4 - y)^2}$

Step 2 · Square Both Sides and Expand

Squaring both sides: $(3 - x)^2 + (6 - y)^2 = (-3 - x)^2 + (4 - y)^2$

Expanding each term using algebraic identities: $9 - 6x + x^2 + 36 - 12y + y^2 = 9 + 6x + x^2 + 16 - 8y + y^2$

Step 3 · Simplify and Find the Relation

Combine constant terms on both sides: $45 + x^2 - 6x + y^2 - 12y = 25 + x^2 + 6x + y^2 - 8y$

Cancelling $x^2$ and $y^2$ from both sides: $45 - 6x - 12y = 25 + 6x - 8y$

Rearranging terms:

\begin{aligned} 45 - 25 &= 6x + 6x - 8y + 12y \\[0.6em] 20 &= 12x + 4y \end{aligned}

Dividing the entire equation by $4$ :

\begin{aligned} \dfrac{20}{4} &= \dfrac{12x}{4} + \dfrac{4y}{4} \\[0.6em] 5 &= 3x + y \end{aligned}

$3x + y = 5$

Answer

$3x + y = 5$

Common Mistakes

Sign Errors in Expansion: Incorrectly expanding $(-3 - x)^2$ as $9 - 6x + x^2$ . Since $(-3 - x)^2 = [-(3+x)]^2 = (3+x)^2$ , the correct expansion is $9 + 6x + x^2$ .
Forgetting to Square Both Sides: Trying to simplify without eliminating the radical signs, leading to algebraic errors.

Question 10

More questions in Exercise 7.1