Question 10

Equidistant: A point is equidistant from two other points when its distance from both is exactly equal. So here, the distance from $(x, y)$ to $(3, 6)$ equals the distance from $(x, y)$ to $(-3, 4)$.

Distance Formula: The distance between two points $\text{A}(x_1, y_1)$ and $\text{B}(x_2, y_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

We set the two distances equal and simplify to find the relation between $x$ and $y$.

Let's find a relation where the distance from point (x, y) to (3, 6) is the same as to (-3, 4).

Step 1 — Set up the distance equation

Let the point be P(x, y). Let the first given point be A(3, 6). Let the second given point be B(-3, 4). The problem states that point P is equidistant from A and B. This means the distance PA must be equal to the distance PB. We use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

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

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

Since $PA = PB$, we can write:

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

Step 2 — Square both sides and expand

Let's square both sides of the equation. This removes the square roots.

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

Now, we expand each squared term using the formula $(a-b)^2 = a^2 - 2ab + b^2$.

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

Step 3 — Simplify and find the relation

Let's combine the constant terms on each side.

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

We can cancel $x^2$ and $y^2$ from both sides.

$45 - 6x - 12y = 25 + 6x - 8y$

Now, let's move all terms involving x and y to one side. We will move them to the right side to keep the x term positive.

$45 - 25 = 6x + 6x - 8y + 12y$

Let's simplify the equation.

$20 = 12x + 4y$

We can divide the entire equation by 4 to simplify it further.

$\frac{20}{4} = \frac{12x}{4} + \frac{4y}{4}$

$5 = 3x + y$

We can write this as:

$\boxed{3x + y = 5}$

Answer

The relation between x and y is $3x + y = 5$.