Question 8

Distance Formula: The distance between two points $\text{P}(x_1, y_1)$ and $\text{Q}(x_2, y_2)$ is given by:

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

This formula is derived from the Pythagoras theorem. Here we know $d = 10$ and all coordinates except $y$, so we set up the equation and solve for $y$.

We will use the distance formula to find the unknown coordinate.

Step 1 — Set up the equation

Let's use the distance formula. The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. We are given points $P(\mathbf{2}, \mathbf{-3})$ and $Q(\mathbf{10}, y)$. The distance is $\mathbf{10}$ units. Let's substitute these values into the formula.

$10 = \sqrt{(\mathbf{10} - \mathbf{2})^2 + (y - (\mathbf{-3}))^2}$

$10 = \sqrt{(\mathbf{8})^2 + (y + \mathbf{3})^2}$

$10 = \sqrt{\mathbf{64} + (y^2 + \mathbf{6}y + \mathbf{9})}$

$10 = \sqrt{y^2 + \mathbf{6}y + \mathbf{73}}$

Now, let's square both sides of the equation. This will remove the square root.

$10^2 = y^2 + \mathbf{6}y + \mathbf{73}$

$\mathbf{100} = y^2 + \mathbf{6}y + \mathbf{73}$

Let's rearrange the terms to form a quadratic equation. We will set it equal to zero.

$y^2 + \mathbf{6}y + \mathbf{73} - \mathbf{100} = \mathbf{0}$

$\boxed{y^2 + \mathbf{6}y - \mathbf{27} = \mathbf{0}}$

Step 2 — Solve for y

We have a quadratic equation. Let's factor it to find the values of $y$. We need two numbers that multiply to $\mathbf{-27}$ and add to $\mathbf{6}$. These numbers are $\mathbf{9}$ and $\mathbf{-3}$.

$y^2 + \mathbf{9}y - \mathbf{3}y - \mathbf{27} = \mathbf{0}$

Let's factor by grouping.

$y(y + \mathbf{9}) - \mathbf{3}(y + \mathbf{9}) = \mathbf{0}$

$(y + \mathbf{9})(y - \mathbf{3}) = \mathbf{0}$

Now, we set each factor to zero to find the possible values for $y$.

$y + \mathbf{9} = \mathbf{0} \quad \text{or} \quad y - \mathbf{3} = \mathbf{0}$

$y = \mathbf{-9} \quad \text{or} \quad y = \mathbf{3}$

Answer

The values of $y$ are $\mathbf{-9}$ or $\mathbf{3}$.