Question 1

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

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

This formula is derived from the Pythagoras theorem, where the distance is the hypotenuse of a right triangle formed by the horizontal and vertical differences between the two points.

We will use the distance formula to find the distance between two points.

Step 1 — Calculate distance for (i)

Let's label the points as $\text{A}(\textbf{2}, \textbf{3})$ and $\text{B}(\textbf{4}, \textbf{1})$. The distance formula is $\sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}$. We substitute the coordinates into the formula.

$\text{Distance AB} = \sqrt{((4 - 2)^2 + (1 - 3)^2)}$

$= \sqrt{((2)^2 + (-2)^2)}$

$= \sqrt{(4 + 4)}$

$= \sqrt{8}$

$\boxed{2\sqrt{2} \text{ units}}$

Step 2 — Calculate distance for (ii)

Let's label the points as $\text{P}(\textbf{-5}, \textbf{7})$ and $\text{Q}(\textbf{-1}, \textbf{3})$. We use the same distance formula. Substitute the coordinates into the formula.

$\text{Distance PQ} = \sqrt{((-1 - (-5))^2 + (3 - 7)^2)}$

$= \sqrt{((-1 + 5)^2 + (-4)^2)}$

$= \sqrt{((4)^2 + (-4)^2)}$

$= \sqrt{(16 + 16)}$

$= \sqrt{32}$

$\boxed{4\sqrt{2} \text{ units}}$

Step 3 — Calculate distance for (iii)

Let's label the points as $\text{M}(\textbf{a}, \textbf{b})$ and $\text{N}(\textbf{-a}, \textbf{-b})$. We apply the distance formula again. Substitute the coordinates into the formula.

$\text{Distance MN} = \sqrt{((-a - a)^2 + (-b - b)^2)}$

$= \sqrt{((-2a)^2 + (-2b)^2)}$

$= \sqrt{(4a^2 + 4b^2)}$

$= \sqrt{(4(a^2 + b^2))}$

$\boxed{2\sqrt{(a^2 + b^2)} \text{ units}}$

Answer

(i) $2\sqrt{2}$ units (ii) $4\sqrt{2}$ units (iii) $2\sqrt{(a^2 + b^2)}$ units