Question:

1. Recall the distance formula
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]

2. Case (i): Points \((2,3)\) and \((4,1)\)
\[ \begin{aligned}
d_{1} &= \sqrt{(4-2)^2 + (1-3)^2} \\
&= \sqrt{2^2 + (-2)^2} \\
&= \sqrt{4+4} \\
&= \sqrt{8} \\
&= 2\sqrt{2}.
\end{aligned} \]

3. Case (ii): Points \((-5,7)\) and \((-1,3)\)
\[ \begin{aligned}
d_{2} &= \sqrt{(-1-(-5))^2 + (3-7)^2} \\
&= \sqrt{4^2 + (-4)^2} \\
&= \sqrt{16+16} \\
&= \sqrt{32} \\
&= 4\sqrt{2}.
\end{aligned} \]

4. Case (iii): Points \((a,b)\) and \((-a,-b)\)
\[ \begin{aligned}
d_{3} &= \sqrt{(-a-a)^2 + (-b-b)^2} \\
&= \sqrt{(-2a)^2 + (-2b)^2} \\
&= \sqrt{4a^2 + 4b^2} \\
&= 2\sqrt{a^2 + b^2}.
\end{aligned} \]

5. Write the final answers
- (i) \(d = 2\sqrt{2}\) units
- (ii) \(d = 4\sqrt{2}\) units
- (iii) \(d = 2\sqrt{a^{2}+b^{2}}\) units