Question
Find the distance between the points:
$P(a + b, a - b)$ and $Q(a - b, a + b)$
$P(a + b, a - b)$ and $Q(a - b, a + b)$
✓
Answer
The given points are $P(a + b, a - b) $and $Q(a - b, a + b)$
Then, $[x_1 = (a + b), y_1 = (a - b)]$ and $[x_2 = (a - b)$, and $y_2 = (a + b)]$
$\therefore\text{PQ}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{a}-\text{b}-\text{a}-\text{b})^2+(\text{a}+\text{b}-\text{a}+\text{b})^2}$
$=\sqrt{(-\text{2b})^2+(\text{2b})^2}$
$=\sqrt{\text{4b}^2+\text{4b}^2}$
$=\sqrt{\text{8b}^2}=2\sqrt{2\text{b}}\text{ units}.$
Then, $[x_1 = (a + b), y_1 = (a - b)]$ and $[x_2 = (a - b)$, and $y_2 = (a + b)]$
$\therefore\text{PQ}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{a}-\text{b}-\text{a}-\text{b})^2+(\text{a}+\text{b}-\text{a}+\text{b})^2}$
$=\sqrt{(-\text{2b})^2+(\text{2b})^2}$
$=\sqrt{\text{4b}^2+\text{4b}^2}$
$=\sqrt{\text{8b}^2}=2\sqrt{2\text{b}}\text{ units}.$
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