Question
Find the distance between the following pair of points:$(\text{a}\sin\alpha, -\text{b}\cos\alpha)$ and $(-\text{a}\cos\alpha, \text{b}\sin\alpha)$
✓
Answer
Here, $\text{x}_1=\text{a}\sin\alpha,\ \text{y}_1=-\text{b}\cos\alpha$ $\text{x}_2=-\text{a}\cos\alpha,\ \text{y}_2=\text{b}\sin\alpha$ Let D be the distance between two points $(\text{a}\sin\alpha, -\text{b}\cos\alpha)$and $(-\text{a}\cos\alpha, \text{b}\sin\alpha)$ The distance between two points is given by,$\text{D}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(-\text{a}\cos\alpha-\text{a}\sin\alpha)^2+(\text{b}\sin\alpha-(-\text{b}\cos\alpha))^2}$
$=\sqrt{\text{a}^2(\cos\alpha+\sin\alpha)^2+\text{b}^2(\sin\alpha+\cos\alpha)^2}$
$=\sqrt{\text{a}^2+\text{b}^2(\cos\alpha+\sin\alpha)^2}$
$=(\sin\alpha+\cos\alpha)\sqrt{\text{a}^2+\text{b}^2}$
Hence, distance between two points is $(\sin\alpha+\cos\alpha)\sqrt{\text{a}^2+\text{b}^2}$
$=\sqrt{(-\text{a}\cos\alpha-\text{a}\sin\alpha)^2+(\text{b}\sin\alpha-(-\text{b}\cos\alpha))^2}$
$=\sqrt{\text{a}^2(\cos\alpha+\sin\alpha)^2+\text{b}^2(\sin\alpha+\cos\alpha)^2}$
$=\sqrt{\text{a}^2+\text{b}^2(\cos\alpha+\sin\alpha)^2}$
$=(\sin\alpha+\cos\alpha)\sqrt{\text{a}^2+\text{b}^2}$
Hence, distance between two points is $(\sin\alpha+\cos\alpha)\sqrt{\text{a}^2+\text{b}^2}$
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