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Co-Ordinate Geometry — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsCo-Ordinate Geometry4 Marks
Question
Find the distance between the points:
$\text{P}(\text{a}\sin\alpha,\text{a}\cos\alpha)$ and $\text{Q}(\text{a}\cos\alpha, -\text{a}\sin\alpha)$
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Answer

The given points are $\text{P}(\text{a}\sin\alpha,\text{a}\cos\alpha)$ and $\text{Q}(\text{a}\cos\alpha, -\text{a}\sin\alpha)$
Then, $(\text{x}_1=\text{a}\sin\alpha,\text{y}_1=\text{a}\cos\alpha)$ and $(\text{x}_2=\text{a}\cos\alpha,\text{ y}_2=-\text{a}\sin\alpha)$
$\therefore\text{PQ}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$=\sqrt{(\text{a}\cos\alpha -\text{a}\sin\alpha)^2+(-\text{a}\sin\alpha -\text{a}\cos\alpha)^2}$
$=\sqrt{\text{a}^2\cos^2\alpha+\text{a}^2\sin^2\alpha+\text{a}^2\cos^2\alpha+\text{a}^2\sin^2\alpha}$
$=\sqrt{\text{a}^2(\cos^2\alpha+\sin^2\alpha)\text{a}^2(\cos^2\alpha+\sin^2\alpha)}$
$=\sqrt{\text{a}^2+\text{a}^2}$
$=\sqrt{\text{2a}^2}=\sqrt{2}\text{a}\text{ units}.$

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