Question

Find the distance between the points (a sin theta, a cos theta) and (a cos theta, -a sin theta).

Answer 1

$\huge\star\underline{\mathcal\color{brown}{HELLO\:MATE}}\star$

$\color{brown}{HERE\:IS\:YR\:ANS}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\underline\color{Green}{ATTACHMENT}$

REFER TO THE ATTACHMENT

$\fbox\color{brown}{HOPE\:IT\:HELPS}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\huge{\mathcal{THANKS}}$

.

$<marquee >☝️ABHI☝️$

Answer 2

SOLUTION

The required distance between the given Points is

$= > \sqrt{(a \: cos \theta - a \: sin \theta) {}^{2} + ( - a \: sin \theta - a \: cos \theta) {}^{2} } \\ = > \sqrt{ {a}^{2} (cos \theta + sin\theta) {}^{2} + ( - a) {}^{2} (sin \theta + cos \theta) {}^{2} } \\ = > \sqrt{ {a}^{2} (cos {}^{2} \theta + sin {}^{2} \theta - 2sin \theta \: cos \theta) + {a}^{2} (cos {}^{2} \theta + sin {}^{2} \theta + 2sin \theta \: cos \theta } \\ = > \sqrt{ {a}^{2} (cos {}^{2} \theta +sin {}^{2} \theta - 2sin \theta cos \theta + cos {}^{2} \theta + sin {}^{2} \theta + 2sin \theta \: cos \theta) } \\ = > \sqrt{ {a}^{2} (2sin {}^{2} \theta + 2cos { }^{2} \theta) } \\ = > \sqrt{2a {}^{2}(sin {}^{2} \theta + cos {}^{2} \theta)} \\ = > \sqrt{2 {a}^{2}.1 } \\ = > a \sqrt{2} units$

hope it helps ✔️