Question

find the distance between the points (a cos theta,a sin theta) and (-a sin theta ,a cos theta?

Answer 1

$\sqrt{2 {a}^{2} }$

Answer 2

Answer:

$\text{The distance is }\sqrt2aunits$

Step-by-step explanation:

Given the two points having coordinates

$(a\cos\theta ,a\sin \theta)$ and $(-a \sin \theta ,a \cos \theta)$

we have to find the distance between these two points.

Using distance formula

Distance between the two points having coordinates $(x_1,y_1) and (x_2, y_2)$

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between the points $(a\cos\theta ,a\sin \theta)$ and $(-a \sin \theta ,a \cos \theta)$

$Distance=\sqrt{(-a\sin\theta-a\cos\theta)^2+(a\cos\theta-a\sin\theta)^2}$

$=\sqrt{a^2\sin^2\theta+a^2\cos^2\theta+2a^2\sin\theta\cos\theta+a^2\cos^2\theta+a^2\sin^2\theta-2a^2\sin\theta\cos\theta}$

$=\sqrt{a^2\sin^2\theta+a^2\cos^2\theta+a^2\cos^2\theta+a^2\sin^2\theta}$

$=\sqrt{2a^2(\sin^2\theta+\cos^2\theta)}$

$=\sqrt{2a^2}=\sqrt2aunits$