Find the distance between the following pair of points:$(a sin \alpha, -b cos \alpha)$ and $(-a cos \alpha, -b sin \alpha)$

Given:

The given pair of points is $(a\ sin \alpha, -b\ cos \alpha)$ and $(-a\ cos \alpha, -b\ sin \alpha)$.

To do:

We have to find the distance between the given pair of points.

Solution:

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

 The distance between the points \( (a \sin \alpha,-b \cos \alpha) \) and \( (-a \cos \alpha,-b \sin \alpha) \) \( =\sqrt{(-a \cos \alpha-a \sin \alpha)^{2}+(b \sin \alpha+b \cos \alpha)^{2}} \)

\( =\sqrt{a^{2}(\cos \alpha+\sin \alpha)^{2}+b^{2}(\sin \alpha+\cos \alpha)^{2}} \)

\( =\sqrt{(\sin \alpha+\cos \alpha)^{2}\left(a^{2}+b^{2}\right)} \)

\( =(\sin \alpha+\cos \alpha) \sqrt{a^{2}+b^{2}} \)

The distance between the given points is $\sqrt{a^{2}+b^{2}}(\sin \alpha+\cos \alpha)$.