The distance between the points (a cos θ + b sin θ, 0)

Question:

The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is

(a) $a^{2}+b^{2}$

(b) $a+b$

(c) $a^{2}-b^{2}$

(d) $\sqrt{a 2+b 2}$

Solution:

We have to find the distance between $\mathrm{A}(a \cos \theta+b \sin \theta, 0)$ and $\mathrm{B}(0, a \sin \theta-b \cos \theta)$.

In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,

$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

So,

$\mathrm{AB}=\sqrt{(a \cos \theta+b \sin \theta-0)^{2}+(0-a \sin \theta+b \cos \theta)^{2}}$

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

But according to the trigonometric identity,

$\sin ^{2} \theta+\cos ^{2} \theta=1$

Therefore,

$\mathrm{AB}=\sqrt{a^{2}+b^{2}}$

So, the answer is (d)

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