Question

the distance between the points (a cos theta + b sin theta , 0) and (0, a sin theta - b cos theta) is
A) a2 + b2
B) a + b
C) a2 - b2
D) root of (a2 + b2)

Answer 1

Answer:

D. $\sqrt{a^2+b^2}$

Step-by-step explanation:

The given points are

$(a\cos(\theta)+b\sin(\theta),0)$

and

$(0,a\sin(\theta)-b\cos(\theta))$

We apply the distance formula to obtain,

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

We expand the bracket to get,

$\sqrt{a^2 cos^2(\theta)+b^2sin^2 \theta+2ab sin \theta cos \theta+a^2 sin^2 \theta+b^2 cos^2 \theta-2absin \theta cos \theta}$

This simplifies to,

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

Recall that,

$\cos^2(\theta)+\sin^2(\theta)=1$

We apply this property to obtain,

$=\sqrt{a^2(1)+b^2(1)}$

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

The correct answer is D

Answer 2

Answer:

D.  $\sqrt{a^2+b^2}$

Step-by-step explanation:

Since, by the distance formula,

The distance between the points $(a cos \theta + b sin \theta , 0)$ and $(0, a sin \theta - b cos \theta)$  is,

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

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

( (x ± y)² = x² ± 2xy + y² )

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

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

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

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

⇒ Option D is correct.