Question

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

(a) a^2 +b^2
(b) a^2 - b^2

$(c) \sqrt{a {}^{2} + b { }^{2} }$
$(d) \sqrt{a {}^{2} - b {}^{2} }$
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Answer 1

Answer is option c.

Distance formula

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

Plug in the values

(x1, y1) = (acos∅ + bsin∅, 0)

(x2, y2) = (0, asin∅ - bcos∅)

$\sf\sqrt{(acos\theta + bsin\theta)^2 + (bcos\theta - asin\theta)^2}$

Squaring and rearranging

$\sf\tiny{\sqrt{a^2cos^2\theta + a^2sin^2\theta + b^2\sin^2\theta + b^2cos^2\theta + 2absin\theta cos\theta - 2absin\theta cos\theta}}$

Now, a²cos²∅ + a²sin²∅ = a²(cos²∅ + sin²∅) = a²

Using this, we get

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

Answer 2

Answer:

Given :-

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

Solution :-

Apply distance formula

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

By putting value

Here,

($\sf x_1, y_1$)= (acos∅ + bsin∅, 0)

$\sf (x_2,y_2)$ = (0, asin∅ - bcos∅)

Now,

D = √{(acos∅ + bsin∅)² + (bcos∅ - asin∅²)}

D = √{a²cos²∅ + a²sin²∅ + b²sin²∅ + b²cos²∅ + 2absin∅cos∅ - 2absin∅cos∅}

D = √{a²+ a²sin²∅ + b²sin²∅ + b² + 2absin∅ - 2absin∅}

Now,

D = √{a²+ a²+ b²+ b² + 2ab - 2ab}

D = √{a² + a² + b² + b²}

D = √{2a² + 2b²}

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