Question

Find distance between p(a sin alpha,a cos alpha) and q(a cos alpha,-a sin alpha)

Answer 1

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Answer 2

Answer:

$\sqrt{2}a$

Step-by-step explanation:

p =$(x_1,y_1)=(a sin \alpha , a cos \alpha)$

q=$(x_2,y_2)=(a cos \alpha, - asin \alpha)$

Distance between p and q

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

$pq=\sqrt{(a cos \alpha-a sin \alpha)^2+(-a sin \alpha-a cos \alpha)^2}$

$pq=\sqrt{(a^2 (cos \alpha-sin \alpha)^2+(a^2( sin \alpha-cos \alpha)^2}$

$pq=\sqrt{(a^2 (cos^2 \alpha+sin^2 \alpha-2 cos \alpha sin \alpha)+(a^2( cos^2 \alpha+sin^2 \alpha+2 cos \alpha sin \alpha)}$

$pq=\sqrt{a^2 (1-2 cos \alpha sin \alpha)+a^2( 1+2 cos \alpha sin \alpha)}$

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

$pq=\sqrt{2}a$

Hence The distance is $\sqrt{2}a$