Question

The distance between the points A(a sin α, a cos α)and B (a cos α, -a sin α)is

Answer 1

Answer:

✓2a

Step-by-step explanation:

given points are

A(a sin α, a cos α) and B (a cos α, -a sin α)

using distance formula,

$AB = \sqrt{ {(acos \alpha - asin \alpha )}^{2} +{( - asin \alpha - acos \alpha )}^{2} }$

Taking a² common, we get

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

on simplifying, -(minus) is taken out since, -1 x -1=1

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

now, we know that

(a+b)²+(a-b)²=2(a²+b²)

$AB = a \times \sqrt{ 2 \times( {sin}^{2} \alpha + {cos}^{2} \alpha ) }$

we know that,

sin²A+cos²A=1

so,

$AB = \sqrt{2} \times a$