Question

Q. (16 Find the distance between the points A (Coso, o)
and B(0, - Sino)​

Answer 1

$\sf \large \: Distance \: between \: two \: points \: (x_1 , y_1) \: and \: ( x_2 , y_2) : \\ \\ \rm \large\longrightarrow \sqrt{{ {(x_2 - x_1)}^{2} + (y_2 - y_1)}^{2} }$

$\rm \: Distance \: between \: two \: points \: (cos \theta,0) \: and \: ( 0 , - sin\theta) \\ \\ \rm \large\longrightarrow \sqrt{{ {(0 - cos \theta)}^{2} + (- sin\theta - 0)}^{2} } \\ \\ \\ \rm \large\longrightarrow \sqrt{{ {( cos \theta)}^{2} + ( - sin\theta )}^{2} }\\ \\ \\ \rm \large\longrightarrow \sqrt{{ { cos ^{2} \theta}+ sin^{2} \theta } }\\ \\ \\ \rm \large\longrightarrow \sqrt{{1} } = 1 \\ \\ \\ \therefore\rm \: Distance \: between \: two \: points \: (cos \theta,0) \: and \: ( 0 , - sin\theta) = 1 \: unit$

Answer : 1 unit

Additional Information:-

$\tt \: Equation \: of \: line \: passing \: through \: points \: (x_1 , y_1) \: and \: (x_2 , y_2) :$

$\tt \longrightarrow y - y_1 = x-x_1\bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg )$

$\tt \rightarrow \: here \: \bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg ) is \: slope \: of \: line$