Question

The distance between the points (a cosθ + b sinθ,0) and (0, a sinθ - b cosθ) is:-
$a²+b² \\ \ \textless \ br /\ \textgreater \ a+b \\ \ \textless \ br /\ \textgreater \ a²-b²$
$\sqrt{ {a}^{2} + {b}^{2} }$
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Answer 1

Answer:

$\huge{\boxed{\rm{\red{answer=√(a²+b²) units}}}}$

Step-by-step explanation:

Given:-

Given points are (a cosθ + b sinθ,0) and

(0, a sinθ - b cosθ)

To find:-

The distance between the given two points

Solution:-

Given points are (a cosθ + b sinθ,0) and

(a cosθ + b sinθ,0) and (0, a sinθ - b cosθ)

Let (x1,y1)=(a cosθ + b sinθ,0)

=>x1=a cosθ + b sinθ and y1=0

(x2,y2)=(0, a sinθ - b cosθ)

=>x2=0 and y2=a sinθ - b cosθ

Using formula:-

If (x1,y1) and (x2,y2) are two points then the distance between them is

√{(x2-x1)²+(y2-y1)²} units

Now,

=>√{(0-(a cosθ + b sinθ)²+(a sinθ - b cosθ-0)²}

=>√{(-a cosθ -b sinθ)²+(a sinθ - b cosθ)²}

=>√(a²cos²θ+b²sin²θ+2abcosθsinθ+a²sin²θ+

b²cos²θ-2absinθcosθ)

=>√{a²(sin²θ+cos²θ)+b²(sin²θ+cos²θ)-(0)sinθcosθ}

=>√{a²(1)+b²(1)}

=>√(a²+b²)

Answer:-

Distance between two points =√(a²+b²) units