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The Distance Between the Points (A Cos θ + B Sin θ, 0) and (0, a Sin θ − B Cos θ) is

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Question

The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is

Options

  •  a2 + b2

  •  a + b

  •  a2 − b2

  • \[\sqrt{a2 + b2}\]

     

MCQ
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Solution

We have to find the distance betweenA` (a cos theta + b sin theta , 0 ) " and " B(0, a sin theta - b cos theta )` . 

In general, the distance between A(x1 , y 1)  and B(x2 , y2) is given by,

`AB = sqrt((x_2 - x_1 )^2 + (y_2 - y_1 )^2)`

So,

`AB = sqrt((a cos theta +  b sin theta - 0)^2 + (0- a sin theta  + b cos theta 
)^2)`

     `= sqrt((a^2 (sin^2 theta + cos^2  theta ) + b^2 (sin^2  theta +  cos^2 theta)`

But according to the trigonometric identity,

`sin^2  theta + cos^2  theta = 1` 

Therefore,

`AB = sqrt(a^2 + b^2)`

 

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Chapter 6: Co-ordinate Geometry - Exercise 6.7 [Page 63]

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R.D. Sharma Mathematics [English] Class 10
Chapter 6 Co-ordinate Geometry
Exercise 6.7 | Q 4 | Page 63
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