English

The Distance Between the Points (Cos θ, 0) and (Sin θ − Cos θ) is - Mathematics

Advertisements
Advertisements

Question

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

Options

  • \[\sqrt{3}\]

     

  • \[\sqrt{2}\]

     

  • 2

  • 1

MCQ

Solution

We have to find the distance between ` A (cos theta , sin theta ) and B ( sin theta , - cos theta ) `. 

In general, the distance between A`(x_1 , y_1) `  and B `(x_2 , y_2)`  is given by,

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

So,

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

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

But according to the trigonometric identity,

`sin^2 theta + cos^2 theta = 1`

Therefore,

AB = `sqrt (2) `

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Co-Ordinate Geometry - Exercise 6.7 [Page 63]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 6 Co-Ordinate Geometry
Exercise 6.7 | Q 1 | Page 63

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.


Name the quadrilateral formed, if any, by the following points, and given reasons for your answers:

A(4, 5) B(7, 6), C (4, 3), D(1, 2)


Find the ratio in which the point (2, y) divides the line segment joining the points A (-2,2) and B (3, 7). Also, find the value of y.


If the points A (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values of a and b.


Find the area of quadrilateral ABCD whose vertices are A(-5, 7), B(-4, -5) C(-1,-6) and D(4,5)


In what ratio does the point C (4,5) divides the join of A (2,3)  and B (7,8) ?


Two points having same abscissae but different ordinate lie on


Show that the points (−4, −1), (−2, −4) (4, 0) and (2, 3) are the vertices points of a rectangle.


 Prove hat the points A (2, 3) B(−2,2) C(−1,−2), and D(3, −1) are the vertices of a square ABCD.


Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 2) are similar triangles.


 ABCD is a parallelogram with vertices  \[A ( x_1 , y_1 ), B \left( x_2 , y_2 \right), C ( x_3 , y_3 )\]   . Find the coordinates  of the fourth vertex D in terms of  \[x_1 , x_2 , x_3 , y_1 , y_2 \text{ and }  y_3\]

   

If the points A (1,2) , O (0,0) and C (a,b) are collinear , then find  a : b.

 

If the distance between the points (4, p) and (1, 0) is 5, then p = 


If (x , 2), (−3, −4) and (7, −5) are collinear, then x =


The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are


A line intersects the y-axis and x-axis at P and Q , respectively. If (2,-5) is the mid-point of PQ, then the coordinates of P and Q are, respectively

 

Find the point on the y-axis which is equidistant from the points (5, −2) and (−3, 2).


Point (0, –7) lies ______.


If the coordinates of the two points are P(–2, 3) and Q(–3, 5), then (abscissa of P) – (abscissa of Q) is ______.


In which quadrant, does the abscissa, and ordinate of a point have the same sign?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×