Home
Class 11
MATHS
Find length of perpendicular from origin...

Find length of perpendicular from origin to the lines passes from points `( cos theta , sin theta ) and ( cos phi , sin phi)`.

Text Solution

Verified by Experts

The correct Answer is:
`| cos ((theta - phi)/( 2))|`
Promotional Banner

Topper's Solved these Questions

  • STRAIGHT LINES

    KUMAR PRAKASHAN|Exercise Text based MCQs|25 Videos
  • STRAIGHT LINES

    KUMAR PRAKASHAN|Exercise Latest Exam MCQs|1 Videos
  • STRAIGHT LINES

    KUMAR PRAKASHAN|Exercise Miscellaneous Exercise : 10|26 Videos
  • STATISTICS

    KUMAR PRAKASHAN|Exercise QUESTION OF MODULE (KNOWLEDGE TEST)|9 Videos
  • TRIGONOMETRIC FUNCTIONS

    KUMAR PRAKASHAN|Exercise QUESTION OF MODULE (KNOWLEDGE TEST)|15 Videos

Similar Questions

Explore conceptually related problems

Find perpendicular distance from the origin to the line joining the points ( cos theta , sin theta ) and ( cos phi, sin phi) .

P_1 and P_2 are the length of perpendicular from origin to the line x sec theta + y cosec theta =a and x cos theta - y sin theta = a cos 2 theta then ….... of the following is valid.

If p and q are the lengths of perpendiculars from the origin to the lines x cos theta - y sin theta = k cos 2 theta and x sec theta + y cosec theta=k respectively, prove that p^(2) + 4q^(2) = k^(2) .

z=i+sqrt(3)=r(cos theta+sin theta)

The length of the perpendicular from the origin to the plane passing through the points veca and containing the line vecr = vecb + lambda vecc is

Equation of the line passing through the point ( a cos^(3) theta , a sin^(3) theta ) and perpendicular to the line x sec theta + y cosec theta = a is x cos theta - y sin theta = cos 2 theta .

Show that (1)/( cos theta ) -cos theta =tan theta .sin theta

Find (dy)/(dx) , if x= a (theta + sin theta), y= a (1- cos theta)

what is the angle? when sin (theta)=4 and cos (theta)=5

If alpha and beta are the solution of a cos theta + b sin theta = c , then

KUMAR PRAKASHAN-STRAIGHT LINES-Practice Work
  1. Prove that the product of the lengths of the perpendiculars drawn from...

    Text Solution

    |

  2. Find equation of line passes from middle of two parallel lines 9x + 6y...

    Text Solution

    |

  3. Find length of perpendicular from origin to the lines passes from poin...

    Text Solution

    |

  4. Find foot of perpendicular from point (2,3) on the line x+y+1=0.

    Text Solution

    |

  5. Find the equation at line passes from point of intersection at lines x...

    Text Solution

    |

  6. Find equation of line passes from point of intersection at lines 4x-3y...

    Text Solution

    |

  7. Find equation of line passes from point of intersection at lines x-y-1...

    Text Solution

    |

  8. If origin is shifted to point (1,-2) then find the new transformed for...

    Text Solution

    |

  9. At which origin will be shifted so, that new coordinate at point (4,5)...

    Text Solution

    |

  10. On which point we shift origin so that new transformed form of the equ...

    Text Solution

    |

  11. Prove that area at triangle will remain same by shifting origin at any...

    Text Solution

    |

  12. By shifting origin at (-2,3) find new transformed form at the equation...

    Text Solution

    |

  13. Find equation of line passes from ( sqrt(3) ,-1) whose perpendicular d...

    Text Solution

    |

  14. Two oppositive verticies of rectangle are (-3,1) and (1,1). Also equat...

    Text Solution

    |

  15. Lines 3x+4y+5=0 and 4x-3y-10=0 intersects at point A. B is the point o...

    Text Solution

    |

  16. Find image of point (-8, 12) with respect to line 4x+7y+ 13=0.

    Text Solution

    |

  17. Two sides of the triangle are along the lines 3x-2y+6=0 and 4x+5y-20=0...

    Text Solution

    |

  18. A(0,-1,-2), B(3,1,4) and C(5,7,1) are vertices of DeltaABD then find t...

    Text Solution

    |

  19. Point (2, 1) has reflection as simple mirror (5, 2). Find equation of ...

    Text Solution

    |

  20. Base of the equilateral triangle is along the line x + y- 2 = 0 and it...

    Text Solution

    |