The orthocentre of the triangle formed by the lines x + y = 1 and

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221.

The orthocentre of the triangle formed by the lines x + y = 1 and 2y2 - xy - 6x2 = 0

  • 43, 43

  • 23, 23

  • 23, - 23

  • 43,  - 43


A.

43, 43

We have x + y = 1and 2y2 - xy - 6x2 = 0 2y2 - 4xy + 3xy - 6x2 = 0 Equation of sides of ABC arex + y = 1, 2y + 3x = 0 and y - 2x = 0

Solving these equations simultaneously, we get A0, 0B13, 23 and C - 2, 3Equation of altitude AD isx - y = 0      ...  iEquation of altitude CF isx + 2y = λSince, this passes through  - 2, 3 - 2 + 6 = λ  λ = 4On solving eqs. i and ii, we getx = 43, y = 43


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222.

The incentre of the triangle formed by the straight lines y = 3x, y = - 3x and y = 3 is  and y = 3 is

  • (0, 2)

  • (1, 2)

  • (2, 0)

  • (2, 1)


223.

The points on the straight line 3x - 4y + 1 = 0 which are at a distance of 5 units from the point (3, 2) are

  •  - 2, - 74 - 3,  - 52

  • 4, 114 - 1,  - 1

  • 1, 122, 54

  • 7, 5, - 1, - 1


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