Let the equations of two sides of a triangle be 3x-2y+6=0 and 4x + 5y- 20 = 0. If the orthocentre of this triangle is at (1,1) then the equation of its third side is

Let the equations of two sides of a triangle be 3x-2y+6=0 and 4x + 5y- 20 = 0. If the orthocentre of this triangle is at (1,1) then the equation of its third side is

  1. A

    26x + 61y + 1675 = 0

  2. B

    122y + 26x + 1675 = 0

  3. C

    26x - 122y - 1675 = 0

  4. D

    122y - 26x - 1675 = 0

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    Solution:

    Let the equation of side AB  3x2y+6=0  

                                                                                                              …(i)

    and, the equation of side AC is 4x + 5y - 20 = 0

                                                                                                              …(ii)

    Solving (i) and (ii), we get vertex  A1023,8423.

    Now, slope of AC=4/5

     Slope of perpendicular BE=5/4

    Also, BE passes through orthocentre H(l, 1)

     Equation of BE is

    (y1)=54(x1)

     4y5x+1=0                                                      …(iii)

    Solving (i) and (iii), we get vertex B13,332.

    Now, slope of AB=3/2

     Slope of perpendicular CF=2/3

    Equation of CF is (y1)=23(x1)

    3y+2x5=0                                                          ….(iv)                             

    Solving (ii) and (iv), we get vertex C352,10

    Equation of side BC is  (y+10)=10+332352+13x352

     (y+10)=1361x35226x122y=1675

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