Question

The equation of two sides of triangle are 3x -2y +6=0 and 4x +5y-20 =0 and ortho centre is 1,1 find equation of third side

Answer 1

    We find slopes and then altitudes of given sides. Then from intersection of altitudes with the sides, we get the vertices. Line joining them is the third side desired.

Let the Equation of side AB =  3x - 2y + 6 = 0        ---- (1)
Let the equation of side AC =   4 x + 5 y - 20 = 0      ---(2)

Solving the above equations, we get vertex A:
    (1)*5 +(2)*2 =>  23 x + 30-40 = 0    =>   x = 10/23
     So    3*10/23 - 2y + 6 = 0              =>    y = 84/23

Orthocenter H = (1,1,)     ---- given.

Slope of AC = -4/5
  Slope of the altitude BHE from B onto AC:  + 5/4
   Altitude passes through H(1,1).
     Equation of altitude BHE :      (y-1)/(x-1) = 5/4
                                   4 y - 4 = 5x - 5
                                    4 y = 5 x - 1          --- (3)

Intersection of BHE with AB is obtained by solving equations (1)  & (3):
           6x+12 = 5 x - 1
           x = -13. 
           So   2 y = 3x + 6 =>      y = - 33/2 

Now Slope of AB = 3/2
   Slope of altitude from C onto AB:  =  -2/3
   Equation of altitude CHF :     (y-1)/(x-1) = -2/3
                                   3y - 3 + 2x - 2 = 0
                                   3 y + 2 x = 5         --- (4)

   Intersection of CHF with AC:
               20 - 5 y = 10 - 6 y
               y = - 10
               x = (5-3y)/2 = 35/2

So  B= (-13, -33/2)   and  C= (35/2, -10).

The third side BC =       (y + 10)/(x - 35/2) = (13/2) / (61/2) = 13/61
Simplify it to get the answer.

Answer 2

Answer:

hope this helps you....