Let the equations of two sides of a triangle be 3 x - 2 y + 6 = 0 and 4 x + 5 y - 20 = 0 . If the…

Let the equations of two sides of a triangle be 3x-2y+6=0 and 4x+5y-20=0. If the orthocenter of this triangle is at 1, 1 then the equation of it's third side is:
  1. 122y+26x+1675=0
  2. 26x-122y-1675=0
  3. 26x+61y+1675=0
  4. 122y-26x-1675=0

Solution

To find equation of BC, first we will find coordinates of B and C.

mAC=32

AC and BB' are perpendicular to each other 

hence mBB'=-23

Equation of line passing through x1,y1 and having slope m is y-y1=mx-x1

For equation of BB'

y-1=-23x-1

2x+3y=5   ........1 

Point of intersection of AB and BB' B 352, -10

Similarly point C-13, -332

Equation of BC is, y-y1=y2-y1x2-x1x-x1

y+10=1361 x-352

   26x-122y-1675=0

Asked in: JEE Main 2019 (09 Jan Shift 2)

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