Find the co-ordinates of the orthocentre of the triangle, the equations of whose sides are x + y = 1, 2x + 3y = 6, 4x y + 4 = 0, without finding the coordinates of its vertices.
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Let equation of
AB be x+y-1 =0---(1)
BC be 2x+3y-6= 0 ---(2)
and AC be 4x-y+4=0 ---(3)
Solving (1) and (2) B = (- 3, 4 )
Solving (1) and (3) A =(-3/5, 8/5)
Equation of BC is 2x+3y=6
Altitude AD is perpendicular to BC,
Therefore Equation of AD is x + y + k = 0
AD is passing through A (-3/5, 8/5)
⇒ (-3/5)+(8/5)+k=0
⇒k = -1
∴ Equation if AD is x + y -1 = 0 ----(4)
Altitude BE is perpendicular to AC.
⇒Let the equation of DE be x – 2y = k
BE is passing through D (- 3, 4 )
⇒-3-8=k
⇒ k = -11
Equation of BE is x – 2y = -11-----(5)
Solving (4) and (5), the point of intersection is (-3, 4)
Therefore the orthocenter of the triangle is (-3, 4)
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