Find the orthocentre of the triangle the equation of whose sides are x+y=1, 2x+3y=6, 4x-y+4=0

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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