find the arthocentre of the triangle whose sides are x+2y=0;4x+3y-5=0;3x+y=0
Answers
Answer:
Given equations are:
x+2y=0......(1)
4x+3y−5=0.......(2)
3x+y=0..........(3)
Solving (1) and (2), vertex A=(0,0)
Solving (1) and (3), vertex B=(2,−1)
Equation of BC is 4x+3y−5=0
AB is perpendicular to BC and passes through A=(0,0)
Equation of AB is 3x−4y=0........(4)
BE is perpendicular to AC
Therefore equation of BE is x−3y=k
BE passes through B=(2,−1)
2+3=k⇒k=5
Equation of BE is x−3y=5........(5)
Solving (4) and (5)
Orthocenter is O(−4,−3)
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Answer:
Given equations are:
x+2y=0......(1)
4x+3y−5=0.......(2)
3x+y=0..........(3)
Solving (1) and (2), vertex A=(0,0)
Solving (1) and (3), vertex B=(2,−1)
Equation of BC is 4x+3y−5=0
AB is perpendicular to BC and passes through A=(0,0)
Equation of AB is 3x−4y=0........(4)
BE is perpendicular to AC
Therefore equation of BE is x−3y=k
BE passes through B=(2,−1)
2+3=k⇒k=5
Equation of BE is x−3y=5........(5)
Solving (4) and (5)
Orthocenter is O(−4,−3)