pls solve this question if you know answer
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Answer:
x2
Step-by-step explanation:
Answered by
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Step-by-step explanation:
Let AB,BC and CA be x+y=1,2x+3y=0 and 4x−y+4=0 respectively
Solving for A, we have y=1−x.
Hence 4x−y+4=0
4x+4=y⇒4x+4=1−x
4x+x=1−4
5x=−3⇒x=−3/5
y=1−x=1+
5
3
=
5
8
A(
5
−3
,
5
8
)
Solving for B,2x+3(1−x)=0
2x+3−3x=0
x=3
y=1−x=1−3=−2 B(3,−2)
Altitude from A passes through A and perpendicular to BC.
Equation is 3x−2y+k=0
3(
5
−3
)−2(
5
8
)+k=0 gives k=5.
Similarly from B, the altitude's equation will be
x+4y+l=0, passing through (3,−2).
3−8+l=0 gives l=5.
The point of intersection of 3x−2y+5=0 and x+4y+5=0 on solving is (
7
−15
,
7
−5
)
which is the orthocentre of the triangle.
(
7
−15
,
7
−5
)
this is ur answer sari na annaya
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