find the values of a so that the point p (alpha²,alpha) is an interior point of the triangle formed by the lines x+y-4=0,3x-7y+8=0,4x-y-31=0?
Answers
Step-by-step explanation:
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Answer:
Let ABC be the triangle of sides AB, BC and CA, whose equation x+y−4=0__(1), 3x−7y+8=0__(2) and 4x−y−31=0__(3) respectively.
on solving (1) and (2) we get,
x=2,y=2 Thus AB and BC intersect at B(2,2).
On solving (2) and (3), we get,
x=9,y=5.
Thus, BC and CA intersect at C(9,5)
On solving (1) and (3) we get,
x=7,y=−3
Thus , AB and CA intersect at A(7,−3).
Now, set P(−3,2) will be the given point.
The given point P(−3,2) will be the inside point of triangle if, ref. image
) A and P lies on same side of BC.
II) B and P lies on same side of AC.
Thus, if A and P lie on the same side of BC. Thus,
(21+21+8)(−9−14+8)>0
⇒50×(−15)>0
⇒−750>0, which is false.
Therefore, the point (−3,2) lies outside triangle ABC
