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Q.

The base of a triangle lies along the line x=p and is of length p . The area of the triangle is p2 . The locus of the third vertex is

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a

(x+p)(x–3p)=0

b

(x–p)(x+3p) =0

c

(x–p)(x–3p)=0

d

(x–p)(x–2p) =0

answer is A.

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Detailed Solution

12ph=p2 h=2p|x1−p|=2plocus of p is x−3p=0   (or)   x+p=0                ⇒h=2p∴     x=3px=−p                ⇒(x+p)(x−3p)=0
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