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The area of the triangle with vertices (p+1,1)(2p+1,3) and (2p+2,2p) is 9 square units. Then the integral value of p =

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a

4

b

1

c

2

d

0

answer is A.

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Detailed Solution

12|p+1112p+1312p+22p1|=9⇒|p+111p20p+12p−10|=18 ⇒2p2−3p−2=18⇒(2p+5)(p−4)=0⇒p=4

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