The area of a triangle with vertices (p, 2 – 2p), (1 – p, 2p) and

1.	The area of a triangle with vertices (p, 2 – 2p), (1 – p, 2p) and (–4 –6, 6 – 2p) is 70 sq. units. Then, the number of possible integral values of p is 1. 0 2. 1 3. 2 4. 3
Answer» Correct option is 2. 1 By using area of triangle formula We get, (8p² + 4p - 4) = \(\pm{}\)140 ⇒ 8p²+ 4p - 144 = 0 ⇒ p = 4 or p = -9/2 Other equation 8p²+ 4p + 136 = 0 does not have real roots So, numbers of integral values = 1

The area of a triangle with vertices (p, 2 – 2p), (1 – p, 2p) and (–4 –6, 6 – 2p) is 70 sq. units. Then, the number of possible integral values of p is 1. 0 2. 1 3. 2 4. 3

Answer»

Correct option is 2. 1

By using area of triangle formula

We get, (8p² + 4p - 4) = \(\pm{}\)140

⇒ 8p²+ 4p - 144 = 0

⇒ p = 4 or p = -9/2

Other equation 8p²+ 4p + 136 = 0 does not have real roots So, numbers of integral values = 1

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The area of a triangle with vertices (p, 2 – 2p), (1 – p, 2p) and (–4 –6, 6 – 2p) is 70 sq. units. Then, the number of possible integral values of p is 1. 0 2. 1 3. 2 4. 3

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