$Δ$ denotes the area of triangle with vertices $(p + 1, 1)$ , $(2 p + 1, 3)$ and $(2 p + 2, 2 p)$ . Match the correct values of $p$ in List I with corresponding values of areas in List II.

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Solution

Area of the given triangle is
$Δ = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} p + 1 & 1 & 1 2 p + 1 & 3 & 1 2 p + 2 & 2 p & 1 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} p + 1 & 1 & 1 p & 2 & 0 p + 1 & 2 p - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{2} (2 p^{2} - 3 p - 2)$
Applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$
A) If $p = 0$
$Δ = ∣ ∣ ∣ \frac{1}{2} (- 2) ∣ ∣ ∣ = 1$

B) $p = \pm 1$
$Δ = ∣ ∣ ∣ \frac{1}{2} (\pm 3) ∣ ∣ ∣ = \frac{3}{2}$
C) If $p = 3$
$Δ = \frac{1}{2} (2 \times 9 - 3 \times 3 - 2) = \frac{7}{2}$

D) If $p = - 3$
$Δ = \frac{1}{2} (2 \times 9 + 3 \times 3 - 2) = \frac{25}{2}$

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Δ denotes the area of triangle with vertices (p+1,1), (2p+1,3) and (2p+2,2p). Match the correct values of p in List I with corresponding values of areas in List II.

$Δ$ denotes the area of triangle with vertices $(p + 1, 1)$ , $(2 p + 1, 3)$ and $(2 p + 2, 2 p)$ . Match the correct values of $p$ in List I with corresponding values of areas in List II.