Three points (p + 1, 1),(2p + 1, 3) and (2p + | vedclass

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(C) For three points to be collinear,the area of the triangle formed by them must be zero,or the determinant of the coordinates must be zero: $\begin{

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$2$

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(C) For three points to be collinear,the area of the triangle formed by them must be zero,or the determinant of the coordinates must be zero: $\begin{vmatrix} p+1 & 1 & 1 \ 2p+1 & 3 & 1 \ 2p+2 & 2p & 1 \end{vmatrix} = 0$ Expanding along the first row: $(p+1)(3 - 2p) - 1((2p+1) - (2p+2)) + 1((2p+1)(2p) - 3(2p+2)) = 0$ $(p+1)(3 - 2p) - 1(-1) + (4p^2 + 2p - 6p - 6) = 0$ $(3p - 2p^2 + 3 - 2p) + 1 + (4p^2 - 4p - 6) = 0$ $(-2p^2 + p + 3) + 1 + (4p^2 - 4p - 6) = 0$ $2p^2 - 3p - 2 = 0$ Factoring the quadratic equation: $2p^2 - 4p + p - 2 = 0$ $2p(p - 2) + 1(p - 2) = 0$ $(2p + 1)(p - 2) = 0$ Thus,$p = 2$ or $p = -1/2$. Given the options,the correct value is $p = 2$.

Three points $(p + 1, 1)$,$(2p + 1, 3)$ and $(2p + 2, 2p)$ are collinear,if $p =$

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