Question

find the area of the triangle formed by the points (p+1,1) , (2p+1,3) and (2p+2,3p) and show that the points are collinear if p=2 or -1/2

Answer 1

Answer:

Is there a mistake in the statement of the problem?

$\text{Area} = \frac12\left|\begin{array}{ccc}1 & 1 & 1 \\ p+1 & 2p+1 & 2p+2 \\ 1 & 3 & 3p\end{array}\right|\\= \frac12\left|\begin{array}{ccc}1 & 0 & 0 \\ p+1 & p & p+1 \\ 1 & 2 & 3p-1\end{array}\right|\\= \frac12\left|\begin{array}{cc}p & p+1 \\ 2 & 3p-1\end{array}\right|\\=\frac12\bigl(p(3p-1) - 2(p+1)\bigr)\\=\frac12(3p^2-p-2p-2)\\=\frac12(3p^2-3p-2)$

The points are collinear when the area of the triangle is zero.

This triangle does not have area zero when p = 2 or p = -1/2.  But I'm pretty sure the working above is correct, so this is why I suspect there's a mistake in the statement of the problem.

I hope this is helpful nonetheless.

Answer 2

Answer:

Step-by-step explanation: