Question

Find the value(s) of p for which the points (3p + 1, p), (p + 2, p – 5) and (p + 1, –p) are collinear ?

Answer 1

Since the given points are collinear, the area of the triangle formed by them must be 0.

\[\Rightarrow \frac{1}{2}\left[ x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right] = 0\]

Here,

\[x_1 = 3p + 1, y_1 = p; x_2 = p + 2, y_2 = p - 5\ \text{and}\ x_3 = p + 1, y_3 = - p\]

\[\therefore \frac{1}{2}\left[ \left( 3p + 1 \right)\left\{ \left( p - 5 \right) - \left( - p \right) \right\} + \left( p + 2 \right)\left\{ \left( - p \right) - \left( p \right) \right\} + \left( p + 1 \right)\left\{ \left( p \right) - \left( p - 5 \right) \right\} \right] = 0\]

\[\Rightarrow \left( 3p + 1 \right)\left( 2p - 5 \right) + \left( p + 2 \right)\left( - 2p \right) + \left( p + 1 \right)\left( 5 \right) = 0\]
\[ \Rightarrow 4 p^2 - 12p = 0\]
\[ \Rightarrow p\left( 4p - 12 \right) = 0\]
\[ \Rightarrow p = 0\ \text{or}\ 4p - 12 = 0\]
\[ \Rightarrow p = 0\ \text{or}\ p = 3\]

So, the values of p are 0 and 3.