Question

find the values of p for which the points (3p+1,p),(p+2,p-5),(p+1,-p) are collinear

Answer 1

If these three points are collinear ,then area of triangle formed by them must be zero.

$$\begin{gathered} Andareaoftriangle=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|=0..............\left(1\right) \\ Nowpointsare(3p+1,p),(p+2,p-5),(p+1,-p) \\ So(x_1,y_1)=(3p+1,p) \\ (x_2,y_2)=(p+2,p-5) \\ (x_3,y_3)=(p+1,-p) \\ Onputiingthevaluesin\left(1\right) \\ \frac{1}{2}\left|(3p+1)(p-5+p)+(p+2)(-p-p)+(p+1)(p-p+5)\right|=0 \\ \Rightarrow \frac{1}{2}\left|(3p+1)(2p-5)+(p+2)(-2p)+(p+1)\left(5\right)\right|=0 \\ \Rightarrow \frac{1}{2}\left|6p^2-13p-5-2p^2-4p+5p+5\right|=0 \\ \Rightarrow \frac{1}{2}\left|4p^2-12p\right|=0 \\ \Rightarrow 4p^2-12p=0 \\ \Rightarrow p^2-3p=0 \\ \Rightarrow p(p-3)=0 \\ \Rightarrow p=0,3 \end{gathered}$$