Question

Find the value of p for which the points A(–5, 1), B(1, p) and C(4, –2) are collinear.

Answer 1

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)\mathrm{is}$ $\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$.

Now, the given points A(–5, 1), B(1, p) and C(4, –2) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=0 \\ \Rightarrow \frac{1}{2}\left|\left(-5\right)\left(p-\left(-2\right)\right)+1\left(-2-1\right)+4\left(1-p\right)\right|=0 \\ \Rightarrow \frac{1}{2}\left|\left(-5\right)\left(p+2\right)+1\left(-3\right)+4\left(1-p\right)\right|=0 \\ \Rightarrow \left|-5p-10-3+4-4p\right|=0 \\ \Rightarrow \left|-9p-9\right|=0 \\ \Rightarrow -9p-9=0 \\ \Rightarrow -9p=9 \\ \Rightarrow p=-1 \end{gathered}$$

Hence, the value of p is –1.