Find the value of p for which the points

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 $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ 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.

Area of triangle $=0$

$\Rightarrow \frac{1}{2}|(-5)(p-(-2))+1(-2-1)+4(1-p)|=0$

$\Rightarrow \frac{1}{2}|(-5)(p+2)+1(-3)+4(1-p)|=0$

$\Rightarrow|-5 p-10-3+4-4 p|=0$

$\Rightarrow|-9 p-9|=0$

$\Rightarrow-9 p-9=0$

$\Rightarrow-9 p=9$

$\Rightarrow p=-1$

Hence, the value of p is –1.