# Find the value of p for which the points

Question:

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

Solution:

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 is –1.