# Show that the point(a+5,a-4),(a-2,a+3) and (a,a) do not lie on a straight line for any value of a.

Toolbox:
• If the area of a triangle is 0,then the three points are collinear.
• The area of triangle $\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}$
Let the points ($x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ be (a+5,a-4),(a-2,a+3),(a,a).
We are asked to prove that the points are not collinear.
For this it is enough to prove that
$\Delta=\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 &y_3 &1\end{vmatrix}\neq 0.$
Let $\Delta=\begin{vmatrix}a+5 & a-4 & 1\\a-2 & a+3 & 1\\a & a & 1\end{vmatrix}$
Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=\begin{vmatrix}7 & -7 & 0\\-2 & 3 & 0\\a & a & 1\end{vmatrix}$
Now expanding along $R_1$ we get,
$\Delta=7(3)-(-7)(-2)+0.$
$\quad=21-14\neq 0.$
Hence the points are not collinear.