Browse Questions

The point (3,-2),(x,2),(8,8) are collinear, find x using determinants.

Toolbox:
• Area of a triangle whose vertices are $(x_1,y_2),(x_2,y_2)$ and $(x_3,y_3)$
• $\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}$
Step 1:
Area of a triangle ABC with vertices ($x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
$\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}$
Let the points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ be (3,-2),(x,2) and (8,8).
Now substituting the values in the determinant we get,
$\Delta=\frac{1}{2}\begin{vmatrix}3 & -2 & 1\\x & 2 & 1\\8 & 8 & 1\end{vmatrix}$
Since the points are collinear,|$\Delta$|=0.
(i.e) 0=$\frac{1}{2}\begin{vmatrix}3 & -2 & 1\\x & 2 & 1\\8 & 8 & 1\end{vmatrix}$
Step 2:
Now expanding along $R_1$ we get,
0=$3(2\times 1-1\times 8)-(-2)(x\times 1-8\times 1)+1(x\times 8-2\times 8)$
On simplifying we get
0=-18+2(x-8)+(x-16)
$\Rightarrow 10x-16-16-18=0.$
$\Rightarrow 10x-56=0.$
(i.e) 10x=|50|.
Since the area of the triangle is |$x_1(y_2-y_3)-x_2(y_3-y_1)+x_3(y_1-y_2)$|
If 10x=-50
x=-5.
If 10x=50
x=5.
Hence x=-5 or 5.