# Application of Determinate in Geometry

### Application of Determinate in Geometry

i. Area of triangle: If (x₁, y₁), (x₂, y₂) and (x₃, y₃) are the verticals of a triangle then Area of the triangle =$$\left| \frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right| \right|$$ (or) = ½ |x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)|.

Example: The area of the triangle with vertices (-2, -3) (3, 2) and (-1, -8).

Solution: Given that (-2, -3) (3, 2) and (-1, -8)

We can use the formula

Area of the triangle =$$\left| \frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right| \right|$$,

Area of the triangle =   $$\left| \frac{1}{2}\left| \begin{matrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \\\end{matrix} \right| \right|$$,

= |½|- 2 (2 + 8) + 3 (3 + 1) + 1 (- 24 + 2) ||

= |½|-20 + 12 -22||

= |½|-30||

= 15 sq. Unit

ii. Condition of Collinearity of Three Points: Let tree points are (x₁, y₁) (x₂, y₂) and (x₃, y₃) then these points will be collinear, if Area of triangle = $$\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right|$$ = 0.

Example: The points (a, b + c), (b, c + a) and (c, a + b).

Solution: Area of triangle = $$\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right|$$ = 0,

Area of triangle = $$\left| \begin{matrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \\\end{matrix} \right|=0$$,

We can apply the column transformations: C₁ → C₁ + C₂.

$$\left| \begin{matrix} a+b+c & b+c & 1 \\ a+b+c & c+a & 1 \\ a+b+c & a+b & 1 \\\end{matrix} \right|=0$$,

$$(a+b+c)\left| \begin{matrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \\\end{matrix} \right|=0$$ (∵ Two columns are similar that matrix det is zero)