Determinants - Area of a Triangle

Find the area of the triangle with vertices $(3, 8)$, $(-4, 2)$, and $(5, 1)$ using determinants.

$\text{Area} = \frac{1}{2} \begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}$ Expanding along the first row: $= \frac{1}{2} [ 3(2 - 1) - 8(-4 - 5) + 1(-4 - 10) ]$ $= \frac{1}{2} [ 3(1) - 8(-9) + 1(-14) ]$ $= \frac{1}{2} [ 3 + 72 - 14 ] = \frac{61}{2} = 30.5 \text{ sq. units}$

Substitute the coordinates into the determinant formula and expand it along any row or column (usually the first row). Take the absolute value of the final result.

Find the value of $k$ if the area of the triangle is $4$ square units and the vertices are $(k, 0)$, $(4, 0)$, and $(0, 2)$.

The area is given as $4$. We use $\pm 4$ for the determinant value: $\frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4$ Expanding along the second column (as it has two zeros): $\frac{1}{2} [ -2(k - 4) ] = \pm 4$ $-(k - 4) = \pm 4$ Case 1: $-(k - 4) = 4 \implies -k + 4 = 4 \implies k = 0$ Case 2: $-(k - 4) = -4 \implies -k + 4 = -4 \implies k = 8$ Therefore, $k = 0, 8$.

When the area is given, it is important to equate the determinant expression to both the positive and negative values of the area to find all possible values of the unknown variable.

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants.

Let $P(x, y)$ be any point on the line $AB$. Then the area of $\Delta PAB$ is $0$. $\begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0$ Expanding along the third row: $0(y - 3) - 0(x - 1) + 1(3x - y) = 0$ $3x - y = 0$ or $y = 3x$

To find the equation of a line passing through two points, assume a general point $(x, y)$ on the line and set the determinant representing the area to zero since the three points are collinear.

To show collinearity, the determinant $\Delta$ must be zero. $\Delta = \begin{vmatrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{vmatrix}$ Applying column operation $C_2 \to C_2 + C_1$: $\Delta = \begin{vmatrix} a & a + b + c & 1 \\ b & a + b + c & 1 \\ c & a + b + c & 1 \end{vmatrix}$ Taking $(a + b + c)$ common from $C_2$: $\Delta = (a + b + c) \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix}$ Since $C_2$ and $C_3$ are identical, the determinant value is $0$. Thus, points $A, B$, and $C$ are collinear.

Collinearity is proven by showing that the area of the triangle formed by these three points is zero. Using determinant properties (identical columns) simplifies the proof.

Let $(x, y)$ be any point on line $PQ$. The equation is: $\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0$ $x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0 \implies -2x + 6y = 0 \implies x - 3y = 0$. Now, for point $R(k, 0)$, Area $= 3$: $\frac{1}{2} \left| \begin{vmatrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ k & 0 & 1 \end{vmatrix} \right| = 3$ $\begin{vmatrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 6$ Expanding along $R_3$: $k(1 - 3) - 0 + 1(9 - 9) = \pm 6 \implies -2k = \pm 6$. So, $k = 3$ or $k = -3$.

We first use the condition of collinearity with a variable point $(x,y)$ to find the line equation. Then, we use the area formula with the given absolute value of 3 to solve for the unknown coordinate $k$.