Area of a Triangle

Area of a Triangle using Determinants

In earlier classes, you learned that the area of a triangle whose vertices are $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ is given by the expression: $$\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

This expression can be written in the form of a determinant as: $$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$

Important Remarks

When applying this determinant formula, you must keep the following points in mind:

1.Absolute Value

Since area is a positive quantity, we always take the absolute value of the determinant for calculating the area.

2. Given Area

If the area is given in a problem (e.g., to find an unknown variable like $k$), use both positive and negative values of the determinant for calculation. i.e., $$\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm \text{Area}$$

3.Collinearity

The area of the triangle formed by three collinear points is zero. Thus, points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ are collinear if and only if $$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$$

Equation of a Line

The determinant formula can also be used to find the equation of a line passing through two given points $(x_1, y_1)$ and $(x_2, y_2)$. Let $(x, y)$ be any general point on the line. Since the three points $(x, y)$, $(x_1, y_1)$, and $(x_2, y_2)$ are collinear, the area of the triangle formed by them must be zero. Therefore, the equation of the line is: $$\begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0$$

Example 1: Finding Area of a Triangle

Find the area of the triangle whose vertices are $(3, 8)$, $(-4, 2)$ and $(5, 1)$.

Solution: Step 1: Identify the coordinates of the vertices.

$(x_1, y_1) = (3, 8)$ $(x_2, y_2) = (-4, 2)$ $(x_3, y_3) = (5, 1)$

Step 2: Set up the determinant for the area. $$\Delta = \frac{1}{2} \begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}$$

Step 3: Expand the determinant along the first row ($R_1$). $$\Delta = \frac{1}{2} \left[ 3 \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} - 8 \begin{vmatrix} -4 & 1 \\ 5 & 1 \end{vmatrix} + 1 \begin{vmatrix} -4 & 2 \\ 5 & 1 \end{vmatrix} \right]$$ $$\Delta = \frac{1}{2} [3(2(1) - 1(1)) - 8(-4(1) - 5(1)) + 1(-4(1) - 5(2))]$$ $$\Delta = \frac{1}{2} [3(2 - 1) - 8(-4 - 5) + 1(-4 - 10)]$$ $$\Delta = \frac{1}{2} [3(1) - 8(-9) + 1(-14)]$$ $$\Delta = \frac{1}{2} [3 + 72 - 14]$$ $$\Delta = \frac{1}{2} [61] = \frac{61}{2}$$

Answer: $\frac{61}{2}$ square units.

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Example 2: Area with Negative Determinant

Find the area of the triangle whose vertices are $(2, 7)$, $(1, 1)$ and $(10, 8)$.

Solution: Step 1: Identify the coordinates.

$(x_1, y_1) = (2, 7)$, $(x_2, y_2) = (1, 1)$, $(x_3, y_3) = (10, 8)$

Step 2: Set up the determinant. $$\Delta = \frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}$$

Step 3: Expand the determinant along $R_1$. $$\Delta = \frac{1}{2} [2(1 - 8) - 7(1 - 10) + 1(8 - 10)]$$ $$\Delta = \frac{1}{2} [2(-7) - 7(-9) + 1(-2)]$$ $$\Delta = \frac{1}{2} [-14 + 63 - 2]$$ $$\Delta = \frac{1}{2} [47] = \frac{47}{2}$$

Answer: $\frac{47}{2}$ square units.

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Example 3: Checking Collinearity

Show that points $A(a, b+c)$, $B(b, c+a)$ and $C(c, a+b)$ are collinear.

Solution: Step 1: For the points to be collinear, the area of the triangle formed by them must be zero. Let's evaluate the determinant. $\Delta = \frac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$

Step 2: Apply properties of determinants to simplify. Let's perform the operation $C_2 \rightarrow C_2 + C_1$ (Add Column 1 to Column 2). $$\Delta = \frac{1}{2} \begin{vmatrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{vmatrix}$$

Step 3: Take out the common factor $(a+b+c)$ from Column 2 ($C_2$). $$\Delta = \frac{1}{2} (a+b+c) \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix}$$

Step 4: Observe that Column 2 ($C_2$) and Column 3 ($C_3$) are identical.

According to properties of determinants, if any two rows or columns are identical, the determinant is zero. $$\Delta = \frac{1}{2} (a+b+c) \times 0 = 0$$

Step 5: Since the area is zero, the points are collinear.

Answer: Proved.

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Example 4: Finding an Unknown Variable Given Area

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

Solution: Step 1: Set up the area formula. Since the area is given, we must use $\pm 4$. $$\frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4$$

Step 2: Multiply both sides by 2. $$\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 8$$

Step 3: Expand the determinant. Expanding along Column 2 ($C_2$) is easiest because it has two zeros. $$-0 + 0 - 2 \begin{vmatrix} k & 1 \\ 4 & 1 \end{vmatrix} = \pm 8$$ $$-2(k(1) - 4(1)) = \pm 8$$ $$-2(k - 4) = \pm 8$$

Step 4: Solve the two cases.

Case 1: $$-2(k - 4) = 8$$ $$k - 4 = -4 \Rightarrow k = 0$$

Case 2: $$-2(k - 4) = -8$$ $$k - 4 = 4 \Rightarrow k = 8$$

Answer: $k = 0, 8$

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Example 5: Finding Equation of a Line

Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants.

Solution: Step 1: Let $P(x, y)$ be any point on the line passing through $A(1, 2)$ and $B(3, 6)$.

Step 2: Since $P, A, B$ are collinear on the same line, the area of the triangle formed by them is zero. $$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$$

Step 3: Multiply by 2 and expand along $R_1$. $$x(2(1) - 6(1)) - y(1(1) - 3(1)) + 1(1(6) - 3(2)) = 0$$ $$x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0$$ $$x(-4) - y(-2) + 1(0) = 0$$ $$-4x + 2y = 0$$

Step 4: Simplify the equation.

Divide by $-2$: $2x - y = 0$ or $y = 2x$

Answer: $y = 2x$

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Example 6: Collinear Points with an Unknown

If the points $(2, -3)$, $(k, -1)$ and $(0, 4)$ are collinear, find the value of $k$.

Solution: Step 1: For collinear points, the area determinant is zero. $$\begin{vmatrix} 2 & -3 & 1 \\ k & -1 & 1 \\ 0 & 4 & 1 \end{vmatrix} = 0$$

Step 2: Expand the determinant along $R_1$. $$2(-1(1) - 4(1)) - (-3)(k(1) - 0(1)) + 1(k(4) - 0(-1)) = 0$$ $$2(-1 - 4) + 3(k - 0) + 1(4k - 0) = 0$$ $$2(-5) + 3k + 4k = 0$$

Step 3: Solve for $k$.

$-10 + 7k = 0$ $\Rightarrow 7k = 10 \Rightarrow k = \frac{10}{7}$

Answer: $k = \frac{10}{7}$

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Example 7: Area with Origin as a Vertex

Find the area of the triangle whose vertices are $(0, 0)$, $(4, 0)$ and $(0, 5)$.

Solution: Step 1: Set up the determinant. $\Delta = \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 5 & 1 \end{vmatrix}$

Step 2: Expand along $R_1$. Since the first two elements are zero, we only calculate the cofactor of the third element. $$\Delta = \frac{1}{2} [0 - 0 + 1(4(5) - 0(0))]$$ $$\Delta = \frac{1}{2} [1(20 - 0)] \Rightarrow \Delta = \frac{1}{2} [20] = 10$$

Answer: $10$ square units.

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Example 8: Area and Unknown Coordinate

If the area of a triangle is $35$ sq units with vertices $(2, -6)$, $(5, 4)$ and $(k, 4)$. Then find the value of $k$.

Solution: Step 1: Set up the equation using the area formula, remembering the $\pm$ sign for given area. $$\frac{1}{2} \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 35$$

Step 2: Multiply by 2. $$\begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix} = \pm 70$$

Step 3: Expand the determinant along $C_3$ (or $R_1$). Let's use $R_1$. $$2(4(1) - 4(1)) - (-6)(5(1) - k(1)) + 1(5(4) - k(4)) = \pm 70$$ $$2(0) + 6(5 - k) + (20 - 4k) = \pm 70$$ $$\Rightarrow 0 + 30 - 6k + 20 - 4k = \pm 70$$ $$50 - 10k = \pm 70$$

Step 4: Solve the two cases.

Case 1: $50 - 10k = 70 \Rightarrow -10k = 20 \Rightarrow k = -2$

Case 2: $50 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12$

Answer: $k = 12, -2$

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Example 9: Equation of line and third point

Find the equation of the line joining $A(3, 1)$ and $B(9, 3)$ using determinants. Also find $k$ if $D(k, 0)$ is a point such that area of triangle $ABD$ is $3$ sq units.

Solution: Step 1: Find the equation of line AB. Let $P(x, y)$ be a point on AB. $$\frac{1}{2} \begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0$$ $$x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0$$ $$-2x - y(-6) + 0 = 0$$ $$-2x + 6y = 0 \Rightarrow x - 3y = 0$$ So, the equation of the line is $x - 3y = 0$.

Step 2: Use the area of triangle ABD to find $k$. $$\frac{1}{2} \begin{vmatrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3$$ $$\begin{vmatrix} 3 & 1 & 1 \\ 9 & 3 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 6$$

Step 3: Expand along $R_3$. $$k(1(1) - 3(1)) - 0 + 1(3(3) - 9(1)) = \pm 6$$ $$k(-2) + (9 - 9) = \pm 6$$ $$-2k = \pm 6$$

Step 4: Solve for $k$.

$k = \frac{\pm 6}{-2} = \mp 3 \Rightarrow k = 3$ or $k = -3$

Answer: Equation of line is $x - 3y = 0$. Value of $k$ is $3, -3$.

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Example 10: Condition for Points on X-axis

Find the area of the triangle formed by points $(a, 0)$, $(b, 0)$, and $(c, d)$. What does this result signify if $d = 0$?

Solution: Step 1: Set up the area determinant. $$\Delta = \frac{1}{2} \begin{vmatrix} a & 0 & 1 \\ b & 0 & 1 \\ c & d & 1 \end{vmatrix}$$

Step 2: Expand along Column 2 ($C_2$) since it has two zeros. $$\Delta = \frac{1}{2} \left[ -0 + 0 - d \begin{vmatrix} a & 1 \\ b & 1 \end{vmatrix} \right]$$ $$\Delta = \frac{1}{2} [-d(a(1) - b(1))]$$ $$\Delta = -\frac{1}{2}d(a - b) = \frac{1}{2}d(b - a)$$ Taking absolute value, Area $= \frac{1}{2} |d(b - a)|$.

(Notice this is exactly $\frac{1}{2} \times \text{base} \times \text{height}$, where base is distance between $(a,0)$ and $(b,0)$ which is $|b-a|$, and height is $|d|$).

Step 3: If $d = 0$, then Area $= \frac{1}{2}(0)(b - a) = 0$.

Significance: If $d=0$, the third point $(c, 0)$ also lies on the x-axis. Since all three points lie on the x-axis, they are collinear, and the area of the triangle is naturally $0$.

Answer: Area $= \frac{1}{2} |d(b - a)|$. If $d=0$, the points are collinear on the x-axis.