Question Statement

Find the area of the triangles:

(i) $A(1,1)$; $B(4,5)$; $C(12,-1)$

(ii) $D(3,1)$; $E(2,3)$; $F(2,2)$

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Background and Explanation

To find the area of a triangle when the coordinates of its vertices are known, we use the determinant method (also called the shoelace formula). This involves constructing a $3\times 3$ matrix from the coordinates and evaluating its determinant. The absolute value of half the determinant gives the area.

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Solution

Part (i): Triangle with vertices $A(1,1)$, $B(4,5)$, and $C(12,-1)$

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ is given by: $\text{Area}=\frac{1}{2}\left|\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$

Substituting the coordinates of $A$, $B$, and $C$:

$\text{Area}=\frac{1}{2}\left|\begin{array}{rrr}1& 1& 1\\ 4& 5& 1\\ 12& -1& 1\end{array}\right|$

Expanding the determinant along the first row using cofactor expansion (multiplying each element by its corresponding $2\times 2$ minor with appropriate sign):

$=\frac{1}{2}\left[1\left|\begin{array}{rr}5& 1\\ -1& 1\end{array}\right|-1\left|\begin{array}{rr}4& 1\\ 12& 1\end{array}\right|+1\left|\begin{array}{rr}4& 5\\ 12& -1\end{array}\right|\right]$

Calculating each $2\times 2$ determinant using the formula $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$:

$$\begin{gathered} =\frac{1}{2}[1(5\cdot 1-(-1)\cdot 1)-1(4\cdot 1-12\cdot 1)+1(4\cdot (-1)-12\cdot 5)] \\ =\frac{1}{2}[1(5+1)-1(4-12)+1(-4-60)] \\ =\frac{1}{2}[6-(-8)+(-64)] \\ =\frac{1}{2}[6+8-64] \\ =\frac{1}{2}(-50) \\ =-25 \end{gathered}$$

Since area is always positive, we take the absolute value:

$\text{Area}=25\text{ square units}$

Part (ii): Triangle with vertices $D(3,1)$, $E(2,3)$, and $F(2,2)$

Using the same determinant formula:

$\text{Area}=\frac{1}{2}\left|\begin{array}{lll}3& 1& 1\\ 2& 3& 1\\ 2& 2& 1\end{array}\right|$

Expanding along the first row:

$=\frac{1}{2}\left[3\left|\begin{array}{ll}3& 1\\ 2& 1\end{array}\right|-1\left|\begin{array}{ll}2& 1\\ 2& 1\end{array}\right|+1\left|\begin{array}{ll}2& 3\\ 2& 2\end{array}\right|\right]$

Evaluating the $2\times 2$ determinants:

$$\begin{gathered} =\frac{1}{2}[3(3\cdot 1-2\cdot 1)-1(2\cdot 1-2\cdot 1)+1(2\cdot 2-2\cdot 3)] \\ =\frac{1}{2}[3(3-2)-1(2-2)+1(4-6)] \\ =\frac{1}{2}[3(1)-1(0)+1(-2)] \\ =\frac{1}{2}[3-0-2] \\ =\frac{1}{2}(1) \\ =\frac{1}{2} \end{gathered}$$

Therefore, the area of the triangle is $\frac{1}{2}$ square units.

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Key Formulas or Methods Used

• Determinant Formula for Triangle Area: $\text{Area}=\frac{1}{2}\left|\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$
• Cofactor Expansion: Expanding a $3\times 3$ determinant along the first row using minors and cofactors (signs alternate: $+$, $-$, $+$)
• $2\times 2$ Determinant: $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$
• Absolute Value Property: Area is always positive, so we take $|\text{result}|$ if the determinant yields a negative value

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Summary of Steps

1. Set up the matrix: Write the coordinates as a $3\times 3$ matrix with $x$ and $y$ coordinates in the first two columns and $1$ in the third column for each vertex.
2. Expand the determinant: Use cofactor expansion along the first row to break the $3\times 3$ determinant into three $2\times 2$ determinants with alternating signs ($+$, $-$, $+$).
3. Calculate minors: Evaluate each $2\times 2$ determinant using the rule $ad-bc$.
4. Sum and simplify: Multiply each minor by its cofactor, sum the results, and multiply by $\frac{1}{2}$.
5. Final answer: Take the absolute value of the result to ensure the area is positive.