Question Statement

Find the area of the following figures using the coordinate geometry method:

(i) Triangle with vertices $A(-3,-1)$, $B(2,3)$, and $C(2,-1)$

(ii) Triangle with vertices $A(1,3)$, $B(-2,-3)$, and $C(4,-3)$

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Background and Explanation

The area of a triangle given the coordinates of its three vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ can be calculated using the determinant method. This involves constructing a $3\times 3$ matrix with the coordinates and a column of 1s, then evaluating half the absolute value of the determinant.

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Solution

Part (i)

Given: Triangle $ABC$ with vertices $A(-3,-1)$, $B(2,3)$, and $C(2,-1)$.

The area is calculated using the formula:

$\text{Area}=\frac{1}{2}\left|\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$

Substituting the coordinates:

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

Expanding along the first row using cofactor expansion:

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

Calculating each $2\times 2$ determinant:

• First minor: $(3)(1)-(1)(-1)=3+1=4$
• Second minor: $(2)(1)-(1)(2)=2-2=0$
• Third minor: $(2)(-1)-(3)(2)=-2-6=-8$

Substituting back:

$=\frac{1}{2}[-3(4)+1(0)+1(-8)]$

$=\frac{1}{2}[-12+0-8]$

$=\frac{1}{2}(-20)=-10$

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

$\text{Area}=|-10|=10\text{ sq. units}$

Part (ii)

Given: Triangle $ABC$ with vertices $A(1,3)$, $B(-2,-3)$, and $C(4,-3)$.

Using the determinant formula:

$\text{Area}=\frac{1}{2}\left|\left|\begin{array}{ccc}1& 3& 1\\ -2& -3& 1\\ 4& -3& 1\end{array}\right|\right|$

Expanding along the first row:

$=\frac{1}{2}\left[1\left|\begin{array}{cc}-3& 1\\ -3& 1\end{array}\right|-3\left|\begin{array}{cc}-2& 1\\ 4& 1\end{array}\right|+1\left|\begin{array}{cc}-2& -3\\ 4& -3\end{array}\right|\right]$

Calculating each $2\times 2$ determinant:

• First minor: $(-3)(1)-(1)(-3)=-3+3=0$
• Second minor: $(-2)(1)-(1)(4)=-2-4=-6$
• Third minor: $(-2)(-3)-(-3)(4)=6+12=18$

Substituting back:

$=\frac{1}{2}[1(0)-3(-6)+1(18)]$

$=\frac{1}{2}[0+18+18]$

$=\frac{1}{2}(36)$

$=18\text{ sq. units}$

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Key Formulas or Methods Used

• Area of Triangle (Determinant Method): For vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$: $\text{Area}=\frac{1}{2}\left|\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$
• Cofactor Expansion: Expansion of $3\times 3$ determinant along first row: $a_{11}{C}_{11}+a_{12}{C}_{12}+a_{13}{C}_{13}$ where $C_{ij}=(-1{)}^{i+j}{M}_{ij}$
• Absolute Value Property: Area is always positive, so take $|\text{determinant value}|$ if negative

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Summary of Steps

1. Identify coordinates of the three vertices of the triangle
2. Construct the $3\times 3$ matrix with $x$ coordinates in column 1, $y$ coordinates in column 2, and 1s in column 3
3. Expand the determinant along the first row (or any row/column) using cofactor expansion
4. Calculate the $2\times 2$ minors for each element in the chosen row
5. Sum the terms with appropriate signs ($+-+$ pattern for first row)
6. Multiply by $\frac{1}{2}$ and take the absolute value to get the final area in square units