Core Concepts: Area of a Triangle and Collinearity

Exercise 7.4 primarily builds upon two fundamental ideas: calculating the area of a triangle whose vertices are given as coordinates, and determining if three given points are collinear. These concepts are interconnected and are powerful tools for solving geometrical problems algebraically.

Area of a Triangle Formula

If the vertices of a triangle are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, the area of triangle ABC can be calculated using the formula:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Important Notes:

• The vertical bars $|...|$ denote the absolute value, as area cannot be negative. If your calculation results in a negative value, simply take its positive counterpart.
• The order of the points matters for consistency in the formula, but the absolute value ensures the final area is positive regardless of the order you list the vertices. A good way to remember the pattern is "$x_1(y_2 - y_3)$, $x_2(y_3 - y_1)$, $x_3(y_1 - y_2)

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  quot; – notice the cyclic shift of indices (1-2-3, 2-3-1, 3-1-2) within the parentheses.

Condition for Collinearity

Three points are said to be collinear if they lie on the same straight line. In coordinate geometry, a powerful way to check for collinearity is by using the area of a triangle concept. If three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear, then the 'triangle' formed by them would be a degenerate triangle, essentially a straight line. The area of such a "triangle" is zero.

So, if Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$, then the points A, B, and C are collinear. Conversely, if the area is not zero, the points are not collinear.

Step-by-Step: Calculating the Area of a Triangle

1. Identify Coordinates — Clearly label the coordinates of the three vertices as $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$. The order usually doesn't affect the final positive area due to the absolute value, but consistency helps prevent calculation errors.
2. Write Down the Formula — Recall and write the area formula: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. This helps in systematic substitution and reduces errors.
3. Substitute Values Carefully — Substitute the respective $x$ and $y$ values into the formula. Pay close attention to signs, especially when subtracting negative numbers. It's often helpful to calculate the terms inside the parentheses first.
4. Perform Calculations — Simplify the expression step by step. First, solve the subtractions inside the parentheses. Then, perform the multiplications. Finally, add/subtract the resulting terms within the absolute value.
5. Apply Absolute Value and Final Division — Take the absolute value of the sum obtained. If the sum is negative, make it positive. Then, multiply the result by $\frac{1}{2}$ to get the final area. Remember to include the unit "square units" in your answer.

Worked Examples

• Example 1: Find the area of the triangle whose vertices are (2, 3), (-1, 0), and (2, -4). Step 1: Label the coordinates. Let $(x_1, y_1) = (2, 3)$ $(x_2, y_2) = (-1, 0)$ $(x_3, y_3) = (2, -4)$ Step 2: Apply the area formula. Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Step 3: Substitute values. Area $= \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$ Step 4: Simplify. Area $= \frac{1}{2} |2(0 + 4) - 1(-7) + 2(3)|$ Area $= \frac{1}{2} |2(4) + 7 + 6|$ Area $= \frac{1}{2} |8 + 7 + 6|$ Area $= \frac{1}{2} |21|$ Area $= \frac{21}{2}$ square units. Final Answer: The area of the triangle is 10.5 square units.
• Example 2: Determine if the points A(1, 5), B(2, 3), and C(-2, -11) are collinear. Step 1: Assume the points form a triangle and calculate its area. Let $(x_1, y_1) = (1, 5)$ $(x_2, y_2) = (2, 3)$ $(x_3, y_3) = (-2, -11)$ Step 2: Apply the area formula. Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Step 3: Substitute values. Area $= \frac{1}{2} |1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3)|$ Step 4: Simplify. Area $= \frac{1}{2} |1(3 + 11) + 2(-16) - 2(2)|$ Area $= \frac{1}{2} |1(14) - 32 - 4|$ Area $= \frac{1}{2} |14 - 32 - 4|$ Area $= \frac{1}{2} |-22|$ Area $= \frac{1}{2} (22) = 11$ Step 5: Check for collinearity. Since the Area is $11 \neq 0$, the points A, B, and C are not collinear. Final Answer: The points A(1, 5), B(2, 3), and C(-2, -11) are not collinear.

Exam Tip: Avoiding Common Mistakes

When working with the area of a triangle formula or collinearity problems, students often make a few common errors. Be mindful of these:

1. Sign Errors: This is the most frequent mistake. Always be extra careful when substituting negative coordinates or subtracting negative numbers. Remember that $y_2 - (-y_3)$ becomes $y_2 + y_3$.
2. Forgetting the $\frac{1}{2}$: The formula includes a $\frac{1}{2}$ factor. Don't forget to multiply your final calculated sum by it.
3. Ignoring Absolute Value: Area must always be a positive quantity. If your calculation results in a negative number, always take its absolute value (positive equivalent). For example, if you get $-20$, the area is $20$ square units.
4. Misinterpreting Collinearity: For three points to be collinear, the area of the triangle formed by them must be exactly zero. If you get any non-zero value, no matter how small, the points are not collinear.
5. Incorrect Order of Terms: While the absolute value often corrects for swapping $x_1$ and $x_2$ terms, it's best practice to stick to the cyclic order of indices (e.g., $(y_2 - y_3)$, $(y_3 - y_1)$, $(y_1 - y_2)$) to ensure accuracy and prevent confusion.

Practice Questions with Solutions

• Q: Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3). A: Step 1: Divide the quadrilateral into two triangles. Let the vertices be A(-4, -2), B(-3, -5), C(3, -2), and D(2, 3). We can divide it into triangle ABC and triangle ADC. Step 2: Calculate the area of triangle ABC using the formula: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. For $\triangle ABC$: $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$. Area(ABC) $= \frac{1}{2} |(-4)(-5 - (-2)) + (-3)(-2 - (-2)) + 3(-2 - (-5))|$ $= \frac{1}{2} |(-4)(-3) + (-3)(0) + 3(3)|$ $= \frac{1}{2} |12 + 0 + 9| = \frac{1}{2} |21| = 10.5$ square units. Step 3: Calculate the area of triangle ADC. For $\triangle ADC$: $A(-4, -2)$, $D(2, 3)$, $C(3, -2)$. Area(ADC) $= \frac{1}{2} |(-4)(3 - (-2)) + 2(-2 - (-2)) + 3(-2 - 3)|$ $= \frac{1}{2} |(-4)(5) + 2(0) + 3(-5)|$ $= \frac{1}{2} |-20 + 0 - 15| = \frac{1}{2} |-35| = 17.5$ square units. Step 4: Add the areas of the two triangles to find the area of the quadrilateral. Area(Quadrilateral ABCD) = Area(ABC) + Area(ADC) = $10.5 + 17.5 = 28$ square units. Final answer: The area of the quadrilateral is 28 square units.
• Q: Find the value of 'k' if the points A(7, -2), B(5, 1), and C(3, k) are collinear. A: Step 1: For three points to be collinear, the area of the triangle formed by them must be zero. Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$ Step 2: Substitute the coordinates A(7, -2), B(5, 1), C(3, k) into the formula. $\frac{1}{2} |7(1 - k) + 5(k - (-2)) + 3(-2 - 1)| = 0$ Step 3: Simplify the expression. $|7 - 7k + 5(k + 2) + 3(-3)| = 0$ $|7 - 7k + 5k + 10 - 9| = 0$ $|-2k + 8| = 0$ Step 4: Solve for k. $-2k + 8 = 0$ $-2k = -8$ $k = 4$ Final answer: The value of k is 4.
• Q: If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find its area. A: Step 1: Divide the quadrilateral ABCD into two triangles, for instance, $\triangle ABC$ and $\triangle ACD$. Step 2: Calculate the area of $\triangle ABC$. Let $A(-5, 7), B(-4, -5), C(-1, -6)$. Area(ABC) $= \frac{1}{2} |(-5)(-5 - (-6)) + (-4)(-6 - 7) + (-1)(7 - (-5))|$ $= \frac{1}{2} |(-5)(1) + (-4)(-13) + (-1)(12)|$ $= \frac{1}{2} |-5 + 52 - 12| = \frac{1}{2} |35| = 17.5$ square units. Step 3: Calculate the area of $\triangle ACD$. Let $A(-5, 7), C(-1, -6), D(4, 5)$. Area(ACD) $= \frac{1}{2} |(-5)(-6 - 5) + (-1)(5 - 7) + 4(7 - (-6))|$ $= \frac{1}{2} |(-5)(-11) + (-1)(-2) + 4(13)|$ $= \frac{1}{2} |55 + 2 + 52| = \frac{1}{2} |109| = 54.5$ square units. Step 4: Add the areas to find the area of the quadrilateral. Area(ABCD) = Area(ABC) + Area(ACD) = $17.5 + 54.5 = 72$ square units. Final answer: The area of the quadrilateral is 72 square units.
• Q: Show that the points (a, b+c), (b, c+a), and (c, a+b) are collinear. A: Step 1: For the points to be collinear, the area of the triangle formed by them must be zero. Let $(x_1, y_1) = (a, b+c)$ $(x_2, y_2) = (b, c+a)$ $(x_3, y_3) = (c, a+b)$ Step 2: Substitute these into the area formula and simplify. Area $= \frac{1}{2} |a((c+a) - (a+b)) + b((a+b) - (b+c)) + c((b+c) - (c+a))|$ Area $= \frac{1}{2} |a(c+a-a-b) + b(a+b-b-c) + c(b+c-c-a)|$ Area $= \frac{1}{2} |a(c-b) + b(a-c) + c(b-a)|$ Area $= \frac{1}{2} |ac - ab + ab - bc + bc - ac|$ Area $= \frac{1}{2} |0| = 0$ Step 3: Conclude based on the area. Since the area of the triangle formed by these points is 0, the points are collinear. Final answer: The points (a, b+c), (b, c+a), and (c, a+b) are collinear.

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