Determinants Exercise 4.3 Class 12 NCERT Solutions

Finding the Area of a Triangle Using Determinants

Traditionally, you might have calculated the area of a triangle using the base-height formula or Heron's formula. However, when the coordinates of the three vertices are given, determinants provide an elegant and efficient method. If the vertices of a triangle are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$, its area can be found using the following determinant formula:

$\text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$

Key Points to Remember:

1. Absolute Value: Since area is a measure of space, it must always be non-negative. Therefore, you must take the absolute value of the determinant's result. If the calculation yields a negative value, discard the negative sign. For example, if the determinant evaluates to $-10$, the area is $10$ square units.
2. Units: Always express the area in "square units" unless specified otherwise.
3. Determinant Expansion: You can expand the determinant along any row or column, typically the first row, but choose the one that simplifies calculations, perhaps by picking a row/column with zeros.

This method is particularly useful because it integrates concepts from algebra (determinants) into geometry, showcasing the interconnectedness of different mathematical branches. It's a standard and often tested concept in your CBSE exams.

Condition for Collinearity of Three Points

The concept of finding the area of a triangle using determinants extends naturally to determining if three points are collinear. What does it mean for three points to be collinear? It simply means they all lie on the same straight line.

Now, imagine three points forming a triangle. If these three points were to move and eventually fall onto a single straight line, what would happen to the triangle they form? The 'triangle' would essentially flatten out, and its area would become zero.

Therefore, the condition for three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ to be collinear is that the area of the triangle formed by them must be zero. Mathematically, this means:

$\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$

Which simplifies to:

$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$

This is a very powerful criterion. If you are given three points and asked to check if they are collinear, simply calculate the determinant. If the determinant's value is zero, the points are collinear. If it's non-zero, they form a triangle. This concept is frequently tested, sometimes asking for an unknown coordinate that makes three points collinear.

Worked Examples

• Example 1: Find the area of the triangle whose vertices are (1, 0), (6, 0), and (4, 3). Step 1: Identify the coordinates of the vertices. Let $(x_1, y_1) = (1, 0)$, $(x_2, y_2) = (6, 0)$, and $(x_3, y_3) = (4, 3)$. Step 2: Set up the determinant for the area formula. Area $= \frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$ Step 3: Expand the determinant. We can expand along the second column (C2) as it has two zeros, simplifying the calculation. Area $= \frac{1}{2} [ -0(6-4) + 0(1-4) - 3(1-6) ]$ Area $= \frac{1}{2} [ 0 + 0 - 3(-5) ]$ Area $= \frac{1}{2} [ 15 ]$ Area $= \frac{15}{2}$ Step 4: Take the absolute value and state the units. Since the result is positive, no change is needed. Area $= \frac{15}{2}$ square units. Final Answer: The area of the triangle is $\frac{15}{2}$ square units.
• Example 2: Find the value of 'k' if the area of the triangle with vertices (k, 0), (4, 0), and (0, 2) is 4 square units. Step 1: Identify the coordinates and the given area. $(x_1, y_1) = (k, 0)$, $(x_2, y_2) = (4, 0)$, $(x_3, y_3) = (0, 2)$. Area $= 4$. Step 2: Set up the determinant equation. Remember that the calculated determinant can be positive or negative, so we consider both cases when the area is given. $\frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4$ Step 3: Expand the determinant. Expanding along the second column (C2): $k(0-2) - 0(4-0) + 1(8-0) = \pm 8$ (multiplying by 2 on both sides) $k(-2) - 0 + 8 = \pm 8$ $-2k + 8 = \pm 8$ Step 4: Solve for 'k' considering both positive and negative cases. Case 1: $-2k + 8 = 8$ $-2k = 0$ $k = 0$ Case 2: $-2k + 8 = -8$ $-2k = -16$ $k = 8$ Final Answer: The possible values for 'k' are 0 and 8.

Exam Tips and Common Mistakes

To score well in questions related to Determinants Ex 4.3, keep these crucial tips in mind and avoid common pitfalls:

1. Don't Forget the Half (1/2): The formula for the area of a triangle includes a factor of $1/2$. A common mistake is to forget this, leading to double the correct area.
2. Absolute Value is a Must: Area cannot be negative. After evaluating the determinant, always take its absolute value. If you get -12, the area is 12 square units. When the area is given (like in Example 2), remember to use $\pm$ on the area value when setting up the equation, as the determinant's value could have been negative before applying the absolute sign.
3. Collinearity implies Area = 0: If three points are collinear, the area of the triangle formed by them is zero. Use this fact directly by setting the determinant (without the $1/2$) equal to zero to find unknown variables or to prove collinearity.
4. Careful with Sign Conventions: When expanding the determinant, especially using cofactor expansion, pay close attention to the signs ($+, -, +$) for the elements in the chosen row or column. A single sign error can lead to a completely wrong answer.
5. Choose Expansion Wisely: If a row or column contains zeros, expanding along that row/column will simplify calculations significantly, reducing the chances of errors.

Practice Questions with Solutions

• Q: Find the area of the triangle with vertices (2, 7), (1, 1), and (10, 8). A: Step 1: Identify 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: Area $= \frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}$ Step 3: Expand the determinant (along R1): Area $= \frac{1}{2} [ 2(1-8) - 7(1-10) + 1(8-10) ]$ Area $= \frac{1}{2} [ 2(-7) - 7(-9) + 1(-2) ]$ Area $= \frac{1}{2} [ -14 + 63 - 2 ]$ Area $= \frac{1}{2} [ 47 ] = \frac{47}{2}$ Step 4: Take absolute value and units. Area $= \frac{47}{2}$ square units. Final answer: The area of the triangle is $\frac{47}{2}$ square units.
• Q: Show that the points A(a, b+c), B(b, c+a), C(c, a+b) are collinear. A: Step 1: Set up the determinant for the area: Area $= \frac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$ Step 2: Apply column operations to simplify the determinant. Apply $C_2 \to C_2 + C_1$. Area $= \frac{1}{2} \begin{vmatrix} a & a+b+c & 1 \\ b & b+c+a & 1 \\ c & c+a+b & 1 \end{vmatrix}$ Step 3: Take out the common factor $(a+b+c)$ from C2. Area $= \frac{1}{2} (a+b+c) \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix}$ Step 4: Observe that C2 and C3 are identical. If two columns (or rows) of a determinant are identical, its value is 0. Area $= \frac{1}{2} (a+b+c) (0) = 0$ Step 5: Conclude based on the area. Since the area of the triangle formed by the points is 0, the points A, B, and C are collinear. Final answer: The points A(a, b+c), B(b, c+a), C(c, a+b) are collinear as the area of the triangle formed by them is 0.
• Q: If the points (2, -3), (k, -1), and (0, 4) are collinear, find the value of k. A: Step 1: For collinear points, the area of the triangle formed by them must be 0. Set up the determinant: $\begin{vmatrix} 2 & -3 & 1 \\ k & -1 & 1 \\ 0 & 4 & 1 \end{vmatrix} = 0$ Step 2: Expand the determinant (along R1). $2(-1-4) - (-3)(k-0) + 1(4k-0) = 0$ $2(-5) + 3k + 4k = 0$ $-10 + 7k = 0$ Step 3: Solve for k. $7k = 10$ $k = \frac{10}{7}$ Final answer: The value of k is $\frac{10}{7}$.
• Q: Find the equation of the line joining A(1, 2) and B(3, 6) using determinants. A: Step 1: Let P(x, y) be any point on the line joining A(1, 2) and B(3, 6). Since A, B, and P are collinear, the area of the triangle formed by them must be 0. Step 2: Set up the determinant for collinear points: $\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$ Step 3: Expand the determinant (along R1). $x(2-6) - y(1-3) + 1(6-6) = 0$ $x(-4) - y(-2) + 1(0) = 0$ $-4x + 2y = 0$ Step 4: Simplify the equation. Divide by 2: $-2x + y = 0$ or $y = 2x$ Final answer: The equation of the line joining A(1, 2) and B(3, 6) is $y = 2x$.

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