Ch 4  ·  Q–
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Chapter 4 Exercise 4.2 Solutions

Determinants

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 4 Exercise 4.2

Class 12 Mathematics Exercise 4.2 NCERT Solutions Determinants Class 12 Mathematics Chapter 4 CBSE Board Exam JEE Main CUET Area of Triangle Collinearity Equation of a Line Properties of Determinants Minors and Cofactors Expansion of Determinants Coordinate Geometry
5 Questions
10–15 min Ideal time
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Q1
NUMERIC3 marks
Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)
📘 Concept & Theory
Concept/Theory
›

The determinant method provides a direct and systematic way to find the area of a triangle when the coordinates of its three vertices are known. If the vertices of a triangle are \(A(x_1,y_1)\), \(B(x_2,y_2)\), and \(C(x_3,y_3)\), then its area is

\[ \Delta= \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} \]

The absolute value is essential because area is always a non-negative quantity, whereas the determinant itself may be positive or negative depending upon the order in which the vertices are written.

For a \(3\times3\) determinant

\[ \begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i \end{vmatrix} = a \begin{vmatrix} e & f\\ h & i \end{vmatrix} - b \begin{vmatrix} d & f\\ g & i \end{vmatrix} + c \begin{vmatrix} d & e\\ g & h \end{vmatrix} \]

Thus, the determinant can be evaluated by expanding along any convenient row or column. Choosing a row or column containing zeros can considerably simplify the calculation.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the three given vertices.

  2. Substitute their coordinates into the determinant formula for the area of a triangle.

  3. Choose a convenient row or column for expansion.

  4. Evaluate each \(2\times2\) minor carefully, preserving the correct signs.

  5. Multiply the determinant by \(\frac{1}{2}\).

  6. Take the absolute value whenever the determinant is negative.

✏️ Solution
(i) (1, 0), (6, 0), (4, 3)
›
Step-by-step Solution  ·  6 steps
  1. Given — \(A(1,0),\quad B(6,0),\quad C(4,3)\)
  2. The area of a triangle whose vertices are \(A(x_1,y_1)\), \(B(x_2,y_2)\), and \(C(x_3,y_3)\) is
    \[\Delta=\dfrac12\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix}\]
  3. Substituting the given coordinates,
    \[ \Delta= \frac{1}{2} \begin{vmatrix} 1 & 0 & 1\\ 6 & 0 & 1\\ 4 & 3 & 1 \end{vmatrix} \]
  4. Since the second column contains two zeros, expanding along the second column is the most convenient approach.
    \[\begin{aligned}\begin{vmatrix}1 & 0 & 1\\6 & 0 & 1\\4 & 3 & 1\end{vmatrix}&=-0\begin{vmatrix}6 & 1\\4 & 1\end{vmatrix}+ 0\begin{vmatrix}1 & 1\\4 & 1\end{vmatrix} -3\begin{vmatrix}1 & 1\\6 & 1\end{vmatrix}\end{aligned}\]
  5. The first two terms are zero because their coefficients are zero. Therefore,
    \[\begin{aligned}\begin{vmatrix}1 & 0 & 1\\6 & 0 & 1\\4 & 3 & 1\end{vmatrix}&=-3\begin{vmatrix}1 & 1\\6 & 1\end{vmatrix}\\ &=-3(1\times1-1\times6)\\&=-3(1-6)\\&=-3(-5)\\&=15\end{aligned}\]
  6. Hence,
    \[\begin{aligned}\Delta&=\frac{1}{2}|15|\\&=\frac{15}{2}\\&=7.5\end{aligned}\]
  7. Therefore, the area of the triangle is
    \[\boxed{7.5\text{ square units}}\]
✏️ Solution
(ii) (2, 7), (1, 1), (10, 8)
›
Step-by-step Solution  ·  8 steps
  1. Given —
    \[A(2,7),\quad B(1,1),\quad C(10,8)\]
  2. Using the determinant formula for the area of a triangle,
    \[\Delta=\frac{1}{2}\begin{vmatrix}2 & 7 & 1\\1 & 1 & 1\\10 & 8 & 1\end{vmatrix}\]
  3. We expand along the first row. The signs of the cofactors in the first row are \(+,-,+\).
    \[\begin{aligned}\begin{vmatrix}2 & 7 & 1\\1 & 1 & 1\\10 & 8 & 1\end{vmatrix} &=2\begin{vmatrix}1 & 1\\8 & 1\end{vmatrix}-7\begin{vmatrix}1 & 1\\10 & 1\end{vmatrix}+ 1\begin{vmatrix}1 & 1\\10 & 8\end{vmatrix}\end{aligned}\]
  4. Now evaluate each \(2\times2\) determinant separately:
    \[\begin{vmatrix}1 & 1\\8 & 1\end{vmatrix}=1\times1-1\times8=1-8=-7\]
  5. \[\begin{vmatrix}1 & 1\\10 & 1\end{vmatrix}=1\times1-1\times10=1-10=-9\]
  6. \[\begin{vmatrix}1 & 1\\10 & 8\end{vmatrix}=1\times8-1\times10=8-10=-2\]
  7. Substituting these values,
    \[\begin{aligned}\begin{vmatrix}2 & 7 & 1\\1 & 1 & 1\\10 & 8 & 1\end{vmatrix}&=2(-7)-7(-9)+1(-2)\\&=-14+63-2\\&=47\end{aligned}\]
  8. Therefore,
    \[\begin{aligned}\Delta&=\frac{1}{2}|47|\\&=\frac{47}{2}\end{aligned}\]
  9. Therefore, the area of the triangle is
    \[\boxed{\frac{47}{2}\text{ square units}}\]
✏️ Solution
(iii) (–2, –3), (3, 2), (–1, –8)
›
Step-by-step Solution  ·  9 steps
  1. Given — \(A(-2,-3),\quad B(3,2),\quad C(-1,-8)\)
  2. Using the determinant formula,
    \[\Delta=\frac{1}{2}\begin{vmatrix}-2 & -3 & 1\\3 & 2 & 1\\-1 & -8 & 1\end{vmatrix}\]
  3. We expand along the first row. The cofactor signs are \(+,-,+\).
    Therefore,
    \[\begin{aligned}\begin{vmatrix}-2 & -3 & 1\\3 & 2 & 1\\-1 & -8 & 1\end{vmatrix} &=(-2)\begin{vmatrix}2 & 1\\-8 & 1\end{vmatrix}-(-3)\begin{vmatrix}3 & 1\\-1 & 1\end{vmatrix}+1\begin{vmatrix}3 & 2\\-1 & -8\end{vmatrix}\end{aligned}\]
  4. Notice that the second term becomes positive because the corresponding entry is \(-3\) and the cofactor sign is negative:
    \[-(-3)=+3\]
  5. Now evaluate the first \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}2 & 1\\-8 & 1\end{vmatrix}&=2\times1-1\times(-8)\\&=2+8\\&=10\end{aligned}\]
  6. Evaluate the second \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}3 & 1\\-1 & 1\end{vmatrix}&=3\times1-1\times(-1)\\&=3+1\\&=4\end{aligned}\]
  7. Evaluate the third \(2\times2\) determinant:
    \[\begin{aligned}\begin{vmatrix}3 & 2\\-1 & -8\end{vmatrix}&=3\times(-8)-2\times(-1)\\&=-24+2\\&=-22\end{aligned}\]
  8. Substituting these values,
    \[\begin{aligned}\begin{vmatrix}-2 & -3 & 1\\3 & 2 & 1\\-1 & -8 & 1\end{vmatrix}&=(-2)(10)+3(4)+1(-22)\\&=-20+12-22\\&=-30\end{aligned}\]
  9. The determinant is negative, but area cannot be negative. Hence, we take its absolute value.
    \[\begin{aligned}\Delta&=\frac{1}{2}|-30|\\&=\frac{1}{2}(30)\\&=15\end{aligned}\]
  10. Therefore, the area of the triangle is
    \[\boxed{15\text{ square units}}\]
💡 Answer
Final Answer
›
\[\boxed{\text{(i) }7.5\text{ square units}}\]
\[\boxed{\text{(ii) }\frac{47}{2}\text{ square units}}\]
\[\boxed{\text{(iii) }15\text{ square units}}\]
🎯 Exam Significance
Exam Significance
›
This problem is important because it tests several fundamental skills from the chapter on Determinants: formation of a determinant from coordinate data, expansion of a determinant, evaluation of minors, sign management, and interpretation of the final result. For CBSE Board examinations, a clear step-by-step presentation is particularly useful because each stage of determinant evaluation can carry marks. For competitive entrance examinations, the same concept frequently appears in shorter forms, especially in questions involving collinearity, coordinate geometry, determinants, and comparison of areas.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. For three coordinate points, the determinant formula gives the area of the triangle directly.

  2. The standard formula is

    \[\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}\]

  3. The absolute value must be used because geometrical area is always non-negative.

  4. The determinant may be positive or negative depending on the order of the vertices.

  5. When expanding a determinant, carefully follow the cofactor sign pattern:

    \[\begin{matrix}+ & - & +\\- & + & -\\+ & - & +\end{matrix}\]

  6. Choose the row or column that makes the expansion easiest, particularly one containing zeros.

  7. Always evaluate the \(2\times2\) minors explicitly to avoid sign and arithmetic errors.

  8. The determinant method for area is closely connected with coordinate geometry and is also useful for testing the collinearity of three points.

  9. For three collinear points, the determinant is zero, and consequently the area of the corresponding triangle is zero.

  10. For board examinations, showing the determinant setup, expansion, minors, arithmetic, and final absolute-value step makes the solution complete and marks-friendly.
↑ Top
1 / 5  ·  20%
Q2 →
Q2
NUMERIC3 marks
Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear
📘 Concept & Theory
Concept/Theory
›

The determinant method provides a powerful criterion for testing whether three points in a coordinate plane are collinear. Three points are said to be collinear if they lie on the same straight line.

If the three points are

\[ A(x_1,y_1),\qquad B(x_2,y_2),\qquad C(x_3,y_3), \]
then the area of the triangle formed by them is

\[ \Delta= \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} \]

If the three points are collinear, they cannot form a triangle with positive area. Therefore,

\[ \Delta=0 \]

Since \(\frac{1}{2}\) and the absolute value do not affect whether the determinant is zero, the condition for collinearity is

\[ \boxed{ \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix}=0 } \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the coordinates of the three given points carefully.

  2. Form the \(3\times3\) determinant using the coordinates and a final column of \(1\)'s.

  3. Since collinearity is to be established, show that this determinant is zero.

  4. Expand the determinant along a convenient row or column.

  5. Evaluate every \(2\times2\) determinant explicitly.

  6. Simplify the resulting algebraic expression carefully.

  7. If the determinant evaluates to zero, conclude that the three points are collinear.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Given — Points are \(A(a,b+c),\qquad B(b,c+a),\qquad C(c,a+b)\)
  2. To show that \(A\), \(B\), and \(C\) are collinear, we use the determinant criterion for collinearity.
    For three points to be collinear, the area of the triangle formed by them must be zero. Therefore,
    \[\Delta=0\]
  3. The area determinant is
    \[\Delta=\frac{1}{2}\begin{vmatrix}a & b+c & 1\\b & c+a & 1\\c & a+b & 1\end{vmatrix}\]
  4. It is sufficient to prove that the determinant itself is zero. Thus, consider
    \[D=\begin{vmatrix}a & b+c & 1\\b & c+a & 1\\c & a+b & 1\end{vmatrix}\]
  5. Step 1: Expand the determinant along the first row
  6. The cofactor signs in the first row are \(+,-,+\). Hence,
    \[\begin{aligned}D&=a\begin{vmatrix}c+a & 1\\a+b & 1\end{vmatrix}-(b+c) \begin{vmatrix}b & 1\\c & 1\end{vmatrix}+1\begin{vmatrix}b & c+a\\c & a+b\end{vmatrix}\end{aligned}\]
  7. Step 2: Evaluate the first \(2\times2\) determinant
  8. \[\begin{aligned}\begin{vmatrix}c+a & 1\\a+b & 1\end{vmatrix}&=(c+a)(1)-1(a+b)\\&=c+a-a-b\\&=c-b\end{aligned}\]
  9. Step 3: Evaluate the second \(2\times2\) determinant
  10. \[\begin{aligned}\begin{vmatrix}b & 1\\c & 1\end{vmatrix}&=b(1)-1(c)\\&=b-c\end{aligned}\]
  11. Step 4: Evaluate the third \(2\times2\) determinant
  12. \[\begin{aligned}\begin{vmatrix}b & c+a\\c & a+b\end{vmatrix}&=b(a+b)-c(c+a)\\&=ab+b^2-c^2-ac\end{aligned}\]
  13. Step 5: Substitute the three results
  14. Therefore,
    \[\begin{aligned}D&=a(c-b)-(b+c)(b-c)+ab+b^2-c^2-ac\end{aligned}\]
  15. Step 6: Expand the remaining algebraic terms
  16. First,
    \[a(c-b)=ac-ab\]
  17. Also,
    \[(b+c)(b-c)=b^2-c^2\]
  18. Hence,
    \[-(b+c)(b-c)=-(b^2-c^2)=-b^2+c^2\]
  19. Substituting these expressions into \(D\),
    \[\begin{aligned}D&=ac-ab-b^2+c^2+ab+b^2-c^2-ac\end{aligned}\]
  20. Step 7: Cancel like terms
  21. Grouping the terms,
    \[\begin{aligned}D&=(ac-ac)+(-ab+ab)+(-b^2+b^2)+(c^2-c^2)\\&=0+0+0+0\\&=0\end{aligned}\]
  22. Therefore,
    \[\boxed{\begin{vmatrix}a & b+c & 1\\b & c+a & 1\\c & a+b & 1\end{vmatrix}=0}\]
  23. Hence the area of the triangle formed by \(A\), \(B\), and \(C\) is
    \[\Delta=\frac{1}{2}|D|=\frac{1}{2}|0|=0\]
    Since the area is zero, the three points lie on the same straight line.
  24. \[\boxed{\text{Therefore, the points }A,\ B,\text{ and }C\text{ are collinear.}}\]
🎯 Exam Significance
Exam Significance
›

This is an important application of determinants in coordinate geometry. In CBSE Board examinations, such questions test whether a student can translate a geometric condition into a determinant condition and then simplify an algebraic determinant systematically. The determinant criterion for collinearity is also useful in competitive entrance examinations, where it can provide a faster alternative to finding slopes or equations of lines.

A particularly important observation is that the condition

\[ \Delta=0 \]
simultaneously represents two equivalent facts: the determinant is zero and the area of the triangle formed by the three points is zero. Hence, determinant methods connect algebraic computation directly with geometric interpretation.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Three points are collinear if and only if the area of the triangle formed by them is zero.

  2. The determinant criterion for collinearity is

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

  3. For symbolic coordinates, careful algebraic simplification is as important as correct determinant expansion.

  4. When expanding a determinant, always use the cofactor sign pattern \(+,-,+\) in the first row.

  5. In this problem, the terms cancel in pairs:

    \[ac-ac=0,\qquad -ab+ab=0,\qquad -b^2+b^2=0,\qquad c^2-c^2=0.\]

  6. A zero determinant means that the three coordinate points do not enclose any non-zero area and therefore are collinear.

  7. The determinant approach is often more efficient than calculating slopes, particularly when the coordinates contain algebraic parameters.

  8. For board examinations, explicitly showing the formation of the determinant, expansion, simplification, and final conclusion provides a complete proof.
← Q1
2 / 5  ·  40%
Q3 →
Q3
NUMERIC3 marks
Find values of k if area of triangle is 4 sq. units and vertices are
(i) (k, 0), (4, 0), (0, 2)
(ii) (–2, 0), (0, 4), (0, k)
📘 Concept & Theory
Concept/Theory
›

The determinant formula for the area of a triangle is particularly useful when one or more coordinates contain an unknown parameter such as \(k\). In such problems, the given area is used to form an equation in \(k\), which is then solved algebraically.

If the vertices of a triangle are

\[ A(x_1,y_1),\qquad B(x_2,y_2),\qquad C(x_3,y_3), \]
then its area is

\[\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}\]

The absolute value cannot be omitted because the determinant can be positive or negative according to the order of the vertices, whereas the geometrical area is always non-negative.

In this question, the area is given as \(4\) square units. Therefore, after forming the determinant, we must impose the condition

\[ \frac{1}{2}|D|=4 \]

where \(D\) denotes the determinant. Hence,

\[ |D|=8 \]

This absolute-value equation generally produces two possible cases:

\[ D=8 \qquad\text{or}\qquad D=-8 \]

This is an important point in the present problem. The original working incorrectly removes the absolute value at an intermediate stage. The correct procedure is to retain the absolute value until the two possible cases have been considered.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given vertices clearly.

  2. Use the determinant formula for the area of a triangle.

  3. Substitute the given area \(4\) square units.

  4. Multiply by \(2\) to obtain an equation involving the absolute value of the determinant.

  5. Expand the determinant using a convenient row or column.

  6. Simplify the determinant completely in terms of \(k\).

  7. Solve the resulting absolute-value equation by considering both cases.

  8. Verify that each obtained value of \(k\) indeed gives an area of \(4\) square units.

✏️ Solution
(i) Vertices \((k,0)\), \((4,0)\), and \((0,2)\)
›
Step-by-step Solution  ·  16 steps
  1. Given — \(A(k,0),\quad B(4,0),\quad C(0,2)\)
    The area of the triangle is given as: \(\Delta=4\text{ square units}\)
  2. Using the determinant formula for the area of a triangle,
    \[\Delta=\frac{1}{2}\begin{vmatrix}k & 0 & 1\\4 & 0 & 1\\0 & 2 & 1\end{vmatrix}\]
  3. Since the area is \(4\),
    \[4=\frac{1}{2}\begin{vmatrix}k & 0 & 1\\4 & 0 & 1\\0 & 2 & 1\end{vmatrix}\]
  4. Multiplying both sides by \(2\),
    \[8=\begin{vmatrix}k & 0 & 1\\4 & 0 & 1\\0 & 2 & 1\end{vmatrix}\]
  5. Step 1: Expand the determinant along the second column
  6. The second column contains two zeros, so expansion along this column is convenient.
  7. The cofactor signs in the second column are \(-,+,-\). Therefore,
    \[\begin{aligned}D&=-0\begin{vmatrix}4 & 1\\0 & 1\end{vmatrix}+0\begin{vmatrix}k & 1\\0 & 1\end{vmatrix}-2\begin{vmatrix}k & 1\\4 & 1\end{vmatrix}\end{aligned}\]
  8. The first two terms are zero. Hence,
    \[D=-2\begin{vmatrix}k & 1\\4 & 1\end{vmatrix}\]
  9. Step 2: Evaluate the \(2\times2\) determinant
  10. \[\begin{aligned}D&=-2(k\times1-1\times4)\\&=-2(k-4)\end{aligned}\]
  11. Therefore,
    \[D=-2(k-4)\]
  12. The area equation was
    \[|D|=8\]
  13. Therefore,
    \[|-2(k-4)|=8\]
  14. Since \(|-2|=2\),
    \[2|k-4|=8\]
  15. Dividing both sides by \(2\),
    \[|k-4|=4\]
  16. Step 3: Consider both cases
  17. For an equation of the form \(|x|=a\), we have
    \[x=a\quad\text{or}\qquad x=-a\]
  18. Therefore,
    \[k-4=4\]
  19. Hence,
    \[k=8\]
  20. Alternatively,
    \[k-4=-4\]
  21. Hence,
    \[k=0\]
✏️ Solution
(ii) Vertices \((-2,0)\), \((0,4)\), and \((0,k)\)
›
Step-by-step Solution  ·  15 steps
  1. Given — \(A(-2,0),\quad B(0,4),\quad C(0,k)\)
    The area of the triangle is \(\Delta=4\text{ square units}\)
  2. Using the determinant formula,
    \[4=\frac{1}{2}\begin{vmatrix}-2 & 0 & 1\\0 & 4 & 1\\0 & k & 1\end{vmatrix}\]
  3. Multiplying both sides by \(2\),
    \[8=\begin{vmatrix}-2 & 0 & 1\\0 & 4 & 1\\0 & k & 1\end{vmatrix}\]
  4. Step 1: Expand the determinant along the first column
  5. The first column contains two zeros, so expansion along the first column is convenient.
  6. The cofactor signs in the first column are \(+,-,+\). Therefore,
    \[\begin{aligned}D&=(-2)\begin{vmatrix}4 & 1\\k & 1\end{vmatrix}-0\begin{vmatrix}0 & 1\\k & 1\end{vmatrix}+0\begin{vmatrix}0 & 1\\4 & 1\end{vmatrix}\end{aligned}\]
  7. The last two terms are zero. Hence,
    \[D=-2\begin{vmatrix}4 & 1\\k & 1\end{vmatrix}\]
  8. Step 2: Evaluate the \(2\times2\) determinant
  9. \[\begin{aligned}D&=-2(4\times1-1\times k)\\&=-2(4-k)\end{aligned}\]
  10. Therefore,
    \[D=-2(4-k)\]
  11. Since the area equation gives
    \[|D|=8\]
  12. we obtain
    \[|-2(4-k)|=8\]
  13. Therefore,
    \[2|4-k|=8\]
  14. Dividing by \(2\),
    \[|4-k|=4\]
  15. Step 3: Consider both cases
  16. First case:
    \[4-k=4\]
  17. Therefore,
    \[k=0\]
  18. Second case:
    \[4-k=-4\]
  19. Therefore,
    \[-k=-8\]
  20. Hence,
    \[k=8\]
💡 Answer
Final Answer
›
\[\boxed{\text{(i) }k=0,\ 8}\]

\[\boxed{\text{(ii) }k=0,\ 8}\]
🎯 Exam Significance
Exam Significance
›
This question is important for CBSE Board examinations because it combines the coordinate-geometry interpretation of determinants with algebraic manipulation. A complete solution should show the determinant formation, expansion, simplification, handling of the absolute value, and both possible values of the parameter.
For competitive entrance examinations, the problem reinforces an important pattern: whenever a geometrical quantity is expressed through a determinant and the determinant contains a parameter, the resulting equation may require an absolute-value analysis. Recognising this quickly prevents the loss of a valid solution.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. When the area of a triangle is known, the determinant formula converts the geometrical condition into an algebraic equation.

  2. Always use the absolute value in the area formula:

    \[ \Delta= \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} \]

  3. If the given area is \(4\), then the absolute value of the determinant must be \(8\), not simply \(8\) without considering its sign.

  4. An equation such as

    \[ |k-4|=4 \]
    has two solutions:
    \[ k-4=4 \quad\text{or}\quad k-4=-4. \]

  5. In both parts, the determinant reduces to an expression involving \(k-4\), leading to the same two possible values of \(k\).

  6. Do not discard the absolute value merely because the determinant happens to have a negative factor.

  7. Verification is useful whenever an absolute-value equation is involved, particularly in a Board examination solution where it confirms that both values satisfy the original geometrical condition.

  8. The final answers for both parts are

    \[ \boxed{k=0,\ 8}. \]
← Q2
3 / 5  ·  60%
Q4 →
Q4
NUMERIC3 marks
(i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
📘 Concept & Theory
Concept/Theory
›

A straight line is uniquely determined by any two distinct points lying on it. The determinant method obtains the equation of this line by introducing a general point \(C(x,y)\) on the line and using the condition that the three points must be collinear.

If

\[A(x_1,y_1),\qquad B(x_2,y_2),\qquad C(x,y)\]
are collinear, then the area of the triangle \(ABC\) is zero. Therefore,

\[\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x & y & 1\end{vmatrix}=0\]

This determinant equation contains the variable coordinates \(x\) and \(y\). On expanding and simplifying it, we obtain the required equation of the straight line.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Take the two given points as fixed points on the required line.

  2. Let \(C(x,y)\) be an arbitrary point on the required line.

  3. Since \(A\), \(B\), and \(C\) are collinear, the area of triangle \(ABC\) is zero.

  4. Form the \(3\times3\) determinant using the coordinates of the three points.

  5. Expand the determinant carefully using the cofactor signs \(+,-,+\).

  6. Simplify the resulting equation in \(x\) and \(y\).

  7. The resulting relation between \(x\) and \(y\) is the equation of the required line.

✏️ Solution
(i) Find equation of line joining (1, 2) and (3, 6) using determinants.
›
Step-by-step Solution  ·  14 steps
  1. Given — Points: \(A(1,2),\quad B(3,6)\)
  2. Let \(C(x,y)\) be any arbitrary point on the line joining \(A\) and \(B\).
  3. Since \(A\), \(B\), and \(C\) are collinear, the area of triangle \(ABC\) is zero.
  4. Therefore,
    \[\frac{1}{2}\begin{vmatrix}1 & 2 & 1\\3 & 6 & 1\\x & y & 1\end{vmatrix}=0\]
  5. Multiplying by \(2\), we get
    \[\begin{vmatrix}1 & 2 & 1\\3 & 6 & 1\\x & y & 1\end{vmatrix}=0\]
  6. Expand the determinant along the first row
  7. The cofactor signs in the first row are \(+,-,+\). Hence,
    \[\begin{aligned}0&=1\begin{vmatrix}6 & 1\\y & 1\end{vmatrix}-2\begin{vmatrix}3 & 1\\x & 1\end{vmatrix}+1\begin{vmatrix}3 & 6\\x & y\end{vmatrix}\end{aligned}\]
  8. Evaluate each \(2\times2\) determinant
  9. For the first minor,
    \[\begin{aligned}\begin{vmatrix}6 & 1\\y & 1\end{vmatrix}&=6(1)-1(y)\\&=6-y\end{aligned}\]
  10. For the second minor,
    \[\begin{aligned}\begin{vmatrix}3 & 1\\x & 1\end{vmatrix}&=3(1)-1(x)\\&=3-x\end{aligned}\]
  11. For the third minor,
    \[\begin{aligned}\begin{vmatrix}3 & 6\\x & y\end{vmatrix}&=3y-6x\end{aligned}\]
  12. Substitute the minors
  13. \[\begin{aligned}0&=(6-y)-2(3-x)+(3y-6x)\end{aligned}\]
  14. Expand the brackets
    \[\begin{aligned}0&=6-y-6+2x+3y-6x\end{aligned}\]
  15. Collect like terms
    \[\begin{aligned}0&=(6-6)+(2x-6x)+(-y+3y)\\&=0-4x+2y\end{aligned}\]
  16. Therefore,
    \[-4x+2y=0\]
  17. Dividing by \(2\),
    \[-2x+y=0\]
  18. Hence,
    \[y-2x=0\]
✏️ Solution
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
›
Step-by-step Solution  ·  14 steps
  1. Given — Points: \(A(3,1),\qquad B(9,3)\)
  2. Let \(C(x,y)\) be an arbitrary point on the required line.
  3. Since \(A\), \(B\), and \(C\) are collinear, the area of triangle \(ABC\) is zero.
  4. Therefore,
    \[\frac{1}{2}\begin{vmatrix}3 & 1 & 1\\9 & 3 & 1\\x & y & 1\end{vmatrix}=0\]
  5. Multiplying by \(2\),
    \[\begin{vmatrix}3 & 1 & 1\\9 & 3 & 1\\x & y & 1\end{vmatrix}=0\]
  6. Expand the determinant along the first row
  7. Using the cofactor signs \(+,-,+\),
    \[\begin{aligned}0&=3\begin{vmatrix}3 & 1\\y & 1\end{vmatrix}-1\begin{vmatrix}9 & 1\\x & 1\end{vmatrix}+1\begin{vmatrix}9 & 3\\x & y\end{vmatrix}\end{aligned}\]
  9. Evaluate each \(2\times2\) determinant
  10. For the first minor,
    \[\begin{aligned}\begin{vmatrix}3 & 1\\y & 1\end{vmatrix}&=3(1)-1(y)\\&=3-y\end{aligned}\]
  11. For the second minor,
    \[\begin{aligned}\begin{vmatrix}9 & 1\\x & 1\end{vmatrix}&=9(1)-1(x)\\&=9-x\end{aligned}\]
  12. For the third minor,
    \[\begin{aligned}\begin{vmatrix}9 & 3\\x & y\end{vmatrix}&=9y-3x\end{aligned}\]
  13. Substitute the minors
  14. \[\begin{aligned}0&=3(3-y)-(9-x)+(9y-3x)\end{aligned}\]
  15. Expand the brackets
    \[\begin{aligned}0&=9-3y-9+x+9y-3x\end{aligned}\]
  16. Collect like terms
    \[\begin{aligned}0&=(9-9)+(x-3x)+(-3y+9y)\\&=0-2x+6y\end{aligned}\]
  17. Therefore,
    \[-2x+6y=0\]
  18. Dividing by \(2\),
    \[-x+3y=0\]
  19. Hence,
    \[x-3y=0\]
💡 Answer
Final Answer
›

\[ \boxed{\text{(i) }y=2x} \]

\[ \boxed{\text{(ii) }x=3y\quad\text{or}\quad y=\frac{x}{3}} \]

🎯 Exam Significance
Exam Significance
›

This is an important application of determinants because it establishes a direct connection between determinants and coordinate geometry. For CBSE Board examinations, the determinant approach provides a systematic derivation of the equation of a line through two given points and allows marks to be awarded at each logical step.

For competitive entrance examinations, this method is especially useful when a problem explicitly asks for a determinant-based solution or when collinearity must be established quickly. It also reinforces the broader principle that a zero determinant represents a zero-area configuration and hence collinearity.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. To find the equation of a line through two given points using determinants, introduce a variable point \(C(x,y)\) on the line.

  2. The three points must be collinear, so the determinant representing the area must be zero:

    \[ \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x & y & 1 \end{vmatrix}=0. \]

  3. Always remember the cofactor sign pattern \(+,-,+\) when expanding along the first row.

  4. Evaluate the \(2\times2\) minors explicitly to reduce the possibility of sign errors.

  5. For part (i), the determinant simplifies to

    \[ y-2x=0, \]
    giving
    \[ y=2x. \]

  6. For part (ii), the determinant simplifies to

    \[ x-3y=0, \]
    giving
    \[ y=\frac{x}{3}. \]

  7. The determinant method is essentially an application of the zero-area condition for three collinear points.

  8. For Board examinations, writing the determinant formation, expansion, simplification, and final equation in separate logical steps makes the derivation complete and easy to evaluate.
← Q3
4 / 5  ·  80%
Q5 →
Q5
NUMERIC3 marks
If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12 (B)–2 (C)–12, –2 (D) 12, –2
📘 Concept & Theory
Concept/Theory
›

The determinant formula for the area of a triangle can be used to determine an unknown coordinate or parameter when the area of the triangle is given.

If the vertices of a triangle are

\[ A(x_1,y_1),\qquad B(x_2,y_2),\qquad C(x_3,y_3), \]
then its area is

\[\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}\]

Since the area is always non-negative, the absolute value of the determinant must be retained throughout the calculation.

In this problem, the area is \(35\) square units. Hence,

\[ \frac{1}{2}|D|=35 \]

where \(D\) is the corresponding determinant. Therefore,

\[ |D|=70 \]

This produces an absolute-value equation, which must be solved by considering both possible signs:

\[ D=70 \qquad\text{or}\qquad D=-70 \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the three vertices and identify the unknown parameter \(k\).

  2. Apply the determinant formula for the area of a triangle.

  3. Substitute the given area \(35\) square units.

  4. Multiply by \(2\) to obtain the absolute value of the determinant as \(70\).

  5. Expand the determinant carefully and simplify it to an expression in \(k\).

  6. Solve the resulting absolute-value equation by considering both cases.

  7. Verify the resulting values of \(k\) in the original area condition.

  8. Select the corresponding option from the given choices.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Given — Vertices: \(A(2,-6),\qquad B(5,4),\qquad C(k,4)\)
    he area of the triangle is given as \(35\text{ square units}.\)
  2. Using the determinant formula for the area of a triangle,
  3. \[\Delta=\frac{1}{2}\begin{vmatrix}2 & -6 & 1\\5 & 4 & 1\\k & 4 & 1\end{vmatrix}\]
  4. Since \(\Delta=35\),
  5. \[35=\frac{1}{2}\begin{vmatrix}2 & -6 & 1\\5 & 4 & 1\\k & 4 & 1\end{vmatrix}\]
  6. Multiplying both sides by \(2\),
    \[70=\begin{vmatrix}2 & -6 & 1\\5 & 4 & 1\\k & 4 & 1\end{vmatrix}\]
  7. Expand the determinant along the first row
  8. The cofactor signs in the first row are \(+,-,+\). Therefore,
    \[\begin{aligned}D&=2\begin{vmatrix}4 & 1\\4 & 1\end{vmatrix}-(-6)\begin{vmatrix}5 & 1\\k & 1\end{vmatrix}+1\begin{vmatrix}5 & 4\\k & 4\end{vmatrix}\end{aligned}\]
  9. Since the second element of the first row is \(-6\), the second term is
    \[-(-6)=+6\]
  10. Hence,
    \[\begin{aligned}D&=2\begin{vmatrix}4 & 1\\4 & 1\end{vmatrix}+6\begin{vmatrix}5 & 1\\k & 1\end{vmatrix}+\begin{vmatrix}5 & 4\\k & 4\end{vmatrix}\end{aligned}\]
  11. Evaluate the first \(2\times2\) determinant
  12. \[\begin{aligned}\begin{vmatrix}4 & 1\\4 & 1\end{vmatrix}&=4(1)-1(4)\\&=4-4\\&=0\end{aligned}\]
  13. Evaluate the second \(2\times2\) determinant
  14. \[\begin{aligned}\begin{vmatrix}5 & 1\\k & 1\end{vmatrix}&=5(1)-1(k)\\&=5-k\end{aligned}\]
  15. Evaluate the third \(2\times2\) determinant
  16. \[\begin{aligned}\begin{vmatrix}5 & 4\\k & 4\end{vmatrix}&=5(4)-4(k)\\&=20-4k\end{aligned}\]
  17. Substitute the minors
  18. Therefore,
    \[\begin{aligned}D&=2(0)+6(5-k)+(20-4k)\\&=0+30-6k+20-4k\\&=50-10k\end{aligned}\]
  19. Hence, the area condition becomes
    \[|50-10k|=70\]
  20. Solve the absolute-value equation
  21. For
    \[|50-10k|=70,\]
    there are two possible cases.
  22. Case 1: \(50-10k=70\)
  23. \[\begin{aligned}50-10k&=70\\-10k&=20\\k&=-2\end{aligned}\]
  24. Case 2: \(50-10k=-70\)
  25. \[\begin{aligned}50-10k&=-70\\-10k&=-120\\k&=12\end{aligned}\]
💡 Answer
Final Answer
›
\[\boxed{k=-2,\ 12}\]
🎯 Exam Significance
Exam Significance
›

This question is a standard application of determinants to coordinate geometry and tests the student's ability to translate a geometrical condition into an algebraic equation. For CBSE Board examinations, the determinant setup, expansion, simplification, and treatment of the absolute value are all important steps.

For competitive entrance examinations, the problem illustrates a frequently tested pattern: a determinant involving a parameter is equated to a prescribed area. Because area involves an absolute value, two values of the parameter may arise. Recognising this structure quickly can substantially reduce calculation time.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The area formula for a triangle with coordinate vertices contains the absolute value of a determinant.

  2. If the given area is \(35\), then the absolute value of the determinant must be \(70\).

  3. For an equation of the form

    \[ |D|=70, \]
    always consider both
    \[ D=70 \quad\text{and}\quad D=-70. \]

  4. In this problem, the determinant simplifies to

    \[ D=50-10k. \]

  5. Therefore,

    \[ |50-10k|=70, \]
    which gives two values of \(k\).

  6. The two valid values are

    \[ k=-2 \quad\text{and}\quad k=12. \]

  7. Never replace an absolute-value equation by only one signed equation, because that may eliminate a valid solution.

  8. The determinant method provides a compact bridge between coordinate geometry and algebra and is highly useful in both Board and competitive examinations.
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NCERT Class 12 Determinants Ex 4.2 Q5 Solution
NCERT Class 12 Determinants Ex 4.2 Q5 Solution — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 12 Mathematics Chapter 4 Determinants Exercise 4.2 Question 5 focuses on an important application of determinants in coordinate geometry: finding the unknown value of k when the area of a triangle is given. In this problem, the vertices of the triangle contain the variable k, while its area is specified as 35 square units. Using the determinant formula for the area of a triangle, the geometrical condition is converted into an algebraic equation in k. Special attention is required…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Determinants — Learning Resources

    📄 Detailed Notes
    🧠 Practice MCQs
    ✔️ True / False
    🎯 Advance MCQs

    Frequently Asked Questions

    The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is ? = 1/2 |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|.

    The determinant may be positive or negative depending on the order of the vertices, but geometric area is always non-negative. Therefore, its absolute value is taken.

    For vertices (2, -6), (5, 4), and (k, 4), the determinant is |2 -6 1; 5 4 1; k 4 1|.

    The area of the triangle is given as 35 square units.

    Since ? = 35, twice the area gives |D| = 70, where D is the determinant. After expansion, this becomes |50 - 10k| = 70.

    Two values of k are obtained because the absolute-value equation |50 - 10k| = 70 has two cases.

    The two cases are 50 - 10k = 70 and 50 - 10k = -70.

    The two values are k = -2 and k = 12.

    Option (D), which gives k = 12, -2, is correct.

    This question tests the application of determinants to coordinate geometry, calculation of triangle area, algebraic simplification, and solving absolute-value equations, making it useful for both Board and entrance examinations.

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