Maharashtra Board Class 11 Maths Part 1 Chapter 6 Determinants Miscellaneous Solutions

Question 1. Evaluate:
(i) \( \begin{vmatrix} 2 & -5 & 7 \\ 5 & 2 & 1 \\ 9 & 0 & 2 \end{vmatrix} \)
Answer: Solution: \( \begin{vmatrix} 2 & -5 & 7 \\ 5 & 2 & 1 \\ 9 & 0 & 2 \end{vmatrix} = 2 \begin{vmatrix} 2 & 1 \\ 0 & 2 \end{vmatrix} - (-5) \begin{vmatrix} 5 & 1 \\ 9 & 2 \end{vmatrix} + 7 \begin{vmatrix} 5 & 2 \\ 9 & 0 \end{vmatrix} \) \( = 2(4 - 0) + 5(10 - 9) + 7(0 - 18) \) \( = 2(4) + 5(1) + 7(-18) \) \( = 8 + 5 - 126 \) \( = -113 \) In simple words: To evaluate the determinant, expand it along the first row. Multiply each element by its corresponding cofactor and sum the results, carefully handling the signs.

🎯 Exam Tip: Pay close attention to the signs when expanding determinants, especially for the middle term which is always subtracted. Double-check all arithmetic calculations to avoid errors.

Question 1. Evaluate:
(ii) \( \begin{vmatrix} 1 & -3 & 12 \\ 0 & 2 & -4 \\ 9 & 7 & 2 \end{vmatrix} \)
Answer: Solution: \( \begin{vmatrix} 1 & -3 & 12 \\ 0 & 2 & -4 \\ 9 & 7 & 2 \end{vmatrix} \) \( = 1 \begin{vmatrix} 2 & -4 \\ 7 & 2 \end{vmatrix} - (-3) \begin{vmatrix} 0 & -4 \\ 9 & 2 \end{vmatrix} + 12 \begin{vmatrix} 0 & 2 \\ 9 & 7 \end{vmatrix} \) \( = 1(4 + 28) + 3(0 + 36) + 12(0 - 18) \) \( = 1(32) + 3(36) + 12(-18) \) \( = 32 + 108 - 216 \) \( = -76 \) In simple words: The determinant is calculated by expanding along the first row, multiplying each element by its cofactor, and then summing these terms.

🎯 Exam Tip: For determinants with zero elements, choose to expand along the row or column containing the most zeros to simplify calculations. Accuracy in signs and basic arithmetic is key.

Question 2. Find the value(s) of x, if
(i) \( \begin{vmatrix} 1 & 4 & 20 \\ 1 & -2 & -5 \\ 1 & 2x & 5x^2 \end{vmatrix} = 0 \)
Answer: Solution: \( \begin{vmatrix} 1 & 4 & 20 \\ 1 & -2 & -5 \\ 1 & 2x & 5x^2 \end{vmatrix} = 0 \) \( \therefore 1(-10x^2 + 10x) - 4(5x^2 + 5) + 20(2x + 2) = 0 \) \( \therefore -10x^2 + 10x - 20x^2 - 20 + 40x + 40 = 0 \) \( \therefore -30x^2 + 50x + 20 = 0 \) \( \therefore 3x^2 - 5x - 2 = 0 ......[\text{Dividing throughout by } (-10)] \) \( \therefore 3x^2 - 6x + x - 2 = 0 \) \( \therefore 3x(x - 2) + 1(x - 2) = 0 \) \( \therefore (x - 2)(3x + 1) = 0 \) \( \therefore x - 2 = 0 \) or \( 3x + 1 = 0 \) \( \therefore x = 2 \) or \( x = -\frac{1}{3} \) In simple words: Set the determinant equal to zero, expand it to get a quadratic equation in x, and then solve the quadratic equation by factoring or using the quadratic formula to find the values of x.

🎯 Exam Tip: When solving for a variable within a determinant, remember that the expansion leads to an algebraic equation. Carefully simplify the equation after expansion and solve for the variable using appropriate algebraic techniques.

Question 2. Find the value(s) of x, if
(ii) \( \begin{vmatrix} 1 & 2x & 4x \\ 1 & 4 & 16 \\ 1 & 1 & 1 \end{vmatrix} = 0 \)
Answer: Solution: \( \begin{vmatrix} 1 & 2x & 4x \\ 1 & 4 & 16 \\ 1 & 1 & 1 \end{vmatrix} = 0 \) \( \therefore 1(4 - 16) - 2x(1 - 16) + 4x(1 - 4) = 0 \) \( \therefore 1(-12) - 2x(-15) + 4x(-3) = 0 \) \( \therefore -12 + 30x - 12x = 0 \) \( \therefore 18x = 12 \) \( \therefore x = \frac{12}{18} \) \( \therefore x = \frac{2}{3} \) In simple words: Expand the given determinant, equate it to zero, and simplify the resulting linear equation to solve for x.

🎯 Exam Tip: Simplify the algebraic expression derived from the determinant expansion carefully. Remember to reduce fractions to their simplest form in the final answer.

Question 3. By using properties of determinants, prove that \( \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} = 0 \).
Answer: Solution: L.H.S. = \( \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \) Applying \( R_1 \to R_1 + R_2 \), we get L.H.S. = \( \begin{vmatrix} x+y+z & y+z+x & z+x+y \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \) Taking \( (x + y + z) \) common from \( R_1 \), we get L.H.S. = \( (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \) \( = (x + y + z)(0) \) \( \dots[\because R_1 \text{ and } R_3 \text{ are identical}] \) \( = 0 = \text{R.H.S.} \) In simple words: By adding the second row to the first row, the elements of the first row become identical to each other. Then, by taking a common factor from the first row, we observe that the first and third rows of the determinant are identical, making its value zero.

🎯 Exam Tip: Remember that if any two rows or columns of a determinant are identical, the value of the determinant is zero. Strategic row/column operations can reveal such properties.

Question 4. Without expanding the determinants, show that
(i) \( \begin{vmatrix} b+c & bc & bc^2 \\ c+a & ca & ca^2 \\ a+b & ab & a^2b^2 \end{vmatrix} = 0 \)
Answer: Solution: L.H.S. = \( \begin{vmatrix} b+c & bc & b^2c^2 \\ c+a & ca & c^2a^2 \\ a+b & ab & a^2b^2 \end{vmatrix} \) Taking bc, ca, ab common from \( R_1, R_2, R_3 \) respectively, we get L.H.S. = \( (bc)(ca)(ab) \begin{vmatrix} \frac{b+c}{bc} & 1 & bc \\ \frac{c+a}{ca} & 1 & ca \\ \frac{a+b}{ab} & 1 & ab \end{vmatrix} \) Taking abc common from \( C_3 \), we get L.H.S. = \( (a^2b^2c^2) (abc) \begin{vmatrix} \frac{1}{c}+\frac{1}{b} & 1 & 1/a \\ \frac{1}{a}+\frac{1}{c} & 1 & 1/b \\ \frac{1}{b}+\frac{1}{a} & 1 & 1/c \end{vmatrix} \) Applying \( C_1 \to C_1 + C_3 \), we get L.H.S. = \( a^3b^3c^3 \begin{vmatrix} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{a} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{b} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1 & \frac{1}{c} \end{vmatrix} \) Taking \( (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \) common from \( C_1 \), we get L.H.S. = \( a^3b^3c^3 (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \begin{vmatrix} 1 & 1 & \frac{1}{a} \\ 1 & 1 & \frac{1}{b} \\ 1 & 1 & \frac{1}{c} \end{vmatrix} \) \( = a^3b^3c^3 (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) (0) \) \( \dots[\because C_1 \text{ and } C_2 \text{ are identical}] \) \( = 0 = \text{R.H.S.} \) In simple words: By factoring out common terms from rows and then from a column, and performing a column operation, the first two columns become identical, proving the determinant is zero without full expansion.

🎯 Exam Tip: When asked to prove a determinant is zero without expansion, look for row/column operations that can make two rows or columns identical, or a row/column entirely zero.

Question 4. Without expanding the determinants, show that
(ii) \( \begin{vmatrix} xa & yb & zc \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} x & y & z \\ a & b & c \\ bc & ca & ab \end{vmatrix} \)
Answer: Solution: L.H.S. = \( \begin{vmatrix} xa & yb & zc \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1 \end{vmatrix} \) Taking a, b, c common from \( C_1, C_2, C_3 \) respectively, we get L.H.S. = \( abc \begin{vmatrix} x & y & z \\ a & b & c \\ 1/a & 1/b & 1/c \end{vmatrix} \) Multiply \( R_3 \) by \( abc \), we get \( = \frac{abc}{abc} \begin{vmatrix} x & y & z \\ a & b & c \\ bc & ca & ab \end{vmatrix} \) (This step seems incorrect in the OCR. The expected operation is to multiply R3 by abc and divide the determinant by abc. Or, the original problem implies this form is reached directly.) Let's re-evaluate the step: L.H.S. = \( \begin{vmatrix} xa & yb & zc \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1 \end{vmatrix} \) Multiply \( R_1 \) by \( bc \), \( R_2 \) by \( ac \), \( R_3 \) by \( ab \). To maintain determinant value, divide by \( (abc)^2 \). This is complex. Let's stick to the OCR steps provided as much as possible, interpreting the output as a transformation. If we factor out 'a' from C1, 'b' from C2, 'c' from C3, the determinant becomes: \( abc \begin{vmatrix} x & y & z \\ a & b & c \\ 1/a & 1/b & 1/c \end{vmatrix} \) Now, to get the desired R.H.S. structure, we need bc, ca, ab in R3. We can achieve this by multiplying R3 by abc and dividing the entire determinant by abc. \( = abc \times \frac{1}{abc} \begin{vmatrix} x & y & z \\ a & b & c \\ (1/a)abc & (1/b)abc & (1/c)abc \end{vmatrix} \) \( = \begin{vmatrix} x & y & z \\ a & b & c \\ bc & ca & ab \end{vmatrix} \) \( = \text{R.H.S.} \) In simple words: By factoring out 'a', 'b', and 'c' from the respective columns of the Left Hand Side determinant and then performing a row operation (multiplying the third row by 'abc' and dividing the whole determinant by 'abc' to maintain equality), we transform it into the Right Hand Side determinant.

🎯 Exam Tip: To show equality between two determinants, often a combination of factoring common terms from rows/columns and applying elementary row/column operations is required. Remember that multiplying a row/column by a scalar 'k' multiplies the determinant by 'k'.

Question 4. Without expanding the determinants, show that
(iii) \( \begin{vmatrix} l & m & n \\ e & d & f \\ u & v & w \end{vmatrix} = \begin{vmatrix} n & f & w \\ l & e & u \\ m & d & v \end{vmatrix} \)
Answer: Solution: L.H.S. = \( \begin{vmatrix} l & m & n \\ e & d & f \\ u & v & w \end{vmatrix} \) Interchanging rows and columns, we get \( \begin{vmatrix} l & e & u \\ m & d & v \\ n & f & w \end{vmatrix} \) Applying \( R_2 \leftrightarrow R_3 \), we get \( - \begin{vmatrix} l & e & u \\ n & f & w \\ m & d & v \end{vmatrix} \) (Determinant changes sign on row interchange) Applying \( R_1 \leftrightarrow R_2 \), we get \( - (- \begin{vmatrix} n & f & w \\ l & e & u \\ m & d & v \end{vmatrix} ) \) (Another sign change) \( = \begin{vmatrix} n & f & w \\ l & e & u \\ m & d & v \end{vmatrix} \) \( = \text{R.H.S.} \) In simple words: First, interchange rows and columns (transpose the matrix) which does not change the determinant's value. Then, perform two row interchanges on the transposed matrix. Each row interchange flips the sign of the determinant, so two such operations restore the original sign and achieve the target arrangement.

🎯 Exam Tip: Transposing a matrix (interchanging rows and columns) does not change its determinant. Interchanging any two rows or columns changes the sign of the determinant. An even number of interchanges restores the original sign.

Question 4. Without expanding the determinants, show that
(iv) \( \begin{vmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{vmatrix} = 0 \)
Answer: Solution: Let \( D = \begin{vmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{vmatrix} \) Taking \( (-1) \) common from \( R_1, R_2, R_3 \), we get \( D = (-1)^3 \begin{vmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{vmatrix} \) Interchanging rows and columns, we get \( D = -1 \begin{vmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{vmatrix} \) \( \therefore D = -1(D) \) \( \therefore 2D = 0 \) \( \therefore D = 0 \) \( \begin{vmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{vmatrix} = 0 \) In simple words: The determinant is of a skew-symmetric matrix of odd order (3x3). For any skew-symmetric matrix of odd order, its determinant is always zero. We can show this by factoring out -1 from each row and then transposing the matrix, which results in D = -D, implying D must be 0.

🎯 Exam Tip: A key property for skew-symmetric matrices (where \( A^T = -A \)) is that their determinant is zero if the order of the matrix is odd. This can save time by avoiding explicit expansion.

Question 5. Solve the following linear equations by Cramer's Rule.
(i) \( 2x - y + z = 1, x + 2y + 3z = 8, 3x + y - 4z = 1 \)
Answer: Solution: Given equations are \( 2x - y + z = 1 \) \( x + 2y + 3z = 8 \) \( 3x + y - 4z = 1 \) \( D = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4 \end{vmatrix} \) \( = 2(-8 - 3) - (-1)(-4 - 9) + 1(1 - 6) \) \( = 2(-11) + 1(-13) + 1(-5) \) \( = -22 - 13 - 5 \) \( = -40 \neq 0 \) \( D_x = \begin{vmatrix} 1 & -1 & 1 \\ 8 & 2 & 3 \\ 1 & 1 & -4 \end{vmatrix} \) \( = 1(-8 - 3) - (-1)(-32 - 3) + 1(8 - 2) \) \( = 1(-11) + 1(-35) + 1(6) \) \( = -11 - 35 + 6 \) \( = -40 \) \( D_y = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 8 & 3 \\ 3 & 1 & -4 \end{vmatrix} \) \( = 2(-32 - 3) - 1(-4 - 9) + 1(1 - 24) \) \( = 2(-35) - 1(-13) + 1(-23) \) \( = -70 + 13 - 23 \) \( = -80 \) \( D_z = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & 8 \\ 3 & 1 & 1 \end{vmatrix} \) \( = 2(2 - 8) - (-1)(1 - 24) + 1(1 - 6) \) \( = 2(-6) + 1(-23) + 1(-5) \) \( = -12 - 23 - 5 \) \( = -40 \) By Cramer's Rule, \( x = \frac{D_x}{D} = \frac{-40}{-40} = 1 \) \( y = \frac{D_y}{D} = \frac{-80}{-40} = 2 \) \( z = \frac{D_z}{D} = \frac{-40}{-40} = 1 \) \( \therefore x = 1, y = 2 \) and \( z = 1 \) are the solutions of the given equations. In simple words: Cramer's Rule solves systems of linear equations by calculating the determinant of the coefficient matrix (D) and the determinants of matrices formed by replacing a column with the constant terms (\(D_x, D_y, D_z\)). The solutions are then found by dividing these by D.

🎯 Exam Tip: When using Cramer's Rule, ensure that the main determinant D is non-zero. If D=0, Cramer's Rule cannot be directly applied, indicating either no unique solution or infinitely many solutions. Be meticulous with determinant calculations.

Question 5. Solve the following linear equations by Cramer's Rule.
(ii) \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -2, \frac{1}{x} - \frac{2}{y} + \frac{1}{z} = 3, \frac{2}{x} - \frac{1}{y} + \frac{3}{z} = -1 \)
Answer: Solution: Let \( \frac{1}{x} = p, \frac{1}{y} = q, \frac{1}{z} = r \) The given equations become \( p + q + r = -2 \) \( p - 2q + r = 3 \) \( 2p - q + 3r = -1 \) \( D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 2 & -1 & 3 \end{vmatrix} \) \( = 1(-6 + 1) - 1(3 - 2) + 1(-1 + 4) \) \( = -5 - 1 + 3 \) \( = -3 \) \( D_p = \begin{vmatrix} -2 & 1 & 1 \\ 3 & -2 & 1 \\ -1 & -1 & 3 \end{vmatrix} \) \( = -2(-6 + 1) - 1(9 + 1) + 1(-3 - 2) \) \( = 10 - 10 - 5 \) \( = -5 \) \( D_q = \begin{vmatrix} 1 & -2 & 1 \\ 1 & 3 & 1 \\ 2 & -1 & 3 \end{vmatrix} \) \( = 1(9 + 1) + 2(3 - 2) + 1(-1 - 6) \) \( = 10 + 2 - 7 \) \( = 5 \) \( D_r = \begin{vmatrix} 1 & 1 & -2 \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix} \) \( = 1(2 + 3) - 1(-1 - 6) - 2(-1 + 4) \) \( = 5 + 7 - 6 \) \( = 6 \) By Cramer's Rule, \( p = \frac{D_p}{D} = \frac{-5}{-3} = \frac{5}{3} \) \( q = \frac{D_q}{D} = \frac{5}{-3} = -\frac{5}{3} \) \( r = \frac{D_r}{D} = \frac{6}{-3} = -2 \) \( \therefore x = \frac{1}{p} = \frac{3}{5}, y = \frac{1}{q} = -\frac{3}{5}, z = \frac{1}{r} = -\frac{1}{2} \) are the solutions of the given equations. In simple words: Transform the given equations by substituting p = 1/x, q = 1/y, r = 1/z into a system of linear equations in p, q, r. Then, apply Cramer's Rule to solve for p, q, and r. Finally, convert these values back to find x, y, and z.

🎯 Exam Tip: For equations involving reciprocals, a substitution method simplifies the system to a standard linear form solvable by Cramer's Rule. Remember to convert back to the original variables for the final answer.

Question 5. Solve the following linear equations by Cramer's Rule.
(iii) \( x - y + 2z = 7, 3x + 4y - 5z = 5, 2x - y + 3z = 12 \)
Answer: Solution: Given equations are \( x - y + 2z = 7 \) \( 3x + 4y - 5z = 5 \) \( 2x - y + 3z = 12 \) \( D = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{vmatrix} \) \( = 1(12 - 5) - (-1)(9 + 10) + 2(-3 - 8) \) \( = 1(7) + 1(19) + 2(-11) \) \( = 7 + 19 - 22 \) \( = 4 \neq 0 \) \( D_x = \begin{vmatrix} 7 & -1 & 2 \\ 5 & 4 & -5 \\ 12 & -1 & 3 \end{vmatrix} \) \( = 7(12 - 5) - (-1)(15 + 60) + 2(-5 - 48) \) \( = 7(7) + 1(75) + 2(-53) \) \( = 49 + 75 - 106 \) \( = 18 \) \( D_y = \begin{vmatrix} 1 & 7 & 2 \\ 3 & 5 & -5 \\ 2 & 12 & 3 \end{vmatrix} \) \( = 1(15 + 60) - 7(9 + 10) + 2(36 - 10) \) \( = 1(75) - 7(19) + 2(26) \) \( = 75 - 133 + 52 \) \( = -6 \) \( D_z = \begin{vmatrix} 1 & -1 & 7 \\ 3 & 4 & 5 \\ 2 & -1 & 12 \end{vmatrix} \) \( = 1(48 + 5) - (-1)(36 - 10) + 7(-3 - 8) \) \( = 1(53) + 1(26) + 7(-11) \) \( = 53 + 26 - 77 \) \( = 2 \) By Cramer's Rule, \( x = \frac{D_x}{D} = \frac{18}{4} = \frac{9}{2} \) \( y = \frac{D_y}{D} = \frac{-6}{4} = -\frac{3}{2} \) \( z = \frac{D_z}{D} = \frac{2}{4} = \frac{1}{2} \) \( \therefore x = \frac{9}{2}, y = -\frac{3}{2} \) and \( z = \frac{1}{2} \) are the solutions of the given equations. In simple words: To solve this system, calculate the main determinant D from coefficients and \( D_x, D_y, D_z \) by replacing columns with constant terms. Then, apply Cramer's Rule to find x, y, and z as ratios of these determinants.

🎯 Exam Tip: Maintain precision in calculations for determinants, especially with negative numbers. Simplify fractions to their lowest terms for the final solution. Remember to verify the main determinant D is not zero.

Question 6. Find the value(s) of k, if the following equations are consistent.
(i) \( 3x + y - 2 = 0, kx + 2y - 3 = 0 \) and \( 2x - y = 3 \)
Answer: Solution: Given equations are \( 3x + y - 2 = 0 \) \( kx + 2y - 3 = 0 \) \( 2x - y = 3 \) i.e. \( 2x - y - 3 = 0 \) Since, these equations are consistent. \( \begin{vmatrix} 3 & 1 & -2 \\ k & 2 & -3 \\ 2 & -1 & -3 \end{vmatrix} = 0 \) \( \therefore 3(-6 - 3) - 1(-3k + 6) - 2(-k - 4) = 0 \) \( \therefore 3(-9) - 1(-3k + 6) - 2(-k - 4) = 0 \) \( \therefore -27 + 3k - 6 + 2k + 8 = 0 \) \( \therefore 5k - 25 = 0 \) \( \therefore k = 5 \) In simple words: For a system of linear equations to be consistent (have at least one solution), the determinant of its coefficient matrix must be zero. By setting the determinant to zero, we can solve for the unknown variable k.

🎯 Exam Tip: The condition for consistency of a system of three linear equations in three variables is that the determinant of the coefficient matrix (including constants as the third column) must be zero. Ensure accurate expansion and algebraic simplification.

Question 6. Find the value(s) of k, if the following equations are consistent.
(ii) \( kx + 3y + 4 = 0, x + ky + 3 = 0, 3x + 4y + 5 = 0 \)
Answer: Solution: Given equations are \( kx + 3y + 4 = 0 \) \( x + ky + 3 = 0 \) \( 3x + 4y + 5 = 0 \) Since, these equations are consistent. \( \begin{vmatrix} k & 3 & 4 \\ 1 & k & 3 \\ 3 & 4 & 5 \end{vmatrix} = 0 \) \( \therefore k(5k - 12) - 3(5 - 9) + 4(4 - 3k) = 0 \) \( \therefore 5k^2 - 12k + 12 + 16 - 12k = 0 \) \( \therefore 5k^2 - 24k + 28 = 0 \) \( \therefore 5k^2 - 10k - 14k + 28 = 0 \) \( \therefore 5k(k - 2) - 14(k - 2) = 0 \) \( \therefore (k - 2)(5k - 14) = 0 \) \( \therefore k - 2 = 0 \) or \( 5k - 14 = 0 \) \( \therefore k = 2 \) or \( k = \frac{14}{5} \) In simple words: For the given system of linear equations to be consistent, the determinant of the coefficient matrix must be zero. Expanding this determinant yields a quadratic equation in k, which, when solved, gives the possible values of k.

🎯 Exam Tip: When the consistency condition leads to a quadratic equation, remember to find all roots for the variable. Factorization or the quadratic formula are common methods. Be careful with algebraic signs and calculations.

Question 7. Find the area of triangles whose vertices are
(i) A(-1, 2), B(2, 4), C(0, 0)
Answer: Solution: Here, A(\(x_1, y_1\)) = A(-1, 2), B(\(x_2, y_2\)) = B(2, 4), C(\(x_3, y_3\)) = C(0, 0) Area of a triangle = \( \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \) \( \therefore \text{A}(\Delta\text{ABC}) = \frac{1}{2} \begin{vmatrix} -1 & 2 & 1 \\ 2 & 4 & 1 \\ 0 & 0 & 1 \end{vmatrix} \) \( = \frac{1}{2} [-1(4 - 0) - 2(2 - 0) + 1(0 - 0)] \) \( = \frac{1}{2} (-4 - 4) \) \( = \frac{1}{2} (-8) \) \( = -4 \) Since, area cannot be negative. \( \therefore \text{A}(\Delta\text{ABC}) = 4 \text{ sq.units} \) In simple words: The area of a triangle with given vertices can be found using the determinant formula. Calculate the determinant, take half of its absolute value, as area cannot be negative.

🎯 Exam Tip: Always remember that area is a scalar quantity and must be non-negative. If the determinant calculation yields a negative value, take its absolute value for the final area. Proper substitution of coordinates is crucial.

Question 7. Find the area of triangles whose vertices are
(ii) P(3, 6), Q(-1, 3), R(2, -1)
Answer: Solution: Here, P(\(x_1, y_1\)) = P(3, 6), Q(\(x_2, y_2\)) = Q(-1, 3), R(\(x_3, y_3\)) = R(2, -1) Area of a triangle = \( \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \) \( \text{A}(\Delta\text{PQR}) = \frac{1}{2} \begin{vmatrix} 3 & 6 & 1 \\ -1 & 3 & 1 \\ 2 & -1 & 1 \end{vmatrix} \) \( = \frac{1}{2} [3(3 + 1) - 6(-1 - 2) + 1(1 - 6)] \) \( = \frac{1}{2} [3(4) - 6(-3) + 1(-5)] \) \( = \frac{1}{2} (12 + 18 - 5) \) \( = \frac{1}{2} (25) \) \( \therefore \text{A}(\Delta\text{PQR}) = \frac{25}{2} \text{ sq.units} \) In simple words: Substitute the given vertex coordinates into the determinant formula for the area of a triangle. Compute the determinant and take half of its value to get the triangle's area.

🎯 Exam Tip: Be methodical in substituting coordinates and evaluating the determinant to avoid errors. The sign of the determinant initially indicates the orientation of the vertices, but for area, only the magnitude matters.

Question 7. Find the area of triangles whose vertices are
(iii) L(1, 1), M(-2, 2), N(5, 4)
Answer: Solution: Here, L(\(x_1, y_1\)) = L(1, 1), M(\(x_2, y_2\)) = M(-2, 2), N(\(x_3, y_3\)) = N(5, 4) Area of a triangle = \( \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \) \( \text{A}(\Delta\text{LMN}) = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ -2 & 2 & 1 \\ 5 & 4 & 1 \end{vmatrix} \) \( = \frac{1}{2} [1(2 - 4) - 1(-2 - 5) + 1(-8 - 10)] \) \( = \frac{1}{2} [1(-2) - 1(-7) + 1(-18)] \) \( = \frac{1}{2} (-2 + 7 - 18) \) \( = \frac{1}{2} (-13) \) \( = -\frac{13}{2} \) Since, area cannot be negative. \( \therefore \text{A}(\Delta\text{LMN}) = \frac{13}{2} \text{ sq.units} \) In simple words: Use the determinant formula with the given vertex coordinates. Compute the determinant value, and then take half of its absolute value to find the area, as area must be positive.

🎯 Exam Tip: Always remember to take the absolute value of the determinant result before dividing by two to ensure the area is positive. Careful expansion of the 3x3 determinant is critical.

Question 8. Find the value of k,
(i) if the area of \(\Delta\text{PQR}\) is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
Answer: Solution: Here, P(\(x_1, y_1\)) = P(k, 0), Q(\(x_2, y_2\)) = Q(4, 0), R(\(x_3, y_3\)) = R(0, 2) Area of a triangle = \( \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \) \(\text{A}(\Delta\text{PQR}) = 4 \text{ sq.units} \) \( \therefore \pm4 = \frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} \) \( \therefore \pm8 = k(0 - 2) - 0 + 1(8 - 0) \) \( \therefore \pm8 = -2k + 8 \) \( \therefore 8 = -2k + 8 \) or \( -8 = -2k + 8 \) \( \therefore -2k = 0 \) or \( 2k = 16 \) \( \therefore k = 0 \) or \( k = 8 \) In simple words: Set up the area formula using the determinant with the given vertices, equating it to ±4 (since area is positive, but the determinant can be negative). Solve the resulting linear equation(s) for k to find its possible values.

🎯 Exam Tip: When the area is given, always use \( \pm \) with the area value in the determinant equation, as the determinant can be positive or negative depending on the order of vertices. This ensures all possible values of the unknown variable are found.

Question 8. Find the value of k,
(ii) if area of \(\Delta\text{LMN}\) is \( \frac{33}{2} \) square units and vertices are L(3, -5), M(-2, k), N(1, 4).
Answer: Solution: Here, L(\(x_1, y_1\)) = L(3, -5), M(\(x_2, y_2\)) = M(-2, k), N(\(x_3, y_3\)) = N(1, 4) Area of a triangle = \( \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \) \(\text{A}(\Delta\text{LMN}) = \frac{33}{2} \text{ sq.units} \) \( \therefore \pm\frac{33}{2} = \frac{1}{2} \begin{vmatrix} 3 & -5 & 1 \\ -2 & k & 1 \\ 1 & 4 & 1 \end{vmatrix} \) \( \therefore \pm33 = 3(k - 4) - (-5)(-2 - 1) + 1(-8 - k) \) \( \therefore \pm33 = 3k - 12 - 15 - 8 - k \) \( \therefore \pm33 = 2k - 35 \) \( \therefore 2k - 35 = 33 \) or \( 2k - 35 = -33 \) \( \therefore 2k = 68 \) or \( 2k = 2 \) \( \therefore k = 34 \) or \( k = 1 \) In simple words: Set the determinant area formula equal to plus or minus the given area, then expand the determinant and solve the resulting linear equations for k. Two possible values for k will emerge.

🎯 Exam Tip: The \( \pm \) sign is crucial when the area is known to account for both orientations of vertices. Ensure careful algebraic simplification to avoid errors in solving for k.