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CBSE - Class 12 Mathematics Determinants Worksheet
EXERCISE 4.2
Worksheet Answers
Solution:
The vertices of the triangle are provided as follows:
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant representation [Per the definition of area in coordinate geometry]:
$\text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
Substituting the given coordinates into the determinant structure:
$\text{Area} = \frac{1}{2} \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix}$
Expanding the determinant along the first row ($R_1$):
$\Delta = -2 \begin{vmatrix} 2 & 1 \\ -8 & 1 \end{vmatrix} - (-3) \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & 2 \\ -1 & -8 \end{vmatrix}$
Calculating the $2 \times 2$ determinants:
Substituting back into the expansion:
$\text{Area} = \frac{1}{2} [ -2(10) + 3(4) + 1(-22) ]$
$\text{Area} = \frac{1}{2} [ -20 + 12 - 22 ]$
$\text{Area} = \frac{1}{2} [ -30 ]$
Since the area must be a positive scalar quantity [By definition of spatial area], we take the absolute value of the determinant result:
$\text{Area} = | -15 | = 15$
Final Solution: 15 \text{ square units}
Solution:
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
[By the Determinant Definition of Area of a Triangle]
Given vertices are $(x_1, y_1) = (2, -6)$, $(x_2, y_2) = (5, 4)$, and $(x_3, y_3) = (k, 4)$. The area is given as $35$ sq units.
Substituting into the determinant form:
$35 = \pm \frac{1}{2} \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix}$
[Note: Since area is an absolute value, we must account for both positive and negative results when removing the modulus bars.]
Expand along the first row:
$\Delta = 2(4(1) - 4(1)) - (-6)(5(1) - k(1)) + 1(5(4) - k(4))$
$\Delta = 2(0) + 6(5 - k) + (20 - 4k)$
$\Delta = 30 - 6k + 20 - 4k$
$\Delta = 50 - 10k$
From the formula $\text{Area} = \frac{1}{2} |\Delta|$:
$35 = \frac{1}{2} |50 - 10k|$
$70 = |50 - 10k|$
This yields two linear equations:
Final Solution: The possible values for $k$ are $12$ or $-2$.
Solution:
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is defined by the determinant formula [Per the Coordinate Geometry definition of area]:
$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$
Given vertices: $(x_1, y_1) = (-2, 0)$, $(x_2, y_2) = (0, 4)$, $(x_3, y_3) = (0, k)$.
Given Area = $4$ sq. units. Therefore:
$4 = \frac{1}{2} \left| \det \begin{pmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{pmatrix} \right|$
$8 = \left| \det \begin{pmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{pmatrix} \right|$ [By multiplying both sides by 2]
Expanding the determinant along the first column (Column 1) for computational efficiency [By Laplace Expansion]:
$\det = -2 \begin{vmatrix} 4 & 1 \\ k & 1 \end{vmatrix} - 0 \begin{vmatrix} 0 & 1 \\ k & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 4 & 1 \end{vmatrix}$
$\det = -2(4(1) - k(1)) - 0 + 0$
$\det = -2(4 - k)$
$\det = -8 + 2k$
Since the determinant is within an absolute value sign, we must account for both positive and negative scenarios [By the Definition of Absolute Value]:
$|2k - 8| = 8$
Case 1:
$2k - 8 = 8$
$2k = 16 \implies k = 8$
Case 2:
$2k - 8 = -8$
$2k = 0 \implies k = 0$
Final Solution: The values of $k$ are $0$ and $8$.
Solution:
The vertices of the triangle are given as $A(k, 0)$, $B(4, 0)$, and $C(0, 2)$. The area of the triangle is given as $\Delta = 4$ sq. units.
The formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ using determinants is:
$\Delta = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$
[Per the definition of Area of a triangle in Coordinate Geometry]
Substituting the given coordinates into the determinant expression:
$4 = \frac{1}{2} \left| \det \begin{pmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{pmatrix} \right|$
Multiplying both sides by 2, we obtain:
$8 = \left| \det \begin{pmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{pmatrix} \right|$
Expanding the determinant along the first column ($C_1$):
$\det = k(0(1) - 1(2)) - 4(0(1) - 1(2)) + 0(0(1) - 0(1))$
$\det = k(-2) - 4(-2) + 0$
$\det = -2k + 8$
[Per the Laplace Expansion Theorem for determinants]
Since the area is an absolute value, we have:
$\left| -2k + 8 \right| = 8$
This leads to two distinct linear equations:
Case 1:
$-2k = 8 - 8$
$-2k = 0 \implies k = 0$
Case 2:
$-2k = -8 - 8$
$-2k = -16 \implies k = 8$
Final Solution: The values of $k$ are $0$ and $8$.
Solution:
Three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are defined to be collinear if and only if the area of the triangle formed by them is equal to zero. The area ($\Delta$) of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant formula:
$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
[Per the Area of a Triangle theorem in Coordinate Geometry]
Given the points $A(a, b + c)$, $B(b, c + a)$, and $C(c, a + b)$, we substitute these into the determinant:
$\Delta = \frac{1}{2} \begin{vmatrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{vmatrix}$
To simplify the determinant, we apply the column operation $C_2 \to C_2 + C_1$:
$\Delta = \frac{1}{2} \begin{vmatrix} a & a + b + c & 1 \\ b & a + b + c & 1 \\ c & a + b + c & 1 \end{vmatrix}$
[By the property: The value of a determinant remains unchanged under the operation $C_i \to C_i + kC_j$]
We observe that column $C_2$ contains a common factor of $(a + b + c)$. Factoring this out:
$\Delta = \frac{1}{2} (a + b + c) \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix}$
Since column $C_2$ and column $C_3$ are now identical (both consist of elements all equal to 1), the determinant evaluates to zero.
[By the property: If any two columns of a determinant are identical, the value of the determinant is $0$]
$\Delta = \frac{1}{2} (a + b + c) \times 0 = 0$
Because the area of the triangle formed by points $A$, $B$, and $C$ is $0$, the points do not enclose any space and must lie on a single straight line.
Final Solution: Since the area of $\Delta ABC = 0$, the points $A(a, b + c)$, $B(b, c + a)$, and $C(c, a + b)$ are collinear.
Solution:
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is determined by the absolute value of the determinant expression [Per the Area of Triangle Determinant Formula]:
$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$
Given the vertices $A(2, 7)$, $B(1, 1)$, and $C(10, 8)$, we map the variables as follows:
Substitute the coordinates into the formula:
$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{pmatrix} \right|$
Expand the determinant using the Laplace expansion method [By Row Expansion Rule]:
$\det = 2 \begin{vmatrix} 1 & 1 \\ 8 & 1 \end{vmatrix} - 7 \begin{vmatrix} 1 & 1 \\ 10 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 10 & 8 \end{vmatrix}$
Calculate the $2 \times 2$ determinants:
Combine the results:
$\det = 2(-7) - 7(-9) + 1(-2)$
$\det = -14 + 63 - 2$
$\det = 47$
$\text{Area} = \frac{1}{2} |47| = 23.5 \text{ square units}$
Final Solution: 23.5 \text{ square units}
Solution:
The vertices of the triangle are given as:
The area $\Delta$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula: $$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$
Substituting the coordinates $(1, 0)$, $(6, 0)$, and $(4, 3)$ into the matrix expression: $$\Delta = \frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$$
We expand the determinant along the second column ($C_2$) to simplify the calculation, as it contains the most zeros: $$\Delta = \frac{1}{2} \left[ -(0) \begin{vmatrix} 6 & 1 \\ 4 & 1 \end{vmatrix} + (0) \begin{vmatrix} 1 & 1 \\ 4 & 1 \end{vmatrix} - (3) \begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix} \right]$$ [Expanding along $C_2$ based on the cofactor expansion theorem]
Calculating the $2 \times 2$ determinant: $$\begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix} = (1 \times 1) - (6 \times 1) = 1 - 6 = -5$$
Substitute the evaluated determinant back into the area equation: $$\Delta = \frac{1}{2} \left[ -3 \times (-5) \right]$$ $$\Delta = \frac{1}{2} \times 15$$ $$\Delta = 7.5 \text{ square units}$$
Final Solution: The area of the triangle is 7.5 square units.
Solution:
The area of a triangle formed by three vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant formula:
$\text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$
If three points are collinear, they lie on the same line, and the area of the triangle formed by these points is zero [Per the property of collinearity in coordinate geometry].
To find the equation of a line passing through $(3, 1)$ and $(9, 3)$, let an arbitrary point on the line be $(x, y)$. The points $(3, 1)$, $(9, 3)$, and $(x, y)$ are collinear. Therefore:
$\begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row ($R_1$):
$x(1 \cdot 1 - 3 \cdot 1) - y(3 \cdot 1 - 9 \cdot 1) + 1(3 \cdot 3 - 9 \cdot 1) = 0$
$x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0$ [By the Expansion of Determinants property]
$-2x - y(-6) + 1(0) = 0$
$-2x + 6y = 0$
Divide the entire equation by $-2$:
$x - 3y = 0$
Final Solution: x - 3y = 0
Solution:
To find the equation of a line passing through two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ using determinants, we utilize the condition that three points are collinear if and only if the area of the triangle formed by them is zero. For an arbitrary point $P(x, y)$ lying on the line joining $(x_1, y_1)$ and $(x_2, y_2)$, the area of the triangle is defined as:
$$\Delta = \frac{1}{2} \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0$$
[By the condition of collinearity of three points in a 2D plane].
Given the points $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 6)$, we substitute these into the determinant formula:
$$\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$$
Expanding along the first row ($R_1$):
$$x \begin{vmatrix} 2 & 1 \\ 6 & 1 \end{vmatrix} - y \begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = 0$$
[By the Laplace expansion method for determinants].
Calculating the $2 \times 2$ determinants:
Substituting these values back into the equation:
$$-4x + 2y + 0 = 0$$
Divide the equation $-4x + 2y = 0$ by $-2$ to obtain the standard linear form:
$$2x - y = 0 \implies y = 2x$$
Final Solution: 2x - y = 0
