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Q2:
Show that points A $(a, b + c)$, B $(b, c + a)$, C $(c, a + b)$ are collinear.
Solution :
Step 1: Theoretical Foundation for Collinearity
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]
Step 2: Substitution of Coordinates
Given the points $A(a, b + c)$, $B(b, c + a)$, and $C(c, a + b)$, we substitute these into the determinant:
$\Delta = \frac{1}{2} \begin{vmatrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{vmatrix}$
Step 3: Determinant Property Application
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 $C_i \to C_i + kC_j$]
Step 4: Factorization and Zero Property
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 $0$]
$\Delta = \frac{1}{2} (a + b + c) \times 0 = 0$
Visual Representation of Linear Alignment
Conclusion
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.
More Questions from Class 12 Mathematics Determinants EXERCISE 4.2
- Q1(i): Find area of the triangle with vertices at the point given in each of the following: (i) $(1, 0), (6, 0), (4, 3)$
- Q1(ii): Find area of the triangle with vertices at the point given in each of the following: (ii) $(2, 7), (1, 1), (10, 8)$
- Q1(iii): Find area of the triangle with vertices at the point given in each of the following: (iii) $(–2, –3), (3, 2), (–1, –8)$
- Q3(i): Find values of $k$ if area of triangle is 4 sq. units and vertices are (i) $(k, 0), (4, 0), (0, 2)$
- Q3(ii): Find values of $k$ if area of triangle is 4 sq. units and vertices are (ii) $(–2, 0), (0, 4), (0, k)$
- Q4(i): (i) Find equation of line joining $(1, 2)$ and $(3, 6)$ using determinants.
- Q4(ii): (ii) Find equation of line joining $(3, 1)$ and $(9, 3)$ using determinants.
- Q5: If area of triangle is 35 sq units with vertices $(2, – 6), (5, 4)$ and $(k, 4)$. Then $k$ is
CBSE Solutions for Class 12 Mathematics Determinants
Chapters in CBSE - Class 12 Mathematics
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