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Q4(i):
(i) Find equation of line joining $(1, 2)$ and $(3, 6)$ using determinants.
Solution :
Step 1: Theoretical Foundation
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].
Step 2: Substitution of Given Coordinates
Given the points $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 6)$, we substitute these into the determinant formula:
$$\begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0$$
Step 3: Expansion of the Determinant
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].
Step 4: Algebraic Evaluation
Calculating the $2 \times 2$ determinants:
- $x(2(1) - 6(1)) = x(2 - 6) = -4x$
- $-y(1(1) - 3(1)) = -y(1 - 3) = -y(-2) = 2y$
- $1(1(6) - 3(2)) = 1(6 - 6) = 0$
Substituting these values back into the equation:
$$-4x + 2y + 0 = 0$$
Step 5: Simplification
Divide the equation $-4x + 2y = 0$ by $-2$ to obtain the standard linear form:
$$2x - y = 0 \implies y = 2x$$
Visual Representation
Final Solution: 2x - y = 0
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) $(–2, –3), (3, 2), (–1, –8)$
- Q2: Show that points A $(a, b + c)$, B $(b, c + a)$, C $(c, a + b)$ are collinear.
- 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(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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