Concept 1: Area of a Triangle using Determinants
In previous classes, we learned that the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the expression:
\[ \text{Area} = \frac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)| \]
This expression can be elegantly represented in the form of a determinant of order $3 \times 3$.
Key Properties and Formulas
- The Formula: The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:
\[ \Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]
- Absolute Value: Since area is a physical quantity, it is always positive. We take the absolute value of the determinant result.
- Given Area: If the area is already provided in a problem, use both positive and negative values of the determinant for calculations (e.g., if Area = 4, set the determinant expression equal to $\pm 4$).
- Order of Vertices: The value of the area remains unchanged regardless of the order in which the vertices are listed in the determinant.
Fig: Representation of Triangle in Cartesian Plane
Solved Examples (Concept 1)
Example 1: Find the area of the triangle whose vertices are $(3, 8)$, $(-4, 2)$, and $(5, 1)$.
Solution:
\[ \Delta = \frac{1}{2} \begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix} \]
Expanding along $R_1$:
\[ \Delta = \frac{1}{2} [3(2-1) – 8(-4-5) + 1(-4-10)] \]
\[ \Delta = \frac{1}{2} [3(1) – 8(-9) + 1(-14)] = \frac{1}{2} [3 + 72 – 14] = \frac{61}{2} = 30.5 \text{ sq. units.} \]
Example 2: Find the area of a triangle with vertices at $(2, 7)$, $(1, 1)$, and $(10, 8)$.
Solution:
\[ \Delta = \frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix} \]
Applying $R_2 \to R_2 – R_1$ and $R_3 \to R_3 – R_1$:
\[ \Delta = \frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ -1 & -6 & 0 \\ 8 & 1 & 0 \end{vmatrix} \]
Expanding along $C_3$:
\[ \Delta = \frac{1}{2} [1(-1 – (-48))] = \frac{1}{2} [47] = 23.5 \text{ sq. units.} \]
Example 3 (HOTS): If the area of a triangle with vertices $(k, 0)$, $(4, 0)$, and $(0, 2)$ is 4 square units, find the value of $k$.
Solution: Given $\Delta = 4$.
\[ \frac{1}{2} \begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4 \]
Expanding along $C_2$:
\[ \frac{1}{2} [-2(k – 4)] = \pm 4 \implies -(k-4) = \pm 4 \]
Case 1: $-k + 4 = 4 \implies k = 0$.
Case 2: $-k + 4 = -4 \implies k = 8$.
Therefore, $k = 0, 8$.
Example 4: Show that the points $A(a, b+c)$, $B(b, c+a)$, and $C(c, a+b)$ are collinear.
Solution: Points are collinear if the area of the triangle is 0.
\[ \Delta = \frac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} \]
Applying $C_2 \to C_2 + C_1$:
\[ \Delta = \frac{1}{2} \begin{vmatrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{vmatrix} \]
Taking $(a+b+c)$ common from $C_2$:
\[ \Delta = \frac{1}{2} (a+b+c) \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix} \]
Since $C_2$ and $C_3$ are identical, the determinant is 0. Hence, the points are collinear.
Example 5: Find the area of a triangle whose vertices are $(1, 0), (6, 0), (4, 3)$.
Solution:
\[ \Delta = \frac{1}{2} \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix} = \frac{1}{2} [-3(1-6)] = \frac{1}{2} [-3(-5)] = 7.5 \text{ sq. units.} \]
Concept 2: Condition of Collinearity and Equation of a Line
Determinants provide a direct method to check if three points lie on a single straight line and to find the equation of a line passing through two given points.
Key Definitions and Formulas
- Condition for Collinearity: Three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear if and only if the area of the triangle formed by them is zero:
\[ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]
- Equation of a Line: To find the equation of a line passing through two points $A(x_1, y_1)$ and $B(x_2, y_2)$, we take an arbitrary point $P(x, y)$ on the line. Since $A, B, P$ are collinear:
\[ \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0 \]
Solved Examples (Concept 2)
Example 1: Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants.
Solution: Let $P(x, y)$ be any point on the line.
\[ \begin{vmatrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0 \]
Expanding along $R_1$: $x(2-6) – y(1-3) + 1(6-6) = 0$
\[ -4x + 2y = 0 \implies 2y = 4x \implies y = 2x. \]
Example 2: Find the value of $x$ such that the points $(x, -1)$, $(2, 1)$, and $(4, 5)$ are collinear.
Solution: For collinearity:
\[ \begin{vmatrix} x & -1 & 1 \\ 2 & 1 & 1 \\ 4 & 5 & 1 \end{vmatrix} = 0 \]
$x(1-5) – (-1)(2-4) + 1(10-4) = 0$
\[ -4x + (-2) + 6 = 0 \implies -4x + 4 = 0 \implies x = 1. \]
Example 3: Find the equation of the line passing through the points $(3, 1)$ and $(9, 3)$ using determinants.
Solution:
\[ \begin{vmatrix} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{vmatrix} = 0 \]
$x(1-3) – y(3-9) + 1(9-9) = 0 \implies -2x + 6y = 0 \implies x – 3y = 0$.
Example 4: If the points $(a, 0)$, $(0, b)$, and $(x, y)$ are collinear, prove that $\frac{x}{a} + \frac{y}{b} = 1$.
Solution: For collinearity:
\[ \begin{vmatrix} a & 0 & 1 \\ 0 & b & 1 \\ x & y & 1 \end{vmatrix} = 0 \implies a(b-y) – 0 + 1(0-bx) = 0 \]
\[ ab – ay – bx = 0 \implies bx + ay = ab \]
Dividing by $ab$: $\frac{x}{a} + \frac{y}{b} = 1$.
Example 5: Using determinants, find the equation of the line joining the origin to $(5, -2)$.
Solution: Vertices are $(0, 0)$ and $(5, -2)$. Let $P(x, y)$ be on the line.
\[ \begin{vmatrix} x & y & 1 \\ 0 & 0 & 1 \\ 5 & -2 & 1 \end{vmatrix} = 0 \]
Expanding along $R_2$: $-0 + 0 – 1(-2x – 5y) = 0 \implies 2x + 5y = 0$.
Self Exercise & HOTS
Find the area of a triangle with vertices $(2, -6)$, $(5, 4)$, and $(k, 4)$. If the area is 35 sq. units, find the value(s) of $k$.
Prove that the area of a triangle with vertices $(t, t-2)$, $(t+2, t+2)$, and $(t+3, t)$ is independent of $t$.
Use determinants to find the value of $k$ so that the points $(k, 2-2k)$, $(-k+1, 2k)$, and $(-4-k, 6-2k)$ are collinear.
Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants. Also, find $k$ if $D(k, 0)$ is a point such that area of $\Delta ABD$ is 3 sq. units.
If the points $(x, -2), (5, 2),$ and $(8, 8)$ are collinear, find $x$ using the determinant method.
The area of a triangle with vertices $(3, 8), (-4, 2),$ and $(5, 1)$ is $\Delta$. If the vertices are shifted to $(4, 9), (-3, 3),$ and $(6, 2)$, what is the new area? Explain why using properties.
Find the equation of the line passing through $(a, b)$ and $(a’, b’)$.
If $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$ are vertices of an equilateral triangle with side $a$, prove that $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = 3a^4$.
Find the area of the quadrilateral whose vertices are $(1, 1), (3, 4), (5, -2),$ and $(4, -7)$ by dividing it into two triangles.
If the points $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ are collinear, show that $\frac{y_2-y_3}{x_2x_3} + \frac{y_3-y_1}{x_3x_1} + \frac{y_1-y_2}{x_1x_2} = 0$ (Assume $x_1, x_2, x_3 \neq 0$).
Find the area of the triangle whose vertices are $(2, 7)$, $(1, 1)$, and $(10, 8)$ using the determinant method.
Show that the points $A(a, b+c)$, $B(b, c+a)$, and $C(c, a+b)$ are collinear using determinants.
Find the values of $k$ if the area of the triangle is 4 sq. units and vertices are $(k, 0)$, $(4, 0)$, and $(0, 2)$.
Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants.
If the area of a triangle with vertices $(2, -6)$, $(5, 4)$, and $(k, 4)$ is 35 sq. units, find the value(s) of $k$.
Use determinants to find the area of the triangle with vertices $(-2, -3)$, $(3, 2)$, and $(-1, -8)$.
Determine if the points $(3, 8)$, $(-4, 2)$, and $(10, 14)$ are collinear.
Find the equation of the line passing through $(3, 1)$ and $(9, 3)$ using the property of determinants.
If the points $(x, -2)$, $(5, 2)$, and $(8, 8)$ are collinear, find the value of $x$.
A triangle has vertices $(1, 0)$, $(6, 0)$, and $(4, 3)$. Find its area. Also, find the area of a triangle whose vertices are doubled: $(2, 0)$, $(12, 0)$, and $(8, 6)$.
Find the value of $k$ such that the points $(1, -1)$, $(2, 1)$, and $(4, k)$ are collinear.
Find the area of the quadrilateral $ABCD$ whose vertices are $A(1, 2)$, $B(6, 2)$, $C(5, 5)$, and $D(2, 4)$ by dividing it into two triangles and using determinants.
Find the equation of a line passing through the origin and the point $(x_1, y_1)$ using determinants.
If the area of the triangle formed by vertices $(x, 4)$, $(2, -6)$, and $(5, 4)$ is 35 sq. units, find $x$.
Prove that the area of the triangle with vertices $(t, t-2)$, $(t+2, t+2)$, and $(t+3, t)$ is independent of $t$.