Question 1: Find the area of the triangle with vertices at the points given:
(i) (1, 0), (6, 0), (4, 3)
Solution:
Using the formula for the area of triangle: \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|
Substituting the points: \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{array} \right|
Expanding:
\text{Area} = \frac{1}{2} \left( 1(0 \times 1 - 1 \times 3) - 0(6 \times 1 - 1 \times 4) + 1(6 \times 3 - 0 \times 4) \right)
\text{Area}= \frac{1}{2} (1(0-3) - 0(6-4) + 1(18-0))
\text{Area}= \frac{1}{2} (-3 + 0 + 18)
\text{Area}= \frac{1}{2} \times 15
\boxed{\text{Area}= 7.5 \text{ square units}}
(ii) (2, 7), (1, 1), (10, 8)
Solution:
Substituting the points: \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{array} \right|
Expanding:
\text{Area}= \frac{1}{2} (2(1 \times 1 - 1 \times 8) - 7(1 \times 1 - 1 \times 10) + 1(1 \times 8 - 10 \times 1))
\text{Area}= \frac{1}{2} (2(1-8) -7(1-10) + 1(8-10))
\text{Area}= \frac{1}{2} (2(-7) -7(-9) + 1(-2))
\text{Area}= \frac{1}{2} (-14 + 63 + -2)
\text{Area}= \frac{1}{2} \times 47
\boxed{ \text{Area}= \frac{47}{2} \text{ square units}}
(iii) (-2, -3), (3, 2), (-1, -8)
Solution:
Substituting the points: \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{array} \right|
Expanding:
\text{Area}= \frac{1}{2} (-2(2 \times 1 - 1 \times -8) - (-3)(3 \times 1 - 1 \times -1) + 1(3 \times -8 - 2 \times -1))
\text{Area}= \frac{1}{2} (-2(2+8) + 3(3+1) + 1(-24+2))
\text{Area}= \frac{1}{2} (-2(10) + 3(4) + (-22))
\text{Area}= \frac{1}{2} (-20 + 12 - 22)
\text{Area}= \frac{1}{2} \times (-30)
\boxed{\text{Area}= 15 \text{ square units}}
Question 2: Show that points A(a, b+c), B(b, c+a), C(c, a+b) are collinear.
Solution:
If points are collinear the area of a triangle formed by them is zero.
\text{Area} = \frac{1}{2} \left| \begin{array}{ccc} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array} \right|
Expand the determinant along the first row
\text{Area} = \frac{1}{2} \left[
a \left( (c+a)(1) - (1)(a+b) \right) - (b+c) \left( b(1) - c(1) \right) + 1 \left( b(a+b) - c(c+a) \right) \right]
\text{Area}= \frac{1}{2} \left[ a(c - b) - (b^2 - c^2) + ab + b^2 - c^2 - ac \right]
Group like terms:
\text{Area}= \frac{1}{2} \left[ a(c - b) + ab - ac - b^2 + c^2 + b^2 - c^2 \right]
Now simplify:
\text{Area} = \frac{1}{2} \left[ a(c - b) + ab - ac \right]
\text{Area} = \frac{1}{2} \left[ ac - ab + ab - ac \right]
\text{Area} = \frac{1}{2} (0)
area comes out to be 0. Hence, points are collinear.
Question 3: Find the values of k if the area of triangle is 4 sq units with vertices:
(i) (k, 0), (4, 0), (0, 2)
Solution:
Using the formula for area: \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array} \right|
Expanding:
\text{Area}= \frac{1}{2} (k(0-2) - 0(4-0) + 1(4 \times 2 - 0 \times 0))
\text{Area}= \frac{1}{2} (k(-2) + 0 + (8))
\text{Area}= \frac{1}{2} (-2k + 8)
Given that area = 4:
\Rightarrow \left| -2k + 8 \right| = 8
Thus,
-2k + 8 = 8 \quad \text{or} \quad -2k + 8 = -8
Solving:
First case: -2k + 8 = 8 \Rightarrow k = 0
Second case: -2k + 8 = -8 \Rightarrow k = 8
\boxed{k =0 \quad \text{or} \quad k =8}
(ii) (-2, 0), (0, 4), (0, k)
Solution:
Using the formula for area: \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array} \right|
Expanding:
\text{Area}= \frac{1}{2} (-2(4-k) - 0(0-1) + 1(0 \times k - 4 \times 0))
\text{Area}= \frac{1}{2} ((-8+2k) + 0 + 0)
\text{Area}= (k-4)
Given that, Area = 4,
\Rightarrow \left| k-4 \right| = 4
Thus,
k-4 = 4 \quad \text{or} \quad k-4 = -4
Solving:
First case: k-4 = 4 \Rightarrow k = 8
Second case: k-4 = -4 \Rightarrow k = 0
\boxed{k =8 \quad \text{or} \quad k =0}
Question 4: Find the equation of the line joining the following points using determinants:
(i) (1,2) and (3,6)
Solution:
Consider a General point P(x,y) on the line
Using the formula for area:
If 3 points lie on a line then the area of triangle formed by them is 0
\therefore \quad \left| \begin{array}{ccc} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array} \right| = 0
Expanding along 1st row:
x(2-6) - y(1-3) + 1(1 \times 6 - 2 \times 3) = 0
x(-4) - y(-2) + (6-6) = 0
-4x + 2y = 0
2y = 4x
\boxed{ Answer: y = 2x}
(ii) (3,1) and (9,3)
Solution:
Consider a General point P(x,y) on the line
Using the formula:
If 3 points lie on a line then the area of triangle formed by them is 0
\therefore \quad \left| \begin{array}{ccc} x & y & 1 \\ 3 & 1 & 1 \\ 9 & 3 & 1 \end{array} \right| = 0
Expanding along 1st row:
x(1-3) - y(3-9) + 1(3 \times 3 - 1 \times 9) = 0
x(-2) - y(-6) + (9-9) = 0
-2x + 6y = 0
\boxed{Answer: x - 3y = 0}
Question 5: If the area of triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4), then find k.
Solution:
\text{Area} = \frac{1}{2} \left| \begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array} \right|
Expanding:
\text{Area} = \frac{1}{2} (2(4-4) - (-6)(5\times1-k\times1) + 1(5 \times 4 - 4 \times k))
\text{Area}= \frac{1}{2} (0 + 6(5-k) + (20 - 4k))
\text{Area}= \frac{1}{2} ((30 - 6k) +(20 - 4k))
\text{Area}= \frac{1}{2} (50 - 10k)
\text{Area}= (25 - 5k)
Given that, Area = 35
\Rightarrow \left| 25 -5k \right| = 35
Thus,
25-5k = 35 \quad \text{or} \quad 25-5k = -35
Solving:
First case: 25-5k = 35 \Rightarrow k = -2
Second case: 25k-5k = -35 \Rightarrow k = 12
\boxed{ Answer: k = -2 \quad \text{or} \quad k =12}