Determinants Exercise 03
Question:
Find values of k if area of triangle is 4 sq. units and vertices are (i) (k,0),(4,0),(0,2) (ii) (2,0),(0,4),(0,k)
Answer:
(i) Let A(k,0), B(4,0) and C(0,2) be the vertices of the triangle.
Using the formula for area of triangle, Area = (1/2) * (x1 * y2  x2 * y1)
4 = (1/2) * (k * 2  4 * 0)
4 = k
Therefore, the value of k is 4.
(ii) Let A(2,0), B(0,4) and C(0,k) be the vertices of the triangle.
Using the formula for area of triangle, Area = (1/2) * (x1 * y2  x2 * y1)
4 = (1/2) * (2 * k  0 * 4)
4 = k
Therefore, the value of k is 4.
Question:
Find area of triangle with vertices at the point given in each of the following (i)(1,0),(6,0),(4,3) (ii)(2,7),(1,1),(10,8) (iii)(−2,−3),(3,2),(−1,−8)
Answer:
(i) Area of triangle with vertices at (1,0), (6,0), (4,3)
Step 1: Calculate the lengths of the sides of the triangle.
Side 1 = 6  1 = 5 units
Side 2 = 4  6 = 2 units
Side 3 = √[(4  1)2 + (3  0)2] = √25 = 5 units
Step 2: Calculate the semiperimeter (s) of the triangle.
Semiperimeter (s) = (5 + 2 + 5) / 2 = 12 / 2 = 6 units
Step 3: Calculate the area of the triangle using Heron’s formula.
Area = √[s(s  a)(s  b)(s  c)]
Area = √[6(6  5)(6  2)(6  5)]
Area = √[6(1)(4)(1)]
Area = √24 = 4.9 units2
(ii) Area of triangle with vertices at (2,7), (1,1), (10,8)
Step 1: Calculate the lengths of the sides of the triangle.
Side 1 = 1  2 = 1 unit
Side 2 = 10  1 = 9 units
Side 3 = √[(10  2)2 + (8  7)2] = √73 = 8.5 units
Step 2: Calculate the semiperimeter (s) of the triangle.
Semiperimeter (s) = (1 + 9 + 8.5) / 2 = 18.5 / 2 = 9.25 units
Step 3: Calculate the area of the triangle using Heron’s formula.
Area = √[s(s  a)(s  b)(s  c)]
Area = √[9.25(9.25  1)(9.25  9)(9.25  8.5)]
Area = √[9.25(8.25)(0.25)(0.75)]
Area = √6.5625 = 2.56 units2
(iii) Area of triangle with vertices at (−2,−3), (3,2), (−1,−8)
Step 1: Calculate the lengths of the sides of the triangle.
Side 1 = 3  (2) = 5 units
Side 2 = (1)  3 = 4 units
Side 3 = √[(3  (2))2 + (2  (3))2] = √25 = 5 units
Step 2: Calculate the semiperimeter (s) of the triangle.
Semiperimeter (s) = (5 + 4 + 5) / 2 = 14 / 2 = 7 units
Step 3: Calculate the area of the triangle using Heron’s formula.
Area = √[s(s  a)(s  b)(s  c)]
Area = √[7(7  5)(7  4)(7  5)]
Area = √[7(2)(3)(2)]
Area = √42 = 6.48 units2
Question:
Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear
Answer:

Find the equation of the line passing through points A and B. y  b = (c + a  b)/(a  b) (x  a)

Find the equation of the line passing through points B and C. y  c = (a + b  c)/(b  c) (x  b)

Equate the equations of the two lines. (c + a  b)/(a  b) (x  a) = (a + b  c)/(b  c) (x  b)

Simplify the equation. (a  b) (x  a) = (b  c) (x  b)

Solve for x. x = (a (b  c) + b (c + a))/(a  b)

Substitute x in either of the equations to get the value of y. y = (c + a  b)/(a  b) (x  a) y = (c + a  b)/(a  b) [(a (b  c) + b (c + a))/(a  b)  a] y = (c + a  b)/(a  b) [b (c + a)/(a  b)  a] y = (c + a  b) [(c + a)/(a  b)  a] y = (c + a  b) (c + a  a) y = (c + a  b) c

Thus, the equation of the line passing through points A, B and C is given by: y = (c + a  b) cx.
Therefore, the points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
Question:
If area of triangle is 35 sq units with vertices (2,−6),(5,4) and (k,4). Then k is A 12 B −2 C −12,−2 D 12,−2
Answer:
Step 1: Calculate the length of the sides of the triangle.
Step 2: Use the formula for the area of a triangle, A = 1/2(base x height).
Step 3: Substitute the lengths of the sides and the coordinates of the vertices into the formula.
Step 4: Solve for k.
Answer: A 12
Question:
(i) Find equation of line joining (1,2) and (3,6) using determinants (ii) Find equation of line joining (3,1) and (9,3) using determinants.
Answer:
(i) Equation of line joining (1,2) and (3,6) using determinants: Let x = x1 and y = y1
D = x1 1 x2 1
D = (x1  x2)
Substituting the given coordinates,
D = (1  3) = 2
Therefore, equation of line joining (1,2) and (3,6) is y  2 = 2(x  1)
y  2 = 2x + 2
y = 2x + 4
(ii) Equation of line joining (3,1) and (9,3) using determinants: Let x = x1 and y = y1
D = x1 1 x2 1
D = (x1  x2)
Substituting the given coordinates,
D = (3  9) = 6
Therefore, equation of line joining (3,1) and (9,3) is y  1 = 6(x  3)
y  1 = 6x + 18
y = 6x + 19