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 semi-perimeter (s) of the triangle.

Semi-perimeter (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 semi-perimeter (s) of the triangle.

Semi-perimeter (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 semi-perimeter (s) of the triangle.

Semi-perimeter (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:

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

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

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

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

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

6. 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

7. 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