Area of a Triangle using Coordinate Geometry | Coordinate Geometry | Assessment | English | Grade 10

The area of a triangle with vertices A(0,4), B ( ─5,3) and C (1, ─1) is

• 15 square units

• 13 square units

• 3 square units

• Cannot be determined

Answer explanation

Point A ( 0, 4) -- x₁ = 0, y₁ = 4 Point B ( ─ 5, 3) --- x₂ = ─ 5, y₂ = 3 Point C (1,─1) --- x₃ = 1, y₃ = ─ 1 Area of ∆ ABC = ½ [ x₁(y₂ ─ y₃) + x₂ ( y₃ ─ y₁) + x₃(y₁ ─ y₂)] Area of ∆ ABC = ½ [ 0 + (─5)(─1─4) + 1(4 ─3)] = ½[25+1] Area of ∆ ABC = 26/2 = 13 square units So Option 2 is the correct answer

Find the area of a triangle whose vertices are (─3,1), ( 0, ─2) and (1,3).

• Area = 18 square units

• Area = 9 square units

• Area = 6 square units

Answer explanation

Point A (─3, 1) --- x₁ = ─ 3, y₁ = 1 Point B (0, ─ 2) --- x₂ = 0, y₂ = ─ 2 Point C (1, 3) --- x₃ = 1, y₃ = 3 Area of ∆ ABC =½ [ x₁(y₂ ─ y₃) + x₂( y₃ ─ y₁) + x₃(y₁ ─ y₂)] Area of ∆ ABC = ½ [ (─ 3)(─2 ─3) + 0 + 1(1 + 2)] Area of ∆ ABC = ½ [15+3] = 18/2 = 9 square units So Option 2 is the correct answer

Are the following points are collinear? L ( 2, ─5) M (1, ─ 3) N ( ─2, 3)

• Yes

• No

• Cannot be determined

Answer explanation

Point L ( 2, ─ 5) -- x₁ = 2, y₁ = ─ 5 Point M (1, ─ 3) --- x₂ = 1 , y₂ = ─ 3 Point N (─2, 3) --- x₃ = ─ 2, y₃ = 3 Area of ∆ LMN =½ [ x₁(y₂ ─ y₃) + x₂ ( y₃ ─ y₁) + x₃(y₁ ─ y₂)] Area of ∆ LMN = ½ [ 2(─3─3) + (1)(3+5) + (─2)(─5+3)] Area of ∆ LMN = ½ [─12 +8 +4] = 0 square units Since the area of the triangle formed by the three points is Zero, so the three points are collinear. So Option 1 is the correct answer

Find the value of k for which the following points are collinear P ( ─ 2, 3) Q (1, 2) R ( 4, k)

• k = 0

• k = 3

• k = 1

• Cannot be determined

Answer explanation

Point P ( ─ 2, 3) -- x₁ = ─ 2, y₁ = 3 Point Q (1, 2) --- x₂ = 1 , y₂ = 2 Point R ( 4, k) --- x₃ = 4, y₃ = k If the area of the triangle formed by the three points is Zero, Then the three points are collinear Here let us assume that Area of ∆ PQR = 0 Area of ∆ PQR =½ [ x₁(y₂ ─ y₃) + x₂ ( y₃ ─ y₁) + x₃(y₁ ─ y₂)] 0 = ½ [ (─2)( 2 ─ k) + (1)(k ─ 3) + (4)(3 ─ 2)] 0 = ½ [─ 4 + 2k + k ─ 3 + 4] 0 = ½ ( 3k ─ 3) 3k = 3 k = 1 So Option 3 is the correct answer

The area of a parallelogram whose vertices are P(─4, ─7), Q ( ─1,2) , R ( 8, 5) and S (5, ─4) is

• 36 square units

• 72 square units

• Cannot be determined

Answer explanation

Point P ( ─ 4, ─ 7) -- x₁ = ─ 4, y₁ = ─ 7 Point Q ( ─ 1, 2) --- x₂ = ─ 1, y₂ = 2 Point R (8, 5) --- x₃ = 8, y₃ = 5 Point S ( 5, ─ 4) --- x₄ = 5, y₄ = ─ 4 Area of parallelogram PQRS = Area of ∆ PQS + Area of ∆ QRS Area of ∆ =½ [ x₁(y₂ ─ y₃) + x₂ ( y₃ ─ y₁) + x₃(y₁ ─ y₂)] Area of ∆ PQS = ½ [ (─4)(2 +4) +(─1)( ─4 + 7) + 5(─7 ─2)] Area of ∆ PQS = ½[ ─24 ─ 3 ─45] = 72/2 = 36 square units ( Negative sign not considered as area cannot be negative) Area of ∆ QRS = ½ [ (─1)(5 +4) + 8( ─4 ─2) + 5(2 ─5)] Area of ∆ QRS = ½ [ ─9 ─ 48 ─15] = 72/2 = 36 square units ( Negative sign not considered as area cannot be negative) Area of parallelogram PQRS = Area of ∆ PQS + Area of ∆ QRS = 36 + 36 = 72 square units So Option 2 is the correct answer