AREA OF TRIANGLE WITH 3 COORDINATES

Problem 1 :

Find the area of a triangle whose vertices are (3, 0), (7, 0) and (8, 4).

Solution:

The vertices of the given triangle A(3, 0), B(7, 0) and C(8, 4).

x₁ = 3, y₁ = 0, x₂ = 7, y₂ = 0, x₃ = 8, y₃ = 4

Area of a  triangle

Area can't be negative.

Therefore, the area of the triangle is 20 square units.

Problem 2 :

The area of a triangle whose vertices are (5, 0), (8, 0) and (8, 4) (in sq.units) is 

A) 20    B) 12    C) 6

Solution:

The vertices of the given triangle A(5, 0), B(8, 0) and C(8, 4)

x₁ = 5, y₁ = 0, x₂ = 8, y₂ = 0, x₃ = 8, y₃ = 4

Area of a  triangle

So, option (C) is correct.

Problem 3 :

The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (7/2, y), find the value of y.

Solution:

Let A(2, 1), B(3, -2) and C(7/2, y)

x₁ = 2, y₁ = 1, x₂ = 3, y₂ = -2, x₃ = 7/2, y₃ = y

Area of a triangle

Problem 4 :

Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

Solution:

The vertices of the given triangle A(1, -1), B(-4, 2k) and C(-k, -5)

x₁ = 1, y₁ = -1, x₂ = -4, y₂ = 2k, x₃ = -k, y₃ = -5

Area of a  triangle

Problem 5 :

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Solution:

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).

Hence, the area of the required triangle is 1 sq. unit.

Problem 6 :

For what type of k, (k > 0), is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) to 53 sq. units?

Solution:

The vertices of the given triangle A(-2, 5), B(k, -4) and C(2k + 1, 10)

x₁ = -2, y₁ = 5, x₂ = k, y₂ = -4, x₃ = 2k + 1, y₃ = 10

Area of a  triangle