**Area of triangle worksheet 1 :**

Area of triangle worksheet 1 is much useful to the students who would like to practice problems on area of triangles using the given three vertices.

1) Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)

2) If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".

3) Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

4) If P(x, y) is any point on the line segment joining the points (a, 0), and (0, b), then prove that x/a + y/b = 1. where a ≠ b.

**Problem 1 :**

Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)

**Solution : **

Plot the given points in a rough diagram as given below and take them in order (counter clock wise)

Let the vertices be A(1, 2), B(-3, 4) and C(-5, -6)

Then, we have

(x₁, y₁) = (1, 2)

(x₂, y₂) = (-3, 4)

(x₃, y₃) = (-5, -6)

Area of triangle ABC is

= (1/2) x { [x₁y₂ + x₂y₃ + x₃y₁] - [x₂y₁ + x₃y₂ + x₁y₃] }

= (1/2) x { [1.4 + (-3).(-6) + (-5).2] - [(-3).2 + (-5).4 + 1.(-6)] }

= (1/2) x { [4 + 18 - 10] - [-6 - 20 -6] }

= (1/2) x { [12] - [-32] }

= (1/2) x { 12 + 32 }

= (1/2) x { 44 }

= 22 square units.

**Hence, are of triangle ABC is 22 square units.**

Let us look at the next problem on "Area of triangle worksheet 1"

**Problem 2 :**

If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".

**Solution :**

Let (x₁, y₁) = (6, 7), (x₂, y₂) = (-4, 1) and (x₃, y₃) = (a, -9)

Given : Area of triangle ABC = 68 square units

(1/2) x { [x₁y₂ + x₂y₃ + x₃y₁] - [x₂y₁ + x₃y₂ + x₁y₃] } = 68

Multiply by 2 on both sides,

{ [x₁y₂ + x₂y₃ + x₃y₁] - [x₂y₁ + x₃y₂ + x₁y₃] } = 136

{ [6 + 36 + 7a] - [-28 + a - 54] } = 136

[42 + 7a] - [a - 82] = 136

42 + 7a -a +82 = 136

6a + 124 = 136

6a = 12

a = 2

**Hence, the value of "a" is 2. **

Let us look at the next problem on "Area of triangle worksheet 1"

**Problem 3 :**

Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

**Solution :**

Using the concept of area of triangle, if the three points A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are collinear, then

x₁y₂ + x₂y₃ + x₃y₁ = x₂y₁ + x₃y₂ + x₁y₃

Comparing the given points to A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃), we get

(x₁, y₁) = (5, -2)

(x₂, y₂) = (4, -1)

(x₃, y₃) = (1, 2)

x₁y₂ + x₂y₃ + x₃y₁ = 5x(-1) + 4x2 + 1x(-2)

x₁y₂ + x₂y₃ + x₃y₁ = -5 + 8 -2

**x₁y₂ + x₂y₃ + x₃y₁ = 1 --------(1)**

x₂y₁ + x₃y₂ + x₁y₃ = 4x(-2) + 1x(-1) + 5x(2)

x₂y₁ + x₃y₂ + x₁y₃ = -8 - 1 + 10

**x₂y₁ + x₃y₂ + x₁y₃ = 1 --------(2)**

From (1) and (2), we get

**x₁y₂ + x₂y₃ + x₃y₁ = x₂y₁ + x₃y₂ + x₁y₃**

**Hence, the three points A, B and C are collinear.**

Let us look at the next problem on "Area of triangle worksheet 1"

**Problem 4 :**

If P(x, y) is any point on the line segment joining the points (a, 0), and (0, b), then prove that x/a + y/b = 1. where a ≠ b.

**Solution :**

Clearly, the points (x, y), (a, 0), and (0, b) are collinear.

Then, area of the triangle = 0

Since, area of the triangle is zero, we have

** x₁y₂ + x₂y₃ + x₃y₁ = x₂y₁ + x₃y₂ + x₁y₃ ------(1)**

(x₁, y₁) = (x, y)

(x₂, y₂) = (a, 0)

(x₃, y₃) = (0, b)

Plugging the above points in (1),

x.0 + a.b + 0.y = a.y + 0.0 + x.b

0 + a.b + 0 = a.y + 0 + x.b

a.b = a.y + x.b

Divide by ab on bothe sides,

1 = y/b + x/a

or

**x/a + y/b = 1**

After having gone through the stuff given above, we hope that the students would have understood "Area of triangle worksheet 1".

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