**Area of triangle when vertices are given :**

Here, we are going to see, how to find area of a triangle when coordinates of the three vertices are given.

Let us consider the triangle given below.

In the above triangle, A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are the vertices.

To find area of the triangle ABC, now we have take the vertices A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) of the triangle ABC in order (counter clockwise direction) and write them column-wise as shown below.

And the diagonal products x₁y₂, x₂y₃ and x₃y₁ as shown in the dark arrows.

Also add the diagonal products x₂y₁, x₃y₂ and x₁y₃ as shown in the dotted arrows.

Now, subtract the latter product from the former product to get area of the triangle ABC.

Hence, area of the triangle ABC.

Three or more points in a plane are said to be collinear, if they lie on the same straight line.

In other words, three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear, if any one of these points lies on the straight line joining the other two points.

Suppose that the three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear, then they can not form a triangle . Hence, the area of triangle ABC = 0

1/2 x { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) } = 0

(or)

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

One can prove that the converse is also true.

Hence, the area of triangle ABC is zero if and only if the points A, B and C are collinear.

**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 when vertices are given"

**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 when vertices are given"**

**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 when vertices are given"

**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**

**Let us look at the next problem on "Area of triangle when vertices are given"**

**Problem 5 :**

If the points (k, -1), (2, 1) and (4, 5) are collinear, then find the value of "k".

**Solution :**

Since the given points are collinear, area of triangle = 0.

Then, we have

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

(x₁, y₁) = (k, -1)

(x₂, y₂) = (2, 1)

(x₃, y₃) = (4, 5)

Plugging the above points in (1),

k.1 + 2.5 + 4.(-1) = 2.(-1) + 4.1 + k.5

k + 10 - 4 = -2 + 4 + 5k

k + 6 = 2 + 5k

4 = 4k

1 = k

**Hence, the value of "k" is 1.**

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

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