Area of the Triangle Formed by Three co-ordinate Points

Here we will discuss about the area of the triangle formed by three co-ordinate points.

How to find the area of the triangle formed by joining the three given points?

(A) In Terms of Rectangular Cartesian Co-ordinates:

Let (x₁, y₁), (x₂, y₂) and (x₃, y₃) be the co-ordinates of the vertices A, B, C respectively of the triangle ABC. We are to find the area of the triangle ABC.

$Area of the triangle formed by three co-ordinate points$

Draw AL, BM and CN perpendiculars from A, B and C respectively on the x-axis.

Then, we have, OL = x₁, OM = x₂, ON = x₃ and AL = y₁, BM = y₂, CN = y₃.

Therefore, LM = OM - OL = x₂ - x₁;

NM = OM - ON = x₂ - x₃;

and LN = ON - OL = x₃ - x₁.

Since the area of a trapezium = $\frac{1}{2}$ × the sum of the parallel sides × the perpendicular distance between them,

Hence, the area of the triangle ABC = ∆ABC

= area of the trapezium ALNC + area of the trapezium CNMB - area of the trapezium ALMB

= $\frac{1}{2}$ ∙ (AL + NC) . LN + $\frac{1}{2}$ ∙ (CN + BM) ∙ NM - $\frac{1}{2}$ ∙ (AL + BM).LM

= $\frac{1}{2}$ ∙ (y₁ + y₃) (x₃ - x₁) + $\frac{1}{2}$ ∙ (y₃ + y₂) (x₂ - x₃) - $\frac{1}{2}$ ∙ (y₁ + y₂) (x₂ - x₁)

= $\frac{1}{2}$ ∙ [x₁ y₂ - y₁ x₂ + x₂ y₃ - y₂ x₃ + x₃ y₁ - y₃ x₁]

= $\frac{1}{2}$[x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)] sq. units.

Note:

(i) The area of the triangle ABC can also be expressed in the following form:

∆ ABC= $\frac{1}{2}$[y₁ (x₂ - x₃) + y₂ (x₃ - x₁) + y₃ (x₁ - x₂)] sq. units.

(ii)The above expression for the area of the triangle ABC will be positive if the vertices A, B, C are taken in the anti-clockwise direction as shown in the given figure;

$Anti-clockwise direction$

on the contrary, the expression for the area of the triangle will be negative if the vertices A, B and C are taken in the clockwise direction as show in the given figure.

$Clockwise direction$

However, in either case the numerical value of the expression would be the same.

Therefore, for any position of the vertices A, B and C we can write,

∆ ABC = $\frac{1}{2}$| x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂) | sq. units.

$short-cut method to find Area of the Triangle$

(iii) The following short-cut method is often used to find the area of the triangle ABC:

Write in three rows the co-ordinates (x₁, y₁), (x₂, y₂) and (x₃, y₃) of the vertices A, B, C respectively and at the last row write again the co-ordinates (x₁, y₁), of the vertex A. Now, take the sum of the product of digits shown by (↘) and from this sum subtract the sum of the products of digits shown by (↗). The required area of the triangle ABC will be equal to half the difference obtained. Thus,

∆ ABC = $\frac{1}{2}$| (x₁ y₂ + x₂ y₃ + x₃ y₁) - (x₂ y₁ + x₃ y₂ + x₁ y₃) | sq. units.

(B) In Terms of Polar Co-ordinates:

Let (r₁, θ₁), (r₂, θ₂) and (r₃, θ₃) be the polar co-ordinates of the vertices A, B, C respectively of the triangle ABC referred to the pole O and initial line OX.

Then, OA = r₁, OB = r₂, OC = r₃

and ∠XOA = θ₁, ∠XOB = θ₂, ∠ XOC = θ₃

Clearly, ∠AOB = θ₁ - θ₂; ∠BOC = θ₃ - θ₂ and ∠COA = θ₁ - θ₃

$Polar Co-ordinates area$

Now, ∆ ABC = ∆ BOC + ∆ COA - ∆ AOB

= $\frac{1}{2}$ OB ∙ OC ∙ sin ∠BOC + $\frac{1}{2}$ OC ∙ OA ∙ sin ∠COA - $\frac{1}{2}$ OA ∙ OB ∙ sin ∠AOB

= $\frac{1}{2}$ [r₂ r₃ sin (θ₃ – θ₂) + r₃ r₁ sin (θ₁ - θ₃) - r₁ r₂ sin (θ₁ - θ₂)] square units

As before, for all positions of the vertices A, B, C we shall have,

∆ABC = $\frac{1}{2}$| r₂ r₃ sin (θ₃ – θ₂) + r₂ r₃ sin (θ₁ - θ₃) - r₁ r₂ sin (θ₁ - θ₂) | square units.

Examples on area of the triangle formed by three co-ordinate points:

Find the area of the triangle formed by joining the point (3, 4), (-4, 3) and (8, 6).

Solution:

We know that, ∆ ABC = $\frac{1}{2}$| (x₁ y₂ + x₂ y₃ + x₃ y₁) - (x₂ y₁ + x₃ y₂ + ₁ y₃) | sq. units.

The area of the triangle formed by joining the given point

= $\frac{1}{2}$| [9 + (-24) + 32] - [-16 + 24 + 18] | sq. units

= $\frac{1}{2}$| 17 - 26 | sq. units

= $\frac{1}{2}$ | – 9 | sq. units

= $\frac{9}{2}$sq. units.

● Co-ordinate Geometry