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 ALBM 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.

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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.

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 Co-ordinate Geometry 





11 and 12 Grade Math 

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