Here we will discuss about the area of the triangle formed by three coordinate points.
(A) In Terms of Rectangular Cartesian Coordinates:
Let (x₁, y₁), (x₂, y₂) and (x₃, y₃) be the coordinates of the vertices A, B, C respectively of the triangle ABC. We are to find the area of the triangle ABC.
Draw AL, BM and CN perpendiculars from A, B and C respectively on the xaxis.
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 anticlockwise direction as shown in the given figure;
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.
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.
(iii) The following shortcut method is often used to find the area of the triangle ABC:
Write in three rows the coordinates (x₁, y₁), (x₂, y₂) and (x₃, y₃) of the vertices A, B, C respectively and at the last row write again the coordinates (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 Coordinates:
Let (r₁, θ₁), (r₂, θ₂) and (r₃, θ₃) be the polar coordinates 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
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 coordinate 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.
`● Coordinate Geometry
11 and 12 Grade Math
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