Area of a Triangle Given 3 Points



Solving the problems on area of a triangle given 3 points with the help of the formula, in the below examples use the formula to find the area of a triangle given 3 points.

The area of a triangle formed by joining the points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

½ |y₁ (x₂ - x₃) + y₂ (x₃ - x₁) + y₃ (x₁ - x₂)| sq. units 


Worked-out problems to find the area of a triangle given 3 points:

1. Find the value of x for which the area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is 12¹/₂ sq. units. 

Solution:
 The area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is 

½ |(- 1 - 4x - 4x) - (- 4x + x + 4)| 

= ½ |- 1 - 8x + 3x - 41 = 1/2 |- 5x - 5| sq. units. 

By problem, ½|-1 - 5x - 5| = 12¹/₂ = 25/2 

Therefore, 5x + 5 = ± 25

or, x + 1 = ± 5 

Therefore, x = 4 or, - 6. 



2. The point A, B, C have respective co-ordinates (3, 4), (-4, 3) and (8, -6). Find the area of ∆ ABC and the length of the perpendicular from A on BC.


Solution:
 The required area of the triangle ABC. 

= ½ |(9 + 24 + 32) - (- 16 + 24 - 18)| sq. unites. 

= ½ |65 + 10| sq. units = 75/2 sq. units. 

Again, BC = distance between the points B and C

= √[(8 + 4)² + (- 6 - 3)²] = √[44 + 81] = √225 = 15 units. 

Let p be the required length of the perpendicular from A on BC then, 

½ ∙ BC ∙ p = area of the triangle ABC

or, ½ ∙ 15 ∙ p = 75/2 

or, p = 5

Therefore, the required length of the perpendicular from A on BC is 5 units. 


3. The point A, B, C, D have respective co-ordinates (-2, -3), (6, -5) , (18, 9) and (0, 12). Find the area of quadrilateral ABC. 

Solution: 
We have, the area of the triangle ABC

= ½ |(10 + 54 - 54) - (- 18 - 90 - 18)| sq. units

= ½ (10 + 126) sq. units

= 68 sq. units. 

Again, area of the triangle ACD

= ½ |(- 18 + 216 + 0) - (- 54 + 0 - 24)|sq. units

= ½ (198 + 78) sq. units 

= 138 sq. units. 

Therefore, the required area of the quadrilateral ABCD

= area of the ∆ ABC + area of the ∆ACD

= (68 + 138) sq. units

= 206 sq. units. 



Alternative Method: 


[This method is analogous with the short-cut method of getting the area of a triangle. Suppose, we want to find the area of the quadrilateral whose vertices have co-ordinates (x₁, y₁), (x₂, y₂), (x₃, y₃) and (x₄, y₄). For this, we write the co-ordinates of the vertices in four rows repeating the first written co-ordinates in the fifth row. Now take the sum of the products of digits shown by (↘) and from this sum subtract the sum of the products of digits shown by (↗). The required area of the quadrilateral will be equal to half of the difference obtained. Thus, the area of the quadrilateral

½ |(x₁y₂ + x₂ y₃ + x₃y₄ + x₄y₁) - (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)| sq. units.

The above method can be used to find the area of a polygon of any number of sides when the co-ordinates of its vertices are given.]

Solution: The required area of the quadrilateral ABCD

= ½ |(10 + 54 + 216 + 0) - (- 18 - 90 + 0 - 24)| sq. units.

= ½ (280 + 132) sq. units.

= ½ × 412 sq. units.

= 206 sq. units.


4. The co-ordinates of the points A, B, C, D are (0, -1), (-1, 2), (15, 2) and (4, -5) respectively. Find the ratio in which AC divides BD.

Solution:
Let us assume that the line-segment AC divides the line -segment BD in the ratio m : n at P. Therefore, P divides the line-segment BD in the ratio m : n. Hence, the co-ordinates of P are.

[(m ∙ 4 + n ∙ (-1))/(m + n), (m ∙ (-5) + n ∙ 2)/(m + n)] + [(4m - n)/(m + n), (5m + 2n)/(m + n)].

Clearly, the points A, C and P are collinear. Therefore, the area of the triangle formed by the point A, C and P must be zero.

Therefore, ½ [( 0 + 15 ∙ (- 5m + 2n)/(m + n) - (4m - n)/(m + n) ) - (- 15 + 2 ∙ (4m - n)/(m + n) + 0)] = 0

or, 15 ∙ (-5m + 2n)/(m + n) - (4m - n)/(m + n) + 15 - 2 ∙ (4m - n)/(m + n)=0

or, - 75m + 30n – 4m + n + 15m + 15n - 8m + 2n = 0.

or, - 72m + 48n = 0

or, 72m = 48n

or, m/n = 2/3.

Therefore, the line-segment AC divides the line-segment BD internally in the ratio 2 : 3.


5. The polar co-ordinates of the vertices of a triangle are (-a, π/6), (a, π/2) and (-2a, - 2π/3) find the area of the triangle.

Solution:
The area of the triangle formed by joining the given points

= ½ |a ∙ (-2a) sin ⁡(- 2π/3 - π/2) + (-2a) (-a) sin (π/6 + 2π/3) - (-a) ∙ a sin (π/6 + π/2)| sq. units. [ using above formula]

= ½ |2a² sin (π + π/6 ) + 2a² sin⁡ (π - π/6) -2a² sin⁡ (π/2 - π/6)|sq. units.

= ½ |-2a² sin⁡ π/6 + 2a² sin⁡ π/6 - a² cos⁡ π/6| sq. units.

= ½ ∙ a² ∙ (√3/2) sq. units = (√3/4) a² sq. units.


6. The centre of a circle is at (2, 6) and a chord of this circle of length 24 units is bisected at (- 1, 2). Find the radius of the circle.

Solution:
Let C (2, 6) be the centre of the circle and its chord AB of length 24 units is bisected at D (- 1, 2).

Therefore, CD² = (2 + 1) ₁ + (6 - 2) ²

= 9 + 16 = 25 and DB = ½ ∙ AB = ½ ∙ 24 = 12

Join CB. Now, D is the mid-point of the chord AB; hence, CD is perpendicular to AB. Therefore, from the triangle BCD we get,

BC² = CD² + BD² = 25 + 12² = 25 + 144 = 169

or, BC = 13

Therefore, the required radius of the circle = 13 units.


7. If the co-ordinates of the vertices of a ∆ ABC be (3, 0), (0, 6) and (6, 9) and if D and E divide AB and AC, respectively internally in the ratio 1 : 2, then show that the area of ∆ ABC = 9 ∙ the area of ∆ ADE.

Solution:
By question D divides AB internally in the ratio 1 : 2; hence, the co-ordinates of D are ((1 ∙ 0 + 2 ∙ 3)/(1 + 2), (1 ∙ 6 + 2 ∙ 0)/(1 + 2)) = (6/3, 6/3) = (2, 2).

Again, E divides AC internally in the ratio 1 : 2; hence, the co-ordinates of E are

((1 ∙ 6 + 2 ∙ 3)/(1 + 2), (1 ∙ 9 + 2 ∙ 0)/(1 + 2)) = (12/3, 9/3) = (4, 3).

Now, the area of the triangle ABC

= ½ |(18 + 0 + 0) - (0 + 36 + 27)| sq. units.

= ½ |18 - 63| sq. units.

= 45/2 sq. units.

And the area of the triangle ADE

= ½ |( 6 + 6 + 0) - (0 + 8 + 9)| sq. units.

= ½ |12 - 17| sq. units.

= 5/2 sq. units.

therefore, area of the ∆ ABC

= 45/2 sq. units = 9 ∙ 5/2 sq. units.

= 9 ∙ area of the ∆ ADE. Proved.

The above worked-out problems on area of a triangle given 3 points are explained step-by-step with the help of the formula.


 Co-ordinate Geometry 






11 and 12 Grade Math 

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