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
Workedout 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, ½1  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 coordinates (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 coordinates (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 shortcut method of getting the area of a triangle. Suppose, we want to find the area of the quadrilateral whose vertices have coordinates (x₁, y₁), (x₂, y₂), (x₃, y₃) and (x₄, y₄). For this, we write the coordinates of the vertices in four rows repeating the first written coordinates 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 coordinates 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 coordinates 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 linesegment AC divides the line segment BD in the ratio m : n at P. Therefore, P divides the linesegment BD in the ratio m : n. Hence, the coordinates 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 linesegment AC divides the linesegment BD internally in the ratio 2 : 3.
5. The polar coordinates 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 midpoint 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 coordinates 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 coordinates 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 coordinates 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 workedout problems on area of a triangle given 3 points are explained stepbystep with the help of the formula.
● Coordinate Geometry
11 and 12 Grade Math
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