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 (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is
½ |y_{1} (x_{2} - x_{3}) + y_{2} (x_{3} - x_{1}) + y_{3} (x_{1} - x_{2})| 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 12^{1}/_{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| = 12^{1}/_{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 (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) and (x_{4}, y_{4}). 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 |(x_{1}y_{2} + x_{2} y_{3} + x_{3}y_{4} + x_{4}y_{1}) - (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})| 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 |2a^{2} sin (π + π/6 ) + 2a^{2} sin (π - π/6) -2a^{2} sin (π/2 - π/6)|sq. units.
= ½ |-2a^{2} sin π/6 + 2a^{2} sin π/6 - a^{2} cos π/6| sq. units.
= ½ ∙ a^{2} ∙ (√3/2) sq. units = (√3/4) a^{2} 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, CD^{2} = (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,
BC^{2} = CD^{2} + BD^{2} = 25 + 12^{2} = 25 + 144 = 169
or, BC = 13
Therefore, the required radius of the circle = 13 units.
6. 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
What is Co-ordinate Geometry?
Rectangular Cartesian Co-ordinates
Polar Co-ordinates
Relation between Cartesian and Polar Co-Ordinates
Distance between Two given Points
Distance between Two Points in Polar Co-ordinates
Division of Line Segment: Internal & External
Area of the Triangle Formed by Three co-ordinate Points
Condition of Collinearity of Three Points
Medians of a Triangle are Concurrent
Apollonius' Theorem
Quadrilateral form a Parallelogram
Problems on Distance Between Two Points
Area of a Triangle Given 3 Points
Worksheet on Quadrants
Worksheet on Rectangular – Polar Conversion
Worksheet on Line-Segment Joining the Points
Worksheet on Distance Between Two Points
Worksheet on Distance Between the Polar Co-ordinates
Worksheet on Finding Mid-Point
Worksheet on Division of Line-Segment
Worksheet on Centroid of a Triangle
Worksheet on Area of Co-ordinate Triangle
Worksheet on Collinear Triangle
Worksheet on Area of Polygon
Worksheet on Cartesian Triangle
11 and 12 Grade Math
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