Q. 37

Using integration, find the area of the triangle whose vertices are A(2,3), B(4,7) and C(6,2).

Answer :

Given,


• A (2,3), B (4,7) and C (6,2) are the 3 vertices of a triangle.



From above figure we can clearly say that, the area between ABC and DEF is the area to be found.


For finding this area, we can consider the lines AB, BC and CA which are the sides of the given triangle. By calculating the area under these lines we can find the area of the complete region.


Consider the line AB,


If (x1,y1) and (x2, y2) are two points, the equation of a line passing through these points can be given by



Using this formula, equation of the line A(2,3) B =(4,7)





y = 2x – 4 + 3


y = 2x-1


Consider the area under AB:



From the above figure, the area under the line AB will be given by,




[ using the formula, and ]


= [42 - 4] – [22 - 2] = 12-2= 10


Area of ABDE = 10 sq. units. ---- (1)


Consider the line BC,


Using this 2-point formula for line, equation of the line B(4,7) and C (6,2)






Consider the area under BC:



From the above figure, the area under the line BC will be given by,




[ using the formula, and ]





Area of BCFE = 9 sq. units. ---- (2)


Consider the line CA,


Using this 2-point formula for line, equation of the line C(6,2)and A(2,3)






Consider the area under CA:



From the above figure, the area under the line CA will be given by,




[ using the formula, and ]




Area of ACFD = 10 sq.units ---- (3)


If we combined, the areas under AB, BC and AC in the below graph, we can clearly say that the area under AC (3) is overlapping the previous twoareas under AB & BC.



Now, the combined area under the rABC is given by


Area under rABC


=Area under AB + Area under BC - Area under AC


From (1), (2) and (3), we get


Area under rABC = 10+9-10 = 9


Therefore, area under rABC = 9 sq.units.


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