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A triangle has two of its vertices at P(a,0) and Q(0,b) and the third vertex R(x,y) is moving along the straight line y=x. If A be the area of triangle, then dA/dx is equal to? I know the answer it is " -(a+b)/2" but I dont know know how to solve the problem. Plz explain the solution....

A triangle has two of its vertices at P(a,0) and Q(0,b) and the third vertex R(x,y) is moving along the straight line y=x. If A be the area of triangle, then dA/dx is equal to?


I know the answer it is " -(a+b)/2"  but I dont know know how to solve the problem.


Plz explain the solution....

Grade:11

1 Answers

vikas askiitian expert
509 Points
10 years ago

let R be any point (X,Y)

since this point  lies on line Y = X so

 R =(X,X)

 p=(a,0)

q=(0,b)

now by determinant method area = 1/2 (determinant)

                                            A   =1/2 [ab - X(a+b)]

differentiating wrt X

                   dA/dx = 1/2 . -(a+b)

                            =-(a+b)/2

 

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