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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....
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
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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