question_answer

A triangle has two of its vertices at \[P(a,0),Q(0,b)\] and the third vertex \[R(x,\,\,y)\] is moving along the straight line \[y=x.\] If A be the area of the triangle. Then \[\frac{dA}{dx}\] is equal to

A) \[\frac{a-b}{2}\]

B) \[\frac{a-b}{4}\]

C) \[=-\left( \frac{a+b}{2} \right)\]

D) \[\frac{a+b}{4}\]

Correct Answer: C

Solution :

[c] Area of \[\Delta PQR=A\] \[=\frac{-1}{2}[x(b-0)+0(0-y)+a(y-b)]\] \[=\frac{-1}{2}(bx+ax-ab)\] \[\therefore \frac{dA}{dx}=\frac{-1}{2}(a+b)\]