Question
A triangle has two of its vertices at $$P\left( {a,\,0} \right),\,Q\left( {0,\,b} \right)$$ and the third vertex $$R\left( {x,\,y} \right)$$ 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.
$$\frac{{a + b}}{4}$$
Answer :
$$ - \left( {\frac{{a + b}}{2}} \right)$$
Solution :
$$\eqalign{ & {\text{Area of }}\Delta PQR = A \cr & = \frac{{ - 1}}{2}\left[ {x\left( {b - 0} \right) + 0\left( {0 - y} \right) + a\left( {y - b} \right)} \right] \cr & = \frac{{ - 1}}{2}\left( {bx + ax - ab} \right) \cr & \therefore \,\frac{{dA}}{{dx}} = \frac{{ - 1}}{2}\left( {a + b} \right) \cr} $$
$$\eqalign{ & {\text{Area of }}\Delta PQR = A \cr & = \frac{{ - 1}}{2}\left[ {x\left( {b - 0} \right) + 0\left( {0 - y} \right) + a\left( {y - b} \right)} \right] \cr & = \frac{{ - 1}}{2}\left( {bx + ax - ab} \right) \cr & \therefore \,\frac{{dA}}{{dx}} = \frac{{ - 1}}{2}\left( {a + b} \right) \cr} $$