Explanation
Since the given points are collinear, they do not form atriangle, which means area of the triangle is Zero.
Area
Distance between the points C$$(a,b) $$ and D $$ ({a}^{2}, ab) = \sqrt { \left( {a}^{2} - a \right) ^{ 2 }+\left(ab - b\right) ^{ 2 } } = (a-1)\sqrt { {a}^{2} + {b}^{2} } $$
Distance between the points A$$(-a,-b) $$ and D $$ ({a}^{2}, ab) = \sqrt { \left( {a}^{2} + a \right) ^{ 2 }+\left(ab + b\right) ^{ 2 } } = (a +1)\sqrt { {a}^{2} + {b}^{2} } $$
Now. $$ AB+BC+CD = \sqrt { {a}^{2} + {b}^{2} } + \sqrt { {a}^{2} + {b}^{2} }+ (a-1)\sqrt { {a}^{2} + {b}^{2} } = (a +1)\sqrt { {a}^{2} + {b}^{2} } = AD $$
Hence, the points are collinear.
Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
So, slope of the line joining OA $$ = \dfrac { -b-0 }{ -a-0 } = \dfrac {b}{a} $$
Slope of the line joining OC $$ = \dfrac { ab-0 }{{a}^{2}-0 } = \dfrac {b}{a} $$
Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
Please disable the adBlock and continue. Thank you.