Explanation
Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.
Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points. Slope of the line passing through points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ $$ $$=$$ $$ $$\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } $$So, slope of the line joining $$ (3,5) , (m,6) = $$ Slope of the line joining $$ (3,5) $$ and $$\left (\dfrac {1}{2}, \dfrac {15}{2}\right ) $$
Therefore, $$ \dfrac { 6 - 5 }{ m - 3 } = \dfrac { \frac {15}{2} - 5 }{ \frac {1}{2} - 3 } $$$$\Rightarrow \dfrac { 1 }{ m - 3 } = -1 $$
$$\Rightarrow m - 3 = -1 $$
$$\Rightarrow m = 2 $$
As the points are collinear, the slope of the line joiningany two points, should be same as the slope of the line joining two otherpoints. Slope of the line passing through points $$\left( { x }_{ 1 },{ y }_{ 1 }\right) $$ and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ $$ = $$ $$\dfrac { { y}_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } $$So, slope of the line joining $$ (p,0) , (0,q) = $$ Slope of the line joining$$ (0,q) $$ and $$ (1,1) $$ $$ \dfrac { q - 0 }{ 0 - p } = \dfrac { 1 - q }{ 1 - 0 } $$$$ - \dfrac { q }{ p } = 1 - q $$
Dividing both sides by $$q$$, $$ - \dfrac { 1 }{ p } = \dfrac { 1 }{ q } - 1 $$
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