Explanation
When three points A, B and C are collinear, slope of linejoining any two points (say AB) = slope of line joining any other twopoints (say BC). 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 } } Slope of the line passing through ( \dfrac {2}{5}, \dfrac {1}{3}) and ( \dfrac {1}{2}, k) = Slope of line passing through (\dfrac {1}{2}, k) and ( \dfrac {4}{5},0)
\dfrac { k- \dfrac{1}{3} }{ \dfrac {1}{2} - \dfrac {2}{5} } = \dfrac { 0 - k}{ \dfrac {4}{5} - \dfrac{1}{2} } On solving, k = \dfrac {1}{4}
{\textbf{Step1: Proving the three points are collinear }}
{\text{Given points are A(a,b + c),B(b,c + a),C(c,a + b)}}
{\text{ Lets the points of the triangle be }}\mathbf{{({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})}}
{\text{[}}{{\text{x}}_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})] = 0
\text{Comparing these with the given points}
[a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)]
=ac - ab + ab - bc + bc - ac
= 0
{\textbf{Hence,The given points are collinear.}}
\textbf{Step -1: Area of triangle in co-ordinate geometry.}
Distance between the points A (-a,-b) and B (0,0) = \sqrt { \left( 0 + a \right) ^{ 2 }+\left( 0 + b \right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} }
Distance between the points B (0,0) and C (a,b) = \sqrt { \left( a-0 \right) ^{ 2 }+\left( b-0 \right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} }
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| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2}({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|
Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.
Hence, substituting the points ({ x }_{ 1 },{ y }_{ 1 }) = (5,1) ; ({ x}_{ 2 },{ y }_{ 2 }) = (1,P) and ({ x }_{ 3 },{ y }_{ 3 }) = (4,2) in the area formula, we get \left| \dfrac { 5(P-2)+1(2-1)+4(1-P) }{ 2 } \right| =0
=> 5P-10+1+4-4P = 0
=> P = 5
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