Question

# If the points $$(0, 4), (4, 0)$$ and $$(5, p)$$ are collinear, then value of $$p$$ is

A
1
B
7
C
6
D
4

Solution

## The correct option is A $$- 1$$The given points are $$A=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,4 \right) , B=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 4,0 \right) , C=\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 5,p \right)$$. They are collinear.$$\therefore ar\Delta ABC=0$$$$\Rightarrow \dfrac { 1 }{ 2 } \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { 1 } \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| =0$$$$\Rightarrow \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & { 1 } \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| =0$$$$\Rightarrow \left| \begin{matrix} 0 & 4 & 1 \\ 4 & 0 & 1 \\ 5 & p & 1 \end{matrix} \right| =0$$$$\Rightarrow 0\left( 0-p \right) +4\left( p-4 \right) +5\left( 4-0 \right) =0$$$$\Rightarrow p=-1$$Mathematics

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