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Question

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


A
1
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B
7
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C
6
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D
4
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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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