Find the relation between $x$ and $y$ if the points $\left( x,y \right),\left( 1,2 \right)$ and $\left( 7,0 \right)$ are collinear .

Hint: Collinear means points will lie on a straight line. And hence, a triangle formed with the help of three collinear points will have zero area. Use this concept to get the required answer of the problem. Area of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is given as
$\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$

Complete step-by-step answer:
Expansion of above determinant along row one is given as
$\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)-{{y}_{1}}\left( {{x}_{2}}-{{x}_{3}} \right)+1\left( {{x}_{2}}{{y}_{3}}-{{x}_{3}}{{y}_{2}} \right) \right]$
So, put the above determinant as ‘0’ to get the relation between $x$ and $y$ .

As we know collinear means lying on a line. It means, if the number of points are collinear, then points will lie on a line. Now, we are given three points in the problem as $\left( x,y \right),\left( 1,2 \right)$ and $\left( 7,0 \right)$ . As we know as triangle is formed with the help of three points and area of triangle with vertices as $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is given as
Area = $\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|..................\left( i \right)$
As, if three points are collinear i.e. lying on a line, then the triangle will not be formed or in other words, we can say that triangle will form but of area zero-unit square.
Hence, the area of a triangle formed by three collinear points will be 0.
So, with the help of equation (i), we get the condition for collinearity of three points as
$\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|=0.................\left( ii \right)$
Now, putting $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ as $\left( x,y \right),\left( 1,2 \right)$ and $\left( 7,0 \right)$ respectively to the equation (ii). We get
$\left| \begin{matrix}
   x & y & 1 \\
   1 & 2 & 1 \\
   7 & 0 & 1 \\
\end{matrix} \right|=0.....................\left( iii \right)$
Now, we can expand the determinant of equation (iii) along row 1. So, we get
$\Rightarrow x\left( 2\times 1-0\times 1 \right)-y\left( 1\times 1-7\times 1 \right)+1\left( 1\times 0-2\times 7 \right)=0$
On simplifying the above equation, we can re-write the above equation as
$\begin{align}
  & \Rightarrow x\left( 2-0 \right)-y\left( 1-7 \right)+1\left( 0-14 \right)=0 \\
 & \Rightarrow 2x-y\left( -6 \right)+1\left( -14 \right)=0 \\
 & \Rightarrow 2x+6y-14=0 \\
\end{align}$
On dividing the whole equation by ‘2’ we can get the above equation as
$\begin{align}
  & \Rightarrow \dfrac{2x}{2}+\dfrac{6y}{2}-\dfrac{14}{2}=0 \\
 & \Rightarrow x+3y-7=0 \\
\end{align}$
Hence, if three points $\left( x,y \right),\left( 1,2 \right)$ and $\left( 7,0 \right)$ are collinear then the relation between $x$ and $y$ is given as
$x+3y-7=0$
So, $x+3y-7=0$ is the answer.

Note: Another approach for this problem would be that we can calculate slope of line with the help of any two points of given points and equate it to the slope calculated by any other two points of the same line because slope of a line will be unique. Slope of a line with $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ lying on it is given as
$slope=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Another approach for writing area of triangle with the given vertices ass $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is:
$\begin{align}
  & =\dfrac{1}{2}\left[ \left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} \\
   {{x}_{2}} & {{y}_{2}} \\
\end{matrix} \right|+\left| \begin{matrix}
   {{x}_{2}} & {{y}_{2}} \\
   {{x}_{3}} & {{y}_{3}} \\
\end{matrix} \right|+\left| \begin{matrix}
   {{x}_{3}} & {{y}_{3}} \\
   {{x}_{1}} & {{y}_{1}} \\
\end{matrix} \right| \right] \\
 & =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}+{{x}_{2}}{{y}_{3}}-{{y}_{2}}{{x}_{3}}+{{x}_{3}}{{y}_{1}}-{{x}_{1}}{{y}_{3}} \right] \\
\end{align}$
Hence, it can be another approach to write the area of the triangle and hence, equating it to 0 to get the required answer.