A graph will help you to visualise things more clearly. Using the ordered pairs, we can easily draw a graph.
The geometrical representation of a degree one polynomial equation \(ax + by + c = 0\) is a straight line, so only it is called a linear equation.
Let us consider the equation \(y = x + 1\). Now, let us graph the equation.
First, to plot the points in the graph, we need the ordered pairs. To find the ordered pairs, let us substitute values for \(x\).
The equation is \(y = x+1\).
When \(x = -2\), \(y = -2+1= -1\)
When \(x = -1\), \(y = -1+1= 0\)
When \(x = 0\), \(y = 0+1= 1\)
When \(x = 1\), \(y = 1+1= 2\)
When \(x = 2\), \(y = 2+1= 3\)
Therefore, the ordered pairs are \((-2, -1)\), \((-1, 0)\), \((0, 1)\), \((1, 2)\), \((2, 3)\).
Now, let us plot these ordered pairs in the graph and join them.
Here, the scale is \(x\)-axis \(1 \ cm =1 \ unit\) and \(y\)-axis \(1 \ cm = 1 \ unit\).
Thus, the points on the line \((-2, -1)\), \((-1, 0)\), \((0, 1)\), \((1, 2)\) and \((2, 3)\) are the solution of the given equation.
Now, let us take the point \((2, 2)\), that does not lies on the line.
Substitute \((2, 2)\) in the equation and check whether it is a solution or not.
\(y = x + 1\)
\(2 = 2 + 1\)
\(2 \ne 3\)
So, \((2, 2)\) is not a solution to the given equation.
Is the point not lies on the line is a solution to the equation?
No, the point not lies on the line is not a solution to the equation.
The graph of the equation of the form \(y = kx\) is a line which always passes through the origin.
From the above discussion, let us list our observations.
1. Every point whose coordinates satisfy the equation lies on the line of the graph.
2. Every point \((a, b)\) on the line gives a solution \(x = a\), \(y = b\) of the given equation.
3. Any point, which does not lie on the line, is not a solution of the given equation.