$ (1,4),(2,3),(3,1),(4,1) \text { and }(5,0) $

Solution:

The standard slope-intercept equation of a. straight line $y=a x+b$. Using the parameter space, we get,

$ b=-a x+y\\ $

$ \begin{array}{|c|c|c|} \hline x & y & \\ \hline 1 & 4 & b=-a+4 \\ \hline 2 & 3 & b=-2 a+3 \\ \hline 3 & 1 & b=-3 a+1 \\ \hline 4 & 1 & b=-4 a+1 \\ \hline 5 & 0 & -b=-5 a \\ \hline \end{array} $

Using these equations we draw 5 lines in the parameter space a b, each representing a point from the x y plane.

If we observe all lines intersect each other at a single point (-1,5) except barring line $b=-3 a+1$.

We take this value of a and b and use it in the equation,

$ \begin{aligned} y &=a x+b \\ \text { i.e. } y &=-x+5 \end{aligned} $

The line passes through all the given points except the (3,1) point

As (3,1) did not lie on a straight line, the line representing this point in the ab plane did not intersect the other lines at the same point.