Four straight lines have the following equations:

- \(3x+8y=59\)
- \(x-2y=1\)
- \(y-4x=3\)
- \(3y+2x=-19\)

How many intersections do you expect?

**Approach 1**

These equations represent straight lines and can therefore be written in the form \(y=mx+c\).

The four equations can be written as:

\[\begin{align} y &=-\frac{3x}{8}-\frac{59}{8} \label{eq:a'}\tag{$a'$}\\ y &=\frac{x}{2}-\frac{1}{2} \label{eq:b'}\tag{$b'$} \\ y &=4x+3 \label{eq:c'}\tag{$c'$}\\ y &=-\frac{2x}{3}-{19}{3} \label{eq:d'}\tag{$d'$} \end{align}\]This shows that none of the lines are parallel. Any two lines that are not parallel will intersect at some point. This means that each of the four lines will intersect with the other three.

Therefore there could be up to six intersections.

**Approach 2**

If two lines intersect then we can solve their equations simultaneously to find the coordinates of the intersection.

Subtracting three times equation \(\eqref{eq:2}\) from equation \(\eqref{eq:1}\) gives \(14y=56\), so \(y=4\). We can find \(x\) by substituting this into equation \(\eqref{eq:2}\): we have \(x-8=1\) and so \(x=9\). We can easily check that the point \((9,4)\) does indeed lie on both lines.

Therefore the coordinates of intersection between lines \(\eqref{eq:1}\) and \(\eqref{eq:2}\) are: \((4,9)\)

Repeating this process for each of the other five pairs yields the following solutions:

Lines \(\eqref{eq:1}\) and \(\eqref{eq:3}\) intersect at: \((1,7)\)

Lines \(\eqref{eq:1}\) and \(\eqref{eq:4}\) intersect at: \((-47,25)\)

Lines \(\eqref{eq:2}\) and \(\eqref{eq:3}\) intersect at: \((-1,-1)\)

Lines \(\eqref{eq:2}\) and \(\eqref{eq:4}\) intersect at: \((-5,-3)\)

Lines \(\eqref{eq:3}\) and \(\eqref{eq:4}\) intersect at: \((-2,-6)\)

Therefore there are six intersections.

This approach has the benefit of providing information useful in answering the next part of the problem.

### Some more points…

Three of the four lines above will enclose a triangle where each point on the triangle has negative coordinates.

Which three lines?

From approach 2 above we can immediately see that the triangle must have vertices at: \((-1,-1)\), \((-5,-3)\), and \((-2,-6)\). These are the points of intersection between lines \(\eqref{eq:2}\), \(\eqref{eq:3}\) and \(\eqref{eq:4}\).

Therefore, the three lines that will enclose a triangle where each point on the triangle has negative coordinates are:

\(x-2y=1\)

\(y-4x=3\)

\(3y+2x=-19\)