### Geometry of Equations

Many ways problem

## Main solutions

How can you guarantee that your four lines will intersect to form a triangle where each point on the triangle has negative coordinates?

Approach 1

We know that all of the vertices of the triangle need to be located in the bottom left quadrant. If we specify the coordinates of the vertices then we can work out the equations of the lines from there.

We could select any three points in the bottom left quadrant. Let’s take $A(-5,-2)$, $B(-4,-9)$ and $C(-0.4,-0.5)$ as an example. To find the equation of the straight line passing through two points we can use the general form of the equation $y-y_{1}=m(x-x_{1})$.

Let’s start with coordinates $A$ and $B$. The gradient of the line is $\frac{y-y_{1}}{x-x_{1}}=\frac{-2+9}{-5+4}=-7$.

Then the equation of the line is $y+2=-7(x+5)$, or $y+7x+37=0$.

Now if we repeat the process with coordinates $B$ and $C$ we get a gradient of $\frac{-9+0.5}{-4+0.4}$, or $\frac{85}{36}$. Then the equation of the line is $y+9=\frac{85}{36}(x+4)$, or $36y=85x+16$.

And finally we repeat the process with coordinates $A$ and $C$ giving a gradient of $\frac{-2+0.5}{-5+0.4}=\frac{15}{46}$. Then the equation of the line is $y+2=\frac{15}{46}(x+5)$ $46y=15x-17$.

That gives us three equations that will intersect to enclose a triangle where each point on the triangle has negative coordinates. We now just need to specify a fourth equation that will not intersect to create another triangle in the bottom left quadrant. This could be done by choosing any line that is restricted to the other three quadrants. For example, $y+x=5$.

Approach 2

We could solve this problem algebraically by considering the constraints on the values of the gradient, m and $y$-intercept, c.

If we start by assuming that we have two straight lines $y=m_{1}x+c_{1}$ and $y=m_{2}x+c_{2}.$ Then solving these two equations simultaneously gives us the coordinates of the point of intersection of the two lines as $\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}},\frac{m_{2}c_{1}-m_{1}c_{2}}{m_{2}-m_{1}}\right).$

Now, we want the coordinates to be negative. If we first consider the $x$-coordinate then we require that either $c_{1}>c_{2}$ and $m_{1}>m_{2}$, or $c_{2}>c_{1}$ and $m_{2}>m_{1}$.

Considering the $y$-coordinate we require that either $m_{2}c_{1}>m_{1}c_{2}$ and $m_{1}>m_{2}$, or $m_{1}c_{2}>m_{2}c_{1}$.

Combining the conditions for both coordinates we have the following:

$c_{1}>c_{2}$ and $m_{1}>m_{2}$ and $m_{2}c_{1}>m_{1}c_{2}$

or

$c_{2}>c_{1}$ and $m_{2}>m_{1}$ and $m_{1}c_{2}>m_{2}c_{1}$.

If we can select two lines that meet these conditions then they will intersect in the third quadrant.

We could then repeat the process to select a third line to intersect with the first two, also in the third quadrant.

Making use of a spreadsheet to select values of $m$ and $c$ that fit the conditions can be helpful.