Negatively Triangular
Negatively Triangular printable sheet
Four straight lines have the following equations:
$$\begin{align} 3x+8y&=59 \\ x-2y&=1 \\ y-4x&=3 \\ 3y+2x &= -19 \end{align}$$
How many points of intersections do you expect there to be?
Three of the four lines enclose a triangle which only has negative co-ordinates for all points on that triangle.
Which three lines?
What is the gradient of each line? Are any of the lines parallel?
Where does each line cross the axes?
Does it help to draw a rough sketch of each line?
How can you use algebra to find out the exact coordinates of the points of intersection?
Patrick from Woodbridge school used a computer to check his answer:
"I used the Mac OS X application "Grapher" to help with this. I used it to plot the lines to check my answers. I did the intersections part by solving every equation against every other in a simultaneous equation, I got the answer of $6$ (I did each algebraically but it would take a lot of space to write up)"
You can view the four graphs using the Desmos graphing calculator here
Labelling the equations $1$ to $4$ in the obvious way, here are Patrick's co-ordinates of the six intersections:
$1$ and $2$: $(9,4)$
$1$ and $3$: $(1,7)$
$1$ and $4$: $(-47,25)$
$2$ and $3$: $(-1,-1)$
$2$ and $4$: $(-5,-3)$
$3$ and $4$: $(-2,-5)$
We can deduce that lines $2, 3$ and $4$ form the triangle.
Note that if we rearrange the equations into the standard form $y = mx + c$, it would be possible to sketch the four lines and estimate roughly where their intersections lie. It might then be sufficient to solve the problem without needing to explicitly calculate any co-ordinates.
Why do this problem?
This problem provides a good opportunity to explore simultaneous linear equations, both graphically and algebraically.
Possible approach
This printable worksheet may be useful: Negatively Triangular.
Start by discussing the question "How many intersections do you expect there to be?" This could be broken down by considering how many intersections two lines would have, then three, then four.
Emphasise that we won't know for sure until we consider the graphs or solve the simultaneous equations, as there may be a pair of lines which don't intersect at all, or three lines which intersect at a single point.
Then set the main challenge - which three lines enclose a triangle which lies in the third quadrant; that is, all of the points in the triangle have a negative x and a negative y coordinate.
Students might choose to sketch the four graphs to get some insight into where the lines intersect. Alternatively they might solve each pair of simultaneous equations to find all six points of intersection. Some students might find it easier to work with the equations in the form $y=mx+c$.
After students have worked on the problem, you may wish to show them the Desmos graph linked from the solution page; this is a good opportunity to show the power of graphing software for gaining insight into graphical situations.
Key questions
What is the gradient of each line? Are any of the lines parallel?
Where does each line cross the axes?
Does it help to draw a rough sketch of each line?
How can you use algebra to find out the exact coordinates of the points of intersection?
Possible support
Which is Cheaper? provides a real-life context for exploring equations of straight lines and graphical interpretations of simultaneous equations.
Possible extension
Students might like to consider making up similar problems of their own, perhaps choosing lines which intersect in one of the other quadrants.
They could also explore Which is Bigger? which also invites students to create sets of equations whose intersections satisfy certain constraints.