Chapter 4 · Geometry and Equations · Concept 2
4-15 Conditions for Not Forming a Triangle
Three lines usually make a triangle. This lesson asks when they do not — and the answer is that there are two quite different ways to fail, which is why almost every question here has more than one answer.
Guiding question
What has to go wrong for three lines to enclose nothing?
Learning outcome. Work out the conditions under which three lines do not form a triangle.
Two ways to fail, and both must be checked
failure 1
two of the lines are parallel
They never meet, so one corner is missing. Set the two gradients equal.
failure 2
all three pass through one point
Find where the two lines WITHOUT the constant cross, then make the third pass through it.
The parallel case has to be checked twice.
The line carrying the constant could be parallel to the first line or to the second. Those are two different values, and both belong in the answer. Add the concurrent case and most of these questions have three answers.
Start by putting all three into y = mx + k. The gradients are then in front of you and the parallel cases take one line each. And check whether the two lines with no constant in them are parallel to each other first — if they are, there is no crossing point and the concurrent case cannot happen at all.
The pink triangle is what the three lines enclose. Slide a and watch it disappear three times — once by collapsing to a point, twice by a corner running off to infinity.
x − y + 1 = 0
2x + y + 2 = 0
0x + 3y − 4 = 0
A triangle, with area 1.33.
Three lines, three crossings, three corners — the ordinary case.
Two of the three answers come from a line going parallel and one from all three meeting at a point. Slide slowly through each triangle marker and watch which kind of failure it is.
x − y + 1 = 0, 2x + y + 2 = 0 and ax + 3y − 4 = 0 do not form a triangle. Find a.
- 1
Put all three into y = mx + k so the gradients are visible.
y = x + 1 · y = −2x − 2 · y = −a3x + 43Every case in this lesson is about gradients, so this step comes first every time.
Work with a partner. Split the work: one takes the two parallel cases, the other the concurrent one. Then put every value you found back into the three lines and say out loud which kind of failure it causes.
The lines x + 2y − 1 = 0, 3x + y + 2 = 0 and mx + 4y − 3 = 0 do not form a triangle. Find the sum of all the possible values of m.
Where would anyone actually use the crossing of lines and the triangles they make — in navigation, building, or design?
Think about what a crossing point means when the lines are not lines.
Finding one value and stopping.
There are usually three: two parallel cases and one concurrent. Almost every lost mark in this lesson is a missing answer rather than a wrong one.
Checking only one parallel case.
The line with the constant can be parallel to EITHER of the other two. That is two separate values.
Using the wrong pair for the crossing point.
Solve the two lines WITHOUT the unknown constant. Using one that contains it leaves letters in the crossing point.
Reading a gradient before rearranging.
ax − y = a − 1 has to become y = ax − a + 1 first. The a appears twice and only one of them is the gradient.
Rounding the crossing point.
In Exercise 1 it is (5/4, 9/8). Keep it exact — a decimal there turns a clean 11/10 into something unrecognisable.
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