Year 11 Extended · Unit 7 Networks and vectors · Lesson 7 of 8

How can you prove something about every triangle at once?

A numerical example shows one triangle. A vector argument uses letters for the sides, so it covers all of them. This lesson asks what is always true of the line joining two midpoints, and how to test for parallel and collinear points.

Geometry and TrigonometryNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From lesson 5: write 2(3, −1) − (1, 4) as a single column.
    Answer(5, −6).
  2. From lesson 6: the midpoint of (0, 6) and (8, 2)?
    Answer(4, 4).
  3. From Year 9: two similar triangles have a scale factor of 2. How do corresponding lengths compare?
    AnswerEvery length in the second is double the matching length in the first.
Words used in this lesson
proof
an argument showing a result is true in every case. For example: using letters for the sides.
collinear
lying on one straight line. For example: three towns on one road.
parallel
in the same direction, never meeting. For example: railway lines.
bisect
to cut into two equal parts. For example: a line through the middle.
base vectors
two vectors from which others are built. For example: OA and OB.

Is the path between two side midpoints always parallel to the third side?

A triangular field in the hills near Cusco has corners at O, A and B. Two walkers start at O and walk halfway along OA and halfway along OB, to points M and N. They join up with a path MN. Does the path always run parallel to the far side AB, however the field is shaped? A numerical example can only show one field. A proof has to cover them all, so we use letters: write a for OA and b for OB.

Predict first: will the path MN always be parallel to AB, and will it be longer, shorter or the same length? Then move the corners in the figure and compare the vectors MN and AB.

8
2
2
6
(−3, 2)vector MN
(−6, 4)vector AB

A = (8, 2) and B = (2, 6): MN = (−3, 2) and AB = (−6, 4), so MN is half of AB.

Show me the steps
  1. A = (8, 2), B = (2, 6): AB = (−6, 4) and MN = (−3, 2), which is half of AB.
  2. A = (4, 0), B = (0, 8): AB = (−4, 8) and MN = (−2, 4), again half of AB, so the line MN is parallel to AB.
  3. A = (6, 4), B = (4, 4): AB = (−2, 0) and MN = (−1, 0), half of AB for a different triangle.
  4. Any A and B: MN = ON − OM = ½ b − ½ a = ½(b − a) = ½ AB, which holds for every triangle at once.

Plan for a vector proof. Choose two vectors from the picture as base vectors, such as a and b. Write every other vector as a combination of them using the triangle rule: AB = b − a. Then compare the two vectors in the claim.

What a comparison proves. If one vector is a scalar multiple of another they are parallel. If they are parallel and share a point, the points are on one straight line (collinear). A numerical example is evidence, but only the letters prove it for all cases.

Two common slips. Treating one numerical example as a proof: it shows the claim is true for that triangle and nothing else. And showing two vectors are parallel but then claiming the points are collinear without a shared point: parallel vectors on different lines never meet.

Worked examples

1. Prove that MN is parallel to AB and half as long, where M and N are the midpoints of OA and OB.

  1. OM = ½a and ON = ½b. So MN = ON − OM = ½b − ½a = ½(b − a).
  2. AB = b − a, so MN = ½ AB: it is a scalar multiple, so it is parallel, and it is half as long.

2. Show that A(1, 2), B(4, 3) and C(10, 5) lie on a straight line.

  1. AB = 3
    1     and BC = 6
    2    .
  2. BC = 2AB, so the vectors are parallel and share the point B: A, B and C are collinear.

3. A parallelogram has vertices O(0, 0), A(6, 1), C(8, 5) and B(2, 4). Show that its diagonals bisect each other.

  1. The midpoint of OC is (4, 2.5). The midpoint of AB is ((6 + 2) ÷ 2, (1 + 4) ÷ 2) = (4, 2.5).
  2. The two midpoints are the same point, so the diagonals bisect each other.

Your go. The first step is given. Are P(2, 1), Q(5, 4) and R(9, 9) collinear?

  1. Find PQ = Q − P and QR = R − Q.
  2. Test whether one is a scalar multiple of the other.
Check the answer

PQ = 3
3 and QR = 4
5. The across ratio is 4 ÷ 3 and the up ratio is 5 ÷ 3, which are not equal, so QR is not a multiple of PQ and the points are not collinear.

Check before you go on

1. AB = 2
3 and CD = 4
6. Which statement must be true?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. In triangle OAB, a = OA = 4
2 and b = OB = 2
6. M and N are the midpoints of OA and OB. What is the top entry of MN?

3. In a triangle, MN = k AB, where AB = 10
−6 and MN = 5
−3. What is k, to one decimal place?

Level 2 · Building

4. PQ = 3
4 and PR = 9
12, so P, Q and R are collinear and PR = k PQ. What is k?

5. In parallelogram ABCD, AB = 6
2 and AD = 1
5. What is the top entry of the diagonal BD?

Level 3 · Stretch

6. The points A(1, 2), B(3, 6) and C(k, 14) are collinear. What is k?

7. In triangle OAB, OA = a and OB = b. P is on OA with OP : PA = 1 : 4, and Q is on OB with OQ : QB = 1 : 4. PQ = k AB. What is k, to one decimal place?

Something to think about

ABCD is any quadrilateral, and P, Q, R and S are the midpoints of AB, BC, CD and DA. Is PQRS always a parallelogram? Use position vectors a, b, c and d.

One way to think about it

P = ½(a + b) and Q = ½(b + c), so PQ = ½(c − a). S = ½(d + a) and R = ½(c + d), so SR = ½(c − a) as well. One pair of opposite sides is equal and parallel, so PQRS is a parallelogram for every quadrilateral, and each of those sides is parallel to the diagonal AC.

Exit ticket

8. OABC is a parallelogram with its vertices in that order, OA = a and OC = c. M is the midpoint of AB. Which expression gives OM?