Topic 3.11 · Applications and Interpretation HL

One line, many equations

r = (1, 2) + t(3, 4) and r = (4, 6) + s(6, 8) name a different point and a different direction vector. They are the same line, so they share every point on it, and nothing on the page says so until you test one.

line 1
(1, 2)the point it goes through
(3, 4)the direction
4x − 3y = −2in cartesian form

Two equations, one line. The second is drawn dashed on top of the first, so look for it rather than at it.

What the equation says

r = a + tb

and it is an instruction, not a formula to rearrange:

  1. a is a point the line passes through, as a position vector. Here (1, 2).
  2. b is a direction. Here (3, 4), meaning 3 right and 4 up.
  3. t is a parameter: a dial. Every value of t gives one point of the line, and every point of the line comes from exactly one t.

So t = 0 gives (1, 2), t = 1 gives (1+3, 2+4) = (4, 6), t = 2 gives (7, 10), and t = −1 gives (−2, −2). The line runs both ways, because t can be negative.

Neither a nor b is unique. Any point on the line will serve as a, and any multiple of the direction will serve as b. That is why

(1, 2) + t(3, 4)   and   (4, 6) + s(6, 8)

are the same line: (4, 6) is on the first line, and (6, 8) is 2(3, 4). Changing the point slides where t = 0 sits; changing the direction's length changes how far one step of t takes you. Neither changes the set of points.

In three dimensions, the same sentence

r = (2, 1, 0) + t(3, −1, 2), which written out one component at a time is the parametric form:

x = 2 + 3t,   y = 1 − t,   z = 2t

At t = 4: x = 2 + 12 = 14, y = −3, z = 8.

Nothing about the method changes with the extra component, which is why the vector form is used at all. In two dimensions a single cartesian equation describes a line; in three dimensions it describes a plane, and a line needs two of them at once, or the chained form (x − 2)/3 = (y − 1)/(−1) = z/2. r = a + tb is the one form that does not change shape between the two.

Is this point on the line?

Two routes, and the second is the one to trust.

Route 1, solve for t. Is (7, 10) on (1, 2) + t(3, 4)?

x: 1 + 3t = 7 gives t = 2.   y: 2 + 4t = 10 gives t = 2.

The same t from both components, so yes. If the components give different values of t, the point is not on the line, and that is the test in full: it is not enough for one component to work.

Route 2, the cartesian form. From x = 1 + 3t and y = 2 + 4t, eliminate t:

4x − 3y = 4(1 + 3t) − 3(2 + 4t) = 4 − 6 + 12t − 12t = −2

The t cancels, which is the point. Now every test is one line of arithmetic:

  1. (7, 10): 28 − 30 = −2. On the line.
  2. (4, 6): 16 − 18 = −2. On the line.
  3. (5, 5): 20 − 15 = 5. Not on the line.

Comparing two equations tells you nothing. To decide whether two lines are the same, do two things:

  1. Are the directions parallel? One must be a scalar multiple of the other. (6, 8) = 2(3, 4), so yes.
  2. Does a point of one lie on the other? (4, 6) gives 4(4) − 3(6) = −2, so yes.

Both yes means the same line. Parallel but a point that fails means two distinct parallel lines, which never meet. Not parallel means they cross, in two dimensions always and in three dimensions only if they happen to; otherwise they are skew.

On the GDC: parameters and point tests

This sub-topic is mostly substitution, which is faster by hand. What the machine is good for is two things: solving the two component equations for t at once, and drawing a parametric line so a claim about where it goes can be seen rather than believed.

When you may use it. Applications. A calculator is allowed in every paper. Reach for it on the simultaneous equations when two lines are claimed to intersect, and leave the point test by hand, because 28 − 30 is quicker than any menu.

TI-Nspire CX II

  1. The point test, one component at a time: nSolve(1+3t=7,t) gives 2 and nSolve(2+4t=10,t) gives 2
  2. Try it on (5, 5) and the two disagree, 1.33 and 0.75, so the point is not on the line. Doing them one at a time is a feature, not a chore: it is the whole test
  3. To draw it: a Graphs page, menu → Graph Entry/Edit → Parametric, with x1(t)=1+3t and y1(t)=2+4t
  4. Set the t range in Window Settings, from -3 to 3, then add the second line as x2(t)=4+6t, y2(t)=6+8t and watch it land on top. The machine fixes the parameter as t and will not let you call it s; on paper the letter is arbitrary, which is the point

Casio fx-CG50

  1. MENU → Graph, then F3 TYPE and F3 again for Param
  2. Enter Xt1=1+3T and Yt1=2+4T, then Xt2=4+6T and Yt2=6+8T
  3. SHIFT F3 V-Window, and set the T range as well as x and y; the T range is on the second page and is the step people miss
  4. F6 DRAW. One line appears, because the second is underneath

The mark people lose. Checking only one component. For (5, 5) on (1,2)+t(3,4): the x equation gives 1 + 3t = 5, so t = 4/3, and a student who stops there announces the point is on the line. The y equation gives 2 + 4t = 5, so t = 0.75. Two different values of t, so the point is not on the line. Both components, every time, or use the cartesian form which cannot be half-checked.

Your turn

1. A line has x = 2 + 3t. Find x when t = 4.

2. The line (1, 2) + t(3, 4) has cartesian form 4x − 3y = −2. Evaluate 4x − 3y at the point (7, 10).

3. Evaluate 4x − 3y at the point (5, 5).

4. For which value of t does (1, 2) + t(3, 4) pass through (7, 10)?

5. Two vector equations look completely different. What shows that they describe the same line?

Question 5. What shows that two different-looking vector equations describe the same line?
Where the marks go

1 markThe direction, taken correctly from two points or from the equation.

1 markThe equation written as a point plus t times a direction.

1 markThe parameter found, or the point test completed on BOTH components.

Writing a direction where a point belongs, or the reverse, is the expensive slip. Label them a and b on the page before you substitute anything.

Want a verdict on your own draft?

These pages are free and stay free, but they are general and your IA is not. Send me your research question, or whatever exists so far, and I will tell you in writing whether the topic has a ceiling on it, where the marks are going, and what to change first. That costs nothing and it comes back within 24 hours.

Written by a serving IB Diploma and Career-related Programme Coordinator and Head of Mathematics, who reads internal assessments across every subject group every year. If you then want the whole draft reviewed properly against all five criteria, that is the paid one, and it is refunded if it does not name at least three specific things to fix.

Send me your question, free

Already have a full draft? Have the whole thing reviewed against all five criteria, $99.