9709 Pure 1 · 1.3 Coordinate Geometry · Session 1 of 4

Straight Lines

You have met most of this before. Today is about doing it the way Cambridge marks it: exact answers, the right form of the equation, and the right gradient for the right line.

Do Now: gradient, midpoint and length, and why this topic matters7 min
Three measurements and three forms of a line10 min
Parallel and perpendicular lines, and where two lines meet8 min
The perpendicular bisector, the Bisector Lab, and a 4-mark finish from June 202414 min
Spot the mistake, practice together, then on your own13 min
Recap and exit ticket5 min

Do Now · 5 min · no notes, no calculator

\(A(1, 2)\) and \(B(5, 10)\). Find the gradient of \(AB\), the midpoint of \(AB\), and the length of \(AB\).

Give the length in exact form. Sketch the two points first.

Write all three, then tap to check.
gradient:  \(\dfrac{10 - 2}{5 - 1} = \dfrac{8}{4} = 2\)
midpoint:  \(\left(\dfrac{1 + 5}{2}, \dfrac{2 + 10}{2}\right) = (3, 6)\)
length:  \(\sqrt{(5 - 1)^2 + (10 - 2)^2} = \sqrt{16 + 64} = \sqrt{80} = \sqrt{16}\sqrt{5} = 4\sqrt{5}\)
Gradient \(2\), midpoint \((3, 6)\), length \(4\sqrt{5}\). A decimal like 8.94 is not exact.

Why this matters · 2 min

Coordinates are a way of describing a position relative to a fixed point, the origin.

Once a place is a pair of numbers, questions about distance, direction and shape become algebra, and a computer can answer them.

What each skill answers

Gradient: how steep something is. A ramp, a roof, a road, the track of a storm.

Length: how far apart two places are.

Midpoint: the place halfway between two others, such as a meeting point.

Perpendicular: the shortest way from a point to a line, so the closest a storm comes to an island.

Where two lines meet: where two roads, or two ships' courses, cross.

Perpendicular bisector: every place that is equally far from two others, such as a shelter that is fair to two towns.

Where you will see it

Maps and GPS. A place is a pair of numbers, latitude and longitude. Apps work out distances and routes from them.

Hurricane tracking. A forecast track is a line of positions. How close it comes to St. Croix is a perpendicular distance.

Games and animation. Every character and every pixel has an \((x, y)\) position. Moving and colliding are distance and line calculations.

3D printers and robots. The machine follows a list of coordinates and moves in straight lines from one point to the next.

Building plans and surveying. Corners are points. Walls that must meet at right angles are perpendicular lines.

The Do Now used the first three skills. By the end of today you can do all six, and the exam asks for them with exact answers.

Direct instruction · 1 of 9 · three measurements

Gradient, midpoint and length, from two points \((x_1, y_1)\) and \((x_2, y_2)\).

Gradient

\(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)

Change in \(y\) divided by change in \(x\). Subtract in the same order on top and bottom.

Midpoint

\(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\)

The average of the two \(x\)-coordinates, and the average of the two \(y\)-coordinates.

Length (distance)

\(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Pythagoras on the across and up distances. Leave it as a surd unless the question asks for a decimal.

Two habits that save marks

Sketch first, every time. A rough sketch tells you whether a gradient should be positive or negative, and catches a sign slip before it costs you.

None of these three formulas is in the formula booklet (MF19). You need them from memory.

Simplify the surd. Look for a square factor: \(\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}\).

Direct instruction · 2 of 9 · the equation of a line

One line, three ways to write it. Work in the first, convert at the end.

The line through \((-2, 4)\) with gradient \(\tfrac{5}{3}\).

1 · Point and gradient

\(y - y_1 = m(x - x_1)\)
\(y - 4 = \tfrac{5}{3}(x + 2)\)

Use this when you know a point and the gradient. It is the form Cambridge's own model answers start from.

2 · Gradient and intercept

\(y = mx + c\)
expand:  \(y - 4 = \tfrac{5}{3}x + \tfrac{10}{3}\)
add 4:  \(y = \tfrac{5}{3}x + \tfrac{22}{3}\)

Use this to read off the gradient \(m\) and where the line crosses the \(y\)-axis, \(c\).

3 · \(ax + by + c = 0\)

form 1, times 3:
\(3y - 12 = 5x + 10\)
all on one side:
\(0 = 5x - 3y + 22\)
\(5x - 3y + 22 = 0\)

Read the question for the form

When a question says "in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers", multiply through to clear every fraction, and make sure the equation ends in "= 0". On a June 2024 question, about half of the students who found the right line lost the last mark because the answer was not in the form asked for.

Check it: put \((-2, 4)\) into the answer. \(5(-2) - 3(4) + 22 = -10 - 12 + 22 = 0\). It works.

Direct instruction · 3 of 9 · parallel and perpendicular

Parallel lines share a gradient. Perpendicular gradients multiply to \(-1\).

parallel:  \(m_1 = m_2\)
perpendicular:  \(m_1 m_2 = -1\),  so  \(m_2 = -\dfrac{1}{m_1}\)
flip it and change the sign:  \(\tfrac{4}{7} \to -\tfrac{7}{4}\).  Check: \(\tfrac{4}{7} \times \left(-\tfrac{7}{4}\right) = -1\)

A line given as \(ax + by + c = 0\): rearrange to \(y = mx + c\) first.

\(4x + 5y - 20 = 0 \;\Rightarrow\; 5y = -4x + 20 \;\Rightarrow\; y = -\tfrac{4}{5}x + 4\)
gradient \(-\tfrac{4}{5}\),  parallel gradient \(-\tfrac{4}{5}\),  perpendicular gradient \(\tfrac{5}{4}\)

To show two lines are perpendicular, say it in words: "the product of the gradients is \(-1\), so the lines are perpendicular."

Which line does this gradient belong to?

curve   line \(AB\)   tangent at \(B\)

A question tells you the gradient of the curve at \(B\) is \(-3\). That is the gradient of the tangent at \(B\), the dotted line. The line \(AB\) is a different line with a different gradient, \(-\tfrac{13}{2}\).

Some students took the gradient of the perpendicular bisector of \(AB\) to be \(\tfrac{1}{3}\), straight from the \(-3\). Before you flip a gradient, ask which line it belongs to.

Direct instruction · 4 of 9 · where two lines meet

The point where two lines cross satisfies both equations, so solve them simultaneously.

Find where \(x + 2y = 4\) and \(3x - y = 19\) meet.

Substitution

make \(y\) the subject of the second:  \(y = 3x - 19\)
put it into the first:  \(x + 2(3x - 19) = 4\)
\(x + 6x - 38 = 4 \;\Rightarrow\; 7x = 42 \;\Rightarrow\; x = 6\)
\(y = 3(6) - 19 = -1\)

Elimination

double the second so the \(y\) terms cancel:  \(6x - 2y = 38\)
add it to the first:  \(x + 6x = 4 + 38 \;\Rightarrow\; 7x = 42 \;\Rightarrow\; x = 6\)
back into the first:  \(6 + 2y = 4 \;\Rightarrow\; y = -1\)
Both routes give \((6, -1)\).

Use substitution when one equation already says \(y = \ldots\) or \(x = \ldots\). Use elimination when both are in the \(ax + by = c\) shape. Check the answer in the equation you did not substitute into: \(3(6) - (-1) = 19\). It works.

Direct instruction · 5 of 9 · the perpendicular bisector

The perpendicular bisector of \(AB\) cuts \(AB\) in half, at a right angle.

Question. \(A\) is \((1, 1)\) and \(B\) is \((5, 3)\). Find the equation of the perpendicular bisector of \(AB\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.

line \(AB\)   perpendicular bisector

"Bisector" means it passes through the midpoint \(M\). "Perpendicular" means it meets \(AB\) at a right angle. So you need a point and a gradient.

1 · a point, the midpoint of \(AB\):
\(M = \left(\dfrac{1 + 5}{2}, \dfrac{1 + 3}{2}\right) = (3, 2)\)
2 · gradient of \(AB\), flip it, change the sign:
\(m_{AB} = \dfrac{3 - 1}{5 - 1} = \dfrac{1}{2}\)
perpendicular: \(-2\),  check \(\tfrac{1}{2} \times (-2) = -1\)
3 · the line, using that point and that gradient:
\(y - 2 = -2(x - 3)\)
4 · the form the question asks for, \(ax + by + c = 0\):
\(y - 2 = -2x + 6 \;\Rightarrow\; 2x + y - 8 = 0\)
\(2x + y - 8 = 0\)

Every point on this line is the same distance from \(A\) as from \(B\). That fact comes back next lesson with circles.

Direct instruction · 6 of 9 · the Bisector Lab

Pick a method, predict, then show the line. Only one method always works.

line \(AB\)    the line your method builds    Amber dot: a point on that line, used to test \(PA = PB\)

Direct instruction · 7 of 9 · June 2024 · Paper 12 · Q10(b)

The curve is \(y = (5 - 2x)^{\frac{3}{2}} + 5\). Find the perpendicular bisector of \(AB\).

The curve is defined for \(x < \tfrac{5}{2}\). Point \(A\) on the curve has \(y\)-coordinate 32. Point \(B\) on the curve is where the gradient of the curve is \(-3\).

Find the equation of the perpendicular bisector of \(AB\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [6]

finding \(B\) needs calculus (later this year), so here are both points:  \(A(-2, 32)\), \(B(2, 6)\)

Check \(A\) and \(B\) are on the curve: \(x = -2\) gives \((5 + 4)^{\frac{3}{2}} + 5 = 27 + 5 = 32\), and \(x = 2\) gives \((5 - 4)^{\frac{3}{2}} + 5 = 1 + 5 = 6\).

Draw the picture first. \(A\) is up and to the left, \(B\) is down and to the right, so the gradient of \(AB\) should come out negative.

The \(-3\) is the gradient of the curve at \(B\). We do not use it in this part.

curve   \(AB\)   perpendicular bisector

The two axes use different scales, so the right angle at \(M\) does not look square here.

Direct instruction · 8 of 9 · Q10(b) · where the last 4 marks come from

Four marks, four lines of working. Here is what each code means.

Reading the codes

M is a method mark. You chose a valid method and put the numbers in. A slip in the arithmetic does not lose it, but writing only an answer with no working can.

A is an accuracy mark. The value or answer is correct. It needs the M mark before it.

B is a mark for a correct fact or value that needs no method. It is not used in this part.

* (the star) marks an M mark that a later mark leans on.

D means dependent. DM1 is awarded only if both starred marks were earned.

The number is how many marks the line is worth. Of the 6 marks, the first two (M1 A1) are for finding \(B\) with calculus. The four on this slide are the perpendicular bisector.

M1* · gradient of \(AB\), then the perpendicular gradient:
\(m_1 = \dfrac{32 - 6}{-2 - 2} = -\dfrac{13}{2}\)
perpendicular gradient \(= -\dfrac{1}{m_1} = \dfrac{2}{13}\)
earn it: use both points, then flip and change the sign
M1* · midpoint of \(AB\):
\(\left(\dfrac{-2 + 2}{2}, \dfrac{32 + 6}{2}\right) = (0, 19)\)
earn it: average the \(x\)-values and the \(y\)-values
DM1 · the line through the midpoint:
\(y - 19 = \dfrac{2}{13}(x - 0)\)
earn it: use your midpoint and gradient. It needs both stars.
A1 · the answer in the form asked:
\(13y - 247 = 2x \;\Rightarrow\; 2x - 13y + 247 = 0\)
earn it: correct equation, integers, "= 0". Multiples score.

Check: the midpoint \((0, 19)\) must be on the answer. \(2(0) - 13(19) + 247 = 0\). It works.

Spot the mistake · 3 min

Same June 2024 question, three answers. What is wrong with each? Tap to check.

\(A(-2, 32)\), \(B(2, 6)\), and the gradient of the curve at \(B\) is \(-3\). Find the perpendicular bisector of \(AB\) as \(ax + by + c = 0\).
A.  "The gradient at \(B\) is \(-3\), so the bisector has gradient \(\tfrac{1}{3}\).  \(y - 19 = \tfrac{1}{3}x\), so \(x - 3y + 57 = 0\)."
which line is \(-3\) the gradient of?  the tangent at \(B\). The line \(AB\) is a different line.
Wrong gradient. The bisector is perpendicular to \(AB\), whose gradient is \(-\tfrac{13}{2}\), so it needs \(\tfrac{2}{13}\).
B.  "Gradient \(AB = -\tfrac{13}{2}\), so the bisector has gradient \(\tfrac{2}{13}\).  \(y - 32 = \tfrac{2}{13}(x + 2)\), so \(2x - 13y + 420 = 0\)."
test the midpoint \((0, 19)\):  \(2(0) - 13(19) + 420 = 173\), so the midpoint is not on this line
Wrong point. The line goes through \(A\). A bisector has to go through the midpoint.
C.  "Gradient \(\tfrac{2}{13}\) through the midpoint \((0, 19)\), so \(y = \tfrac{2}{13}x + 19\)."
right line, but read the question:  it asks for \(ax + by + c = 0\) with integers
Wrong form. It loses the A1. Multiply by 13 and move everything to one side: \(2x - 13y + 247 = 0\).

These are the three mistakes the June 2024 examiner report describes for this question.

Practice · together · 6 min

Sketch first. Give every equation as \(ax + by + c = 0\) with integers. Tap to check.

1.  \(A(-3, 1)\), \(B(5, 7)\). Find the length \(AB\), the midpoint of \(AB\), and the equation of the line \(AB\).
length:  across \(5 - (-3) = 8\), up \(7 - 1 = 6\), so \(\sqrt{8^2 + 6^2} = \sqrt{100} = 10\)
midpoint:  \(\left(\tfrac{-3 + 5}{2}, \tfrac{1 + 7}{2}\right) = (1, 4)\)
gradient:  \(\tfrac{7 - 1}{5 - (-3)} = \tfrac{6}{8} = \tfrac{3}{4}\)
\(y - 1 = \tfrac{3}{4}(x + 3) \Rightarrow 4y - 4 = 3x + 9\)
\(10\);  \((1, 4)\);  \(3x - 4y + 13 = 0\)
2.  Find the equation of the line through \((2, -1)\) perpendicular to \(2x + 3y = 6\).
rearrange:  \(3y = -2x + 6 \Rightarrow y = -\tfrac{2}{3}x + 2\), gradient \(-\tfrac{2}{3}\)
flip and change sign:  \(\tfrac{3}{2}\)
\(y + 1 = \tfrac{3}{2}(x - 2) \Rightarrow 2y + 2 = 3x - 6\)
\(3x - 2y - 8 = 0\)
3.  Find the equation of the perpendicular bisector of \(P(1, 5)\) and \(Q(7, -3)\).
midpoint:  \(\left(\tfrac{1 + 7}{2}, \tfrac{5 + (-3)}{2}\right) = (4, 1)\)
gradient \(PQ\):  \(\tfrac{-3 - 5}{7 - 1} = -\tfrac{8}{6} = -\tfrac{4}{3}\), so perpendicular \(\tfrac{3}{4}\)
\(y - 1 = \tfrac{3}{4}(x - 4) \Rightarrow 4y - 4 = 3x - 12\)
\(3x - 4y - 8 = 0\)

Practice · on your own · 4 min, finish at home

Your turn. Tap to check when you have a final answer.

4.  \(A(0, 1)\), \(B(4, 3)\), \(C(2, 7)\). Show that angle \(ABC\) is a right angle, and find the area of triangle \(ABC\).
gradients:  \(AB = \tfrac{3 - 1}{4 - 0} = \tfrac{1}{2}\),  \(BC = \tfrac{7 - 3}{2 - 4} = -2\).  \(\tfrac{1}{2} \times (-2) = -1\), so \(AB \perp BC\) and angle \(ABC = 90^\circ\).
lengths:  \(AB = \sqrt{4^2 + 2^2} = 2\sqrt{5}\),  \(BC = \sqrt{(-2)^2 + 4^2} = 2\sqrt{5}\)
area \(= \tfrac{1}{2} \times\) base \(\times\) height.  \(AB\) and \(BC\) meet at the right angle, so use them: \(\tfrac{1}{2} \times 2\sqrt{5} \times 2\sqrt{5} = \tfrac{1}{2} \times 20\)
Area \(= 10\)
5.  The lines \(2x + y = 7\) and \(x - y = -1\) meet at \(P\). Find \(P\), then the line through \(P\) perpendicular to \(x - y = -1\).
second line:  \(y = x + 1\);  into the first: \(2x + x + 1 = 7 \Rightarrow 3x = 6 \Rightarrow x = 2\), so \(y = 2 + 1 = 3\)
gradient of \(y = x + 1\) is 1:  perpendicular gradient \(-1\)
\(y - 3 = -(x - 2) \Rightarrow y - 3 = -x + 2\)
\(P(2, 3)\);  \(x + y - 5 = 0\)
6.  \(y = 2x + 3\) meets the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\). Find the length \(AB\) and the perpendicular bisector of \(AB\).
\(y = 0\):  \(x = -\tfrac{3}{2}\), so \(A\left(-\tfrac{3}{2}, 0\right)\);  \(x = 0\):  \(B(0, 3)\)
\(AB = \sqrt{\tfrac{9}{4} + 9} = \sqrt{\tfrac{45}{4}} = \tfrac{3}{2}\sqrt{5}\)
midpoint  \(\left(-\tfrac{3}{4}, \tfrac{3}{2}\right)\);  \(AB\) has gradient 2, so  \(-\tfrac{1}{2}\)
\(y - \tfrac{3}{2} = -\tfrac{1}{2}\left(x + \tfrac{3}{4}\right)\);  multiply by 8: \(8y - 12 = -4x - 3\)
\(AB = \tfrac{3}{2}\sqrt{5}\);  \(4x + 8y - 9 = 0\)

Finished early? Prove that perpendicular lines have gradients that multiply to \(-1\). Hint: turn a triangle with sides 1 across and \(m\) up through \(90^\circ\).

Direct instruction · 9 of 9 · recap

Four things to carry out of here.

Gradient, midpoint and length are not in the formula booklet. Know them, sketch first, and leave lengths as simplified surds.

Work in \(y - y_1 = m(x - x_1)\), and convert at the end. "Integers" means clear every fraction and finish with "= 0".

Perpendicular: flip and change the sign, so \(m_1 m_2 = -1\). First check which line the gradient you were given belongs to.

Perpendicular bisector = midpoint + perpendicular gradient. Two pieces of information, one line.

Exit ticket · 5 min

Schoology, Paper 1 → 1.3 Coordinate Geometry.

Four questions. Do the working on paper first. For the last one, give the equation in the form \(ax + by + c = 0\) with integers.

Submit before you leave. An attempt left open scores nothing.

Homework · before next class

A short reading and a few questions on topics we have finished.

Stuck on a workbook question? Write down where you got stuck and bring it. That still counts as done.

Line Lab: type in any two points and check every step of your answer  Lines Sprint: 12 quick moves against the clock

Hurricane Closest Approach: how close does the storm come?  Equidistance Painter: a preview of next lesson