Lesson 39 of 42 · Quick revision

Revision: geometry, counting, probability and statistics

Lines, circles, parabolas, areas and volumes, the counting matrix, probability rules, averages and compound percentages on one sheet, with drills.

About 30 minutes

In this lesson

  • Recall the line, circle and parabola formulas and read centres, radii and vertices from equations.
  • Choose between permutations, arrangements with repeats and combinations by asking whether order matters.
  • Use inclusion–exclusion and the complement rule for "or" and "at least one" probabilities.
  • Work averages through totals and chain percentage changes as multipliers.

The idea

How to use this sheet

These formulas come up across the mathematics questions, and without a calculator the fastest route is usually the formula plus a quick sanity check. Read each block, cover it, and say it back; then do the drills below without looking.

Lines in the plane

For points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2):

Quantity Formula
Distance d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Midpoint M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
Slope m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (for x1≠x2x_1 \neq x_2)
Line through a point with slope mm y−y1=m(x−x1)y - y_1 = m(x - x_1)
Parallel lines m1=m2m_1 = m_2
Perpendicular lines (neither vertical) m1m2=−1m_1 m_2 = -1
Distance from (x0,y0)(x_0, y_0) to ax+by+c=0ax + by + c = 0 ∣ax0+by0+c∣a2+b2\frac{\lvert ax_0 + by_0 + c \rvert}{\sqrt{a^2 + b^2}}

The line through A(1,2)A(1, 2) and B(3,8)B(3, 8) has slope 8−23−1=3\frac{8 - 2}{3 - 1} = 3, so y−2=3(x−1)y - 2 = 3(x - 1), that is y=3x−1y = 3x - 1; both points satisfy it. The distance from (1,2)(1, 2) to 3x+4y−1=03x + 4y - 1 = 0 is ∣3+8−1∣5=2\frac{\lvert 3 + 8 - 1 \rvert}{5} = 2. A line written as ax+by+c=0ax + by + c = 0 has slope −ab-\frac{a}{b} when b≠0b \neq 0.

Circles

A circle with centre (h,k)(h, k) and radius rr is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. In the expanded form x2+y2+ax+by+c=0x^2 + y^2 + ax + by + c = 0:

centre=(−a2,−b2)r2=a24+b24−c\text{centre} = \left(-\frac{a}{2}, -\frac{b}{2}\right) \qquad r^2 = \frac{a^2}{4} + \frac{b^2}{4} - c

It is a real circle only if r2>0r^2 > 0, and only if x2x^2 and y2y^2 have equal coefficients and there is no xyxy term. Completing the square gives the same results when you forget the formulas.

Parabolas

For y=ax2+bx+cy = ax^2 + bx + c with a≠0a \neq 0, the vertex has xV=−b2ax_V = -\frac{b}{2a}; substitute it to get yVy_V. The axis of symmetry is the vertical line x=xVx = x_V. The parabola opens upwards if a>0a > 0 (the vertex is a minimum) and downwards if a<0a < 0. For y=2x2−8x+5y = 2x^2 - 8x + 5: xV=84=2x_V = \frac{8}{4} = 2 and yV=2(4)−16+5=−3y_V = 2(4) - 16 + 5 = -3, so the vertex is (2,−3)(2, -3), a minimum.

Areas and volumes

Figure Formula
Triangle 12×base×height\frac{1}{2} \times \text{base} \times \text{height}
Equilateral triangle, side ss 34s2\frac{\sqrt{3}}{4}s^2
Trapezoid 12(b1+b2)h\frac{1}{2}(b_1 + b_2)h
Rhombus, diagonals d1d_1, d2d_2 12d1d2\frac{1}{2}d_1 d_2
Circle area πr2\pi r^2, circumference 2πr2\pi r
Sector with angle θ\theta degrees arc θ360⋅2πr\frac{\theta}{360} \cdot 2\pi r, area θ360⋅πr2\frac{\theta}{360} \cdot \pi r^2
Cylinder V=πr2hV = \pi r^2 h
Cone V=13πr2hV = \frac{1}{3}\pi r^2 h
Sphere V=43πr3V = \frac{4}{3}\pi r^3

A sector's arc and area are the same fraction of the full circumference and the full area, because both are set by the angle.

The counting matrix

Ask one question first: does the order of the chosen items matter?

Model Order matters? Formula Example
Arranging all nn items yes n!n! 5 books on a shelf: 5!=1205! = 120
Ordered choice of kk from nn yes (ranked slots) n!(n−k)!\frac{n!}{(n - k)!} a representative and a deputy from 6: 6×5=306 \times 5 = 30
Arranging items with repeats yes, but identical items are interchangeable n!r1! r2!⋯\frac{n!}{r_1! \, r_2! \cdots} the letters of BOCCONI: 7!2! 2!=1260\frac{7!}{2! \, 2!} = 1260
Unordered choice of kk from nn no (a group) (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n - k)!} 4 delegates from 12: (124)=495\binom{12}{4} = 495

Independent stages multiply (the product rule); separate cases that cannot happen together add (the sum rule).

Probability

With equally likely outcomes, P(A)=favourabletotalP(A) = \frac{\text{favourable}}{\text{total}}. Two rules do most of the work:

  • Inclusion–exclusion: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). The overlap is counted inside both P(A)P(A) and P(B)P(B), so subtract it once.
  • Complement: P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none}). For four tosses of a fair coin, P(no heads)=(12)4=116P(\text{no heads}) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}, so P(at least one head)=1516P(\text{at least one head}) = \frac{15}{16}.

For independent events, P(A∩B)=P(A) P(B)P(A \cap B) = P(A)\,P(B).

Statistics

  • Work through totals: total == mean ×\times count. Class A (20 students, mean 80) and class B (30 students, mean 70) together have mean 1600+210050=370050=74\frac{1600 + 2100}{50} = \frac{3700}{50} = 74, not the plain average 75 of the two means.
  • Median: the middle value of the sorted data (the mean of the two middle values when the count is even). It is resistant to outliers: in a set of three or more values, making the largest value much larger moves the mean but leaves the median where it is.
  • Shifting and scaling: adding kk to every value adds kk to the mean and median and leaves the range, variance and standard deviation unchanged. Multiplying every value by kk multiplies the mean by kk, the standard deviation by ∣k∣\lvert k \rvert and the variance by k2k^2.

Compound percentages

A rise of p%p\% is the multiplier 1+p1001 + \frac{p}{100}, a fall is 1−p1001 - \frac{p}{100}, and successive changes multiply. Up 20% then down 20% is 1.2×0.8=0.961.2 \times 0.8 = 0.96: a net fall of 4%, not 0%. To undo a change, divide by its multiplier: after a 15% discount a laptop costs €680, so the original price was 6800.85=68 00085=800\frac{680}{0.85} = \frac{68\,000}{85} = 800 euros (dividing top and bottom by 17 gives 40005\frac{4000}{5}). Adding 15% to 680 gives 782, which is wrong.

Worked examples

Example 1: Centre and radius from the expanded form

Find the centre and radius of the circle x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0.

  1. Read a=−6a = -6, b=4b = 4, c=−12c = -12. The centre is (−a2,−b2)=(3,−2)\left(-\frac{a}{2}, -\frac{b}{2}\right) = (3, -2).

  2. The radius: r2=364+164−(−12)=9+4+12=25r^2 = \frac{36}{4} + \frac{16}{4} - (-12) = 9 + 4 + 12 = 25, so r=5r = 5.

  3. Check by completing the square: (x−3)2−9+(y+2)2−4−12=0(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0, so (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25.

Answer

Centre (3,−2)(3, -2), radius 5. The usual slips are the signs of the centre, (−3,2)(-3, 2), and subtracting 12 instead of adding it, which gives r2=1r^2 = 1.

Example 2: From a sector to the whole circle

A sector has an arc of length 8π8\pi, and its area is 49\frac{4}{9} of the area of the full circle. Find the area of the full circle.

  1. The sector's area and arc are the same fraction of the whole, because both are fixed by the angle. So the arc is 49\frac{4}{9} of the circumference: 8π=49C8\pi = \frac{4}{9}C, and C=18πC = 18\pi.

  2. From 2πr=18π2\pi r = 18\pi, the radius is r=9r = 9.

  3. The full area is πr2=81π\pi r^2 = 81\pi.

  4. Second route: a sector's area is 12×r×arc=12×9×8π=36π\frac{1}{2} \times r \times \text{arc} = \frac{1}{2} \times 9 \times 8\pi = 36\pi, and 36π÷49=81π36\pi \div \frac{4}{9} = 81\pi.

Answer

81π81\pi. Stopping at the sector's own area, 36π36\pi, answers a different question.

Example 3: Arrangements with repeated letters

How many different arrangements are there of the letters of BOCCONI?

  1. Count the letters: B, O, C, C, O, N, I, so 7 letters with O twice and C twice; the other three appear once.

  2. If all 7 were different there would be 7!=50407! = 5040 arrangements. Swapping the two Os, or the two Cs, gives the same word, so divide by 2!2! for each: 50402×2=1260\frac{5040}{2 \times 2} = 1260.

  3. Second route: choose 2 of the 7 positions for the Cs ((72)=21\binom{7}{2} = 21), 2 of the remaining 5 for the Os ((52)=10\binom{5}{2} = 10), and arrange B, N, I in the last 3 (3!=63! = 6): 21×10×6=126021 \times 10 \times 6 = 1260.

Answer

  1. Using 7!=50407! = 5040 counts each word four times; dividing by 2!2! only once gives 2520.

Example 4: Inclusion–exclusion with a 40-card deck

A deck has 40 cards: four suits (coins, cups, swords and clubs) of 10 cards each, and each suit has one ace. One card is drawn at random. What is the probability that it is a coin or an ace?

  1. P(coin)=1040P(\text{coin}) = \frac{10}{40} and P(ace)=440P(\text{ace}) = \frac{4}{40}.

  2. The ace of coins is in both groups: P(coin and ace)=140P(\text{coin and ace}) = \frac{1}{40}.

  3. Inclusion–exclusion: 1040+440−140=1340\frac{10}{40} + \frac{4}{40} - \frac{1}{40} = \frac{13}{40}.

  4. Second route, by counting: the 10 coins plus the 3 aces of the other suits make 13 favourable cards out of 40.

Answer

1340\frac{13}{40}. Adding without subtracting the overlap gives 1440\frac{14}{40}, which counts the ace of coins twice.

Common traps

  • Perpendicular slope: only half the change

    A perpendicular slope is the negative reciprocal: slope 23\frac{2}{3} becomes −32-\frac{3}{2}. Changing only the sign (−23-\frac{2}{3}) or only flipping the fraction (32\frac{3}{2}) gives lines that are not perpendicular. Check by multiplying: the product must be −1-1.

  • Circle signs and the radius squared

    (x+3)2+(y−1)2=16(x + 3)^2 + (y - 1)^2 = 16 has centre (−3,1)(-3, 1), because x+3=x−(−3)x + 3 = x - (-3), and radius 4, not 16. In the expanded form, r2=a24+b24−cr^2 = \frac{a^2}{4} + \frac{b^2}{4} - c: a negative cc makes r2r^2 larger.

  • Ordered or unordered?

    "Choose a president and a secretary" is ordered (the roles differ); "choose a committee of two" is not. Using (nk)\binom{n}{k} for distinct roles undercounts by k!k!, and using n!(n−k)!\frac{n!}{(n - k)!} for a group overcounts by k!k!. Ask whether swapping two chosen people gives a different outcome.

  • Counting the overlap twice

    "Or" in probability includes both. Adding P(A)P(A) and P(B)P(B) without subtracting P(A∩B)P(A \cap B) is right only when the events cannot happen together. If the sum comes out above 1, you have certainly double-counted.

  • Averaging averages

    The mean of two groups is the plain average of their means only when the groups are the same size. Otherwise convert each mean into a total, add the totals, and divide by the total count.

Check yourself

Answer each question before you open its answer.

Question 1: A perpendicular line through a point

Mathematics: analytical geometry

Which line passes through (2,−1)(2, -1) and is perpendicular to 2x−3y+6=02x - 3y + 6 = 0?

  1. A.y=−32x+2y = -\frac{3}{2}x + 2
  2. B.y=23x−73y = \frac{2}{3}x - \frac{7}{3}
  3. C.y=−23x+13y = -\frac{2}{3}x + \frac{1}{3}
  4. D.y=32x−4y = \frac{3}{2}x - 4
Show the answer

Answer: A. y=−32x+2y = -\frac{3}{2}x + 2

Working

Rewrite the given line as y=23x+2y = \frac{2}{3}x + 2, so its slope is 23\frac{2}{3}. A perpendicular slope is −32-\frac{3}{2}. Through (2,−1)(2, -1): y+1=−32(x−2)y + 1 = -\frac{3}{2}(x - 2), so y=−32x+3−1=−32x+2y = -\frac{3}{2}x + 3 - 1 = -\frac{3}{2}x + 2. Check: at x=2x = 2, y=−3+2=−1y = -3 + 2 = -1. All four lines pass through the point, so the slope alone decides.

Why the other options are wrong

  • B. This keeps the given slope 23\frac{2}{3}, so it is the parallel line through the point.
  • C. This only changes the sign of the slope; the product 23×(−23)\frac{2}{3} \times \left(-\frac{2}{3}\right) is not −1-1.
  • D. This flips the fraction but keeps the sign; the product 23×32\frac{2}{3} \times \frac{3}{2} is +1+1, not −1-1.

What carries over

For a perpendicular line, flip the slope and change its sign, then check that the product of the slopes is −1.

Question 2: Roles and a committee

Mathematics: discrete mathematics

A club has 9 members. It elects a president and a treasurer (two different people), and then picks a 2-person events committee from the 7 members left. In how many ways can this be done?

  1. A.126
  2. B.756
  3. C.1512
  4. D.3024
Show the answer

Answer: C. 1512

Working

The two roles are ordered: 9×8=729 \times 8 = 72 ways. The committee is an unordered pair from the remaining 7: (72)=21\binom{7}{2} = 21. The stages are independent, so multiply: 72×21=151272 \times 21 = 1512.

Why the other options are wrong

  • A. This is (94)\binom{9}{4}: it treats all four people as one unordered group and ignores that president and treasurer are different roles.
  • B. This is (92)×(72)\binom{9}{2} \times \binom{7}{2}: it treats the president and treasurer as an unordered pair, so it halves the count.
  • D. This is 9×8×7×69 \times 8 \times 7 \times 6: it treats the two committee places as ordered, so each committee is counted twice.

What carries over

Split a selection into stages, decide for each stage whether order matters, and multiply the stages.

Question 3: At least one six

Mathematics: probability

A fair die is rolled three times. What is the probability of getting at least one six?

  1. A.75216\frac{75}{216}
  2. B.91216\frac{91}{216}
  3. C.12\frac{1}{2}
  4. D.125216\frac{125}{216}
Show the answer

Answer: B. 91216\frac{91}{216}

Working

Use the complement. The probability of no six in one roll is 56\frac{5}{6}, so in three independent rolls it is (56)3=125216\left(\frac{5}{6}\right)^3 = \frac{125}{216}. Then P(at least one six)=1−125216=91216P(\text{at least one six}) = 1 - \frac{125}{216} = \frac{91}{216}. Check by cases: exactly one six 75216\frac{75}{216}, exactly two 3×5216=152163 \times \frac{5}{216} = \frac{15}{216}, three 1216\frac{1}{216}; total 91216\frac{91}{216}.

Why the other options are wrong

  • A. This is the probability of exactly one six, 3×16×25363 \times \frac{1}{6} \times \frac{25}{36}; it leaves out two and three sixes.
  • C. This adds 16\frac{1}{6} three times, which counts the outcomes with more than one six more than once.
  • D. This is the probability of no six at all; it still has to be subtracted from 1.

What carries over

For "at least one", find the probability of none and subtract it from 1.

Key points

  • Perpendicular slopes multiply to −1; parallel slopes are equal.

  • In x2+y2+ax+by+c=0x^2 + y^2 + ax + by + c = 0 the centre is (−a2,−b2)\left(-\frac{a}{2}, -\frac{b}{2}\right) and r2=a24+b24−cr^2 = \frac{a^2}{4} + \frac{b^2}{4} - c; the parabola's vertex is at x=−b2ax = -\frac{b}{2a}.

  • Counting: order matters gives arrangements, order does not gives combinations, and repeated items divide by the factorial of each repeat.

  • Probability: subtract the overlap for "or", and use 1 minus "none" for "at least one".

  • Averages go through totals, and percentage changes chain as multipliers.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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