📐 Coordinate Geometry — Revision Notes

Walkthrough 1 — length of a line segment Basic

1. Length of PQ = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
   A distance between two points is Pythagoras' Theorem written in coordinates: the segment is the hypotenuse of a right-angled triangle whose other two sides run along the axes. Quoting the formula is the first line of working, and it carries a mark.
2. = $\sqrt{{(7-(\mathrm{-5}))}^2+{(\mathrm{-2}-3)}^2}$
   Take (x₁, y₁) = P and (x₂, y₂) = Q, and substitute. Each negative coordinate goes in inside its own brackets, so the double minus in 7 − (−5) stays visible.
3. = $\sqrt{{12}^2+{(\mathrm{-5})}^2}$ = $\sqrt{144+25}$
   12 is the horizontal change and −5 the vertical change: Q is 12 to the right of P and 5 below it. Squaring makes both positive, which is why the order of the two points does not affect the length.
4. = $\sqrt{169}$
   144 + 25 = 169. The addition happens inside the root, because the theorem squares the two sides and adds those; taking the roots first would give 12 + 5 = 17, which is the length of the path round the triangle, not across it.
5. = 13 units
   169 is a perfect square, so the length is exact — nothing to round and no decimal point. Write units: a length is a measurement even when the axes carry no scale.

Walkthrough 2 — when the length is not a whole number Basic

1. CD = $\sqrt{{(4-(\mathrm{-2}))}^2+{(\mathrm{-2}-5)}^2}$
   Straight into the distance formula, with (x₁, y₁) = C and (x₂, y₂) = D. Each substituted value keeps its own brackets so that 4 − (−2) cannot collapse to 4 − 2.
2. = $\sqrt{6^2+{(\mathrm{-7})}^2}$ = $\sqrt{36+49}$
   6 across and 7 down. (−7)² = 49, not −49 — the square of a negative number is positive, so a downward step contributes exactly as much length as an upward one of the same size.
3. = $\sqrt{85}$ = 9.2195…
   85 is not a perfect square, so this length is not exact. Copy several figures off the calculator before doing anything else: the rounding is a separate, final step, and rounding here would be rounding twice.
4. = 9.22 units (to 3 s.f.)
   The first three significant figures are 9, 2 and 1; the next figure is 9, so the last one is rounded up to 2. State the accuracy in brackets so the marker can see you knew the answer was not exact.

Walkthrough 3 — a distance that is given, a coordinate that is not Intermediate

1. ${(9-1)}^2+{(4-k)}^2={10}^2$
   The distance formula has been used with both sides squared. Squaring clears the root immediately and turns the whole question into ordinary algebra; it is safe here because both sides are lengths, and lengths are never negative.
2. 64 + ${(4-k)}^2$ = 100
   8² = 64 and 10² = 100. The bracket (4 − k)² is deliberately left un-expanded — expanding it would produce a k² that only has to be undone again.
3. ${(4-k)}^2=36$
   Subtracting 64 from both sides leaves the square by itself, which is the one shape you are allowed to take a square root of in a single step.
4. 4 − k = 6  or  4 − k = −6
   36 has two square roots, 6 and −6, so the bracket could be either. The plural in "possible values" is the question saying that both of them count.
5. k = −2  or  k = 10
   4 − k = 6 gives k = −2, and 4 − k = −6 gives k = 10. Both are genuine: A is 8 units across from B either way, so it can sit 6 below B's level or 6 above it.

Walkthrough 4 — gradient from two points Basic

1. Gradient of AB = $\frac{y_2-y_1}{x_2-x_1}$
   Gradient measures steepness as vertical change divided by horizontal change. Writing the formula down first fixes which difference goes on top: the y-difference.
2. = $\frac{8-(\mathrm{-1})}{2-(\mathrm{-4})}$
   Taking (x₁, y₁) = A and (x₂, y₂) = B, both of A's coordinates are the ones being subtracted — and both are negative, so both subtractions turn into additions.
3. = $\frac{9}{6}$
   8 − (−1) = 9 is the rise and 2 − (−4) = 6 is the run: the line climbs 9 units for every 6 units it travels to the right.
4. = $\frac{3}{2}$
   Cancel the common factor 3. A gradient is left as a fraction in lowest terms rather than 1.5, because $\frac{3}{2}$ reads as "up 3 for every 2 across". It is positive, which agrees with a line that rises to the right.

Walkthrough 5 — a negative gradient Basic

1. Gradient of PQ = $\frac{\mathrm{-5}-7}{5-(\mathrm{-3})}$
   Straight into (y₂ − y₁) ÷ (x₂ − x₁) with (x₁, y₁) = P. Both of P's coordinates are the subtracted ones, top and bottom.
2. = $\frac{\mathrm{-12}}{8}$
   −5 − 7 = −12: the line drops 12 units while travelling 8 units to the right. A negative on top over a positive underneath makes the whole fraction negative.
3. = −$\frac{3}{2}$
   Divide top and bottom by 4, and write the minus sign once, in front of the fraction. A negative gradient is the algebra saying what the picture would: this line falls from left to right.

Walkthrough 6 — the gradient is given, a coordinate is not Intermediate

1. $\frac{h-6}{5-h}=\mathrm{-2}$
   The gradient formula is an equation, not just a recipe. Putting the two points into it and setting the result equal to the given gradient leaves h as the only unknown.
2. h − 6 = −2(5 − h)
   Multiplying both sides by (5 − h) clears the fraction. The whole of that bracket is being multiplied by −2, so the bracket stays.
3. h − 6 = −10 + 2h
   Expanding: −2 × 5 = −10 and −2 × (−h) = +2h. That second sign is the one most often dropped.
4. h = 4
   Collecting: −6 + 10 = 2h − h, so 4 = h. Check it — the two points become (4, 6) and (5, 4), and (4 − 6) ÷ (5 − 4) = −2, as required.

Walkthrough 7 — three collinear points Intermediate

1. Since A, B and C lie on one straight line, gradient of AC = gradient of AB.
   "Collinear" is a geometric word with an algebraic meaning: one line, therefore one gradient, whichever pair of the three points you measure it from. Saying so turns the word into something you can compute with.
2. Gradient of AC = $\frac{\mathrm{-7}-5}{7-(\mathrm{-2})}$ = $\frac{\mathrm{-12}}{9}$ = −$\frac{4}{3}$
   A and C are the two points with no unknown in them, so this gradient comes out as a plain number. Cancelling it to −$\frac{4}{3}$ now keeps the arithmetic small later.
3. Gradient of AB = $\frac{k-5}{1-(\mathrm{-2})}$ = $\frac{k-5}{3}$
   The same formula with B in place of C, and the same point A subtracted, so the two gradients are directly comparable. The unknown k stays where the formula puts it, on top.
4. $\frac{k-5}{3}=-\frac{4}{3}$
   This is the first line written in symbols. Both sides happen to be over 3, which is a small gift: the numerators can be compared directly.
5. k = 1
   Multiplying both sides by 3 gives k − 5 = −4, so k = 1. Check: B is (1, 1), and the gradient of AB is (1 − 5) ÷ 3 = −$\frac{4}{3}$, matching AC.

Walkthrough 8 — two parallel line segments Advanced

1. Since AB is parallel to CD, gradient of AB = gradient of CD.
   Parallel lines never meet, which is only possible if they rise at exactly the same rate. So "parallel" is the question handing you an equation between two gradients — the one fact the rest of the working hangs on.
2. Gradient of CD = $\frac{4-(\mathrm{-5})}{7-1}$ = $\frac{9}{6}$ = $\frac{3}{2}$
   C and D are both fully known, so this side of the equation is just a number. Cancelling $\frac{9}{6}$ to $\frac{3}{2}$ keeps the numbers small.
3. Gradient of AB = $\frac{k-3}{4-(\mathrm{-2})}$ = $\frac{k-3}{6}$
   The same formula for the other segment. The unknown k is part of the y-difference, so it stays on top; the bottom is a number because both x-coordinates are known.
4. $\frac{k-3}{6}=\frac{3}{2}$
   The first line, now in symbols. Nothing has been solved yet — this is the equation the question was really asking you to build.
5. k = 12
   Multiplying both sides by 6 gives k − 3 = 9, so k = 12. Check: B is (4, 12), and the gradient of AB is (12 − 3) ÷ 6 = $\frac{3}{2}$, the same as CD.

Walkthrough 9 — equation from a gradient and a point Basic

1. y = −3x + c
   Every line that is not vertical can be written y = mx + c, and m is the gradient — which was given. Only c is still missing, so the whole question has come down to finding one number.
2. −1 = −3(4) + c
   The point lies on the line, so its coordinates satisfy the equation: put x = 4 and y = −1 into it. The bracket around the 4 stops −3 × 4 from being misread as −34.
3. −1 = −12 + c
   −3 × 4 = −12. The equation now has one unknown and plain numbers everywhere else.
4. c = 11, so y = −3x + 11
   Adding 12 to both sides gives c = 11. The question asked for the equation, not for c, so the last line is the whole equation with y on the left.

Walkthrough 10 — equation from two points Intermediate

1. m = $\frac{\mathrm{-5}-7}{4-(\mathrm{-2})}$ = $\frac{\mathrm{-12}}{6}$ = −2
   Two points give a gradient, and the gradient is the m of y = mx + c. Nothing about the equation can be written down until this number exists, so it is the first line.
2. y = −2x + c
   Putting the gradient into the general form narrows the answer from "some line" to "one of the family of parallel lines with gradient −2". The value of c is what picks out the single member of that family.
3. 7 = −2(−2) + c
   A lies on the line, so x = −2 and y = 7 satisfy the equation. Either of the two given points would do; A is used here, which leaves B free for a check at the end.
4. 7 = 4 + c, so c = 3
   −2 × (−2) = +4 — two negatives multiply to a positive — and subtracting 4 from both sides leaves c = 3.
5. y = −2x + 3
   The equation written out in full. Check it against the point that was not used: −2(4) + 3 = −5, which is exactly B's y-coordinate, so the line really does pass through both.

Walkthrough 11 — gradient and intercepts from ax + by = k Intermediate

1. 3y = −4x + 24
   A gradient and a y-intercept can only be read off the form y = mx + c, so the first job is to make y the subject. Subtracting 4x from both sides starts that.
2. y = −$\frac{4}{3}$x + 8
   Dividing by 3 leaves y alone. Every term is divided — the 4x and the 24 — because dividing only some of them would change the equation.
3. Gradient = −$\frac{4}{3}$; l crosses the y-axis at (0, 8)
   Comparing with y = mx + c reads the constants off: m = −$\frac{4}{3}$ and c = 8. The question asked for coordinates, so the 8 becomes the point (0, 8): every point on the y-axis has x = 0.
4. When y = 0:   4x = 24
   Every point on the x-axis has y = 0, so substituting y = 0 is what finds the crossing. Using the original 4x + 3y = 24 rather than the rearranged version keeps the numbers whole.
5. x = 6, so l crosses the x-axis at (6, 0)
   24 ÷ 4 = 6. Again the answer is a point, so both coordinates are written — and this time it is the second one that is zero.

Walkthrough 12 — the equation of a parallel line Advanced

1. 3y = −2x + 12, so y = −$\frac{2}{3}$x + 4
   The gradient of the given line is hidden while y is not the subject, so it is rearranged first. Every term is divided by 3.
2. Gradient of 2x + 3y = 12 is −$\frac{2}{3}$
   Comparing with y = mx + c reads the gradient straight off. Only the gradient of the given line has been taken; its y-intercept of 4 belongs to that line alone.
3. Parallel → same gradient, so y = −$\frac{2}{3}$x + c
   Parallel lines have equal gradients, so the required line has the same m. Every line parallel to the given one has this equation for some value of c, and the problem is down to a single unknown.
4. −1 = −$\frac{2}{3}$(6) + c
   The required line passes through (6, −1), so those coordinates satisfy its equation. Substituting them is the only piece of information that has not been used yet, and it is what fixes c.
5. −1 = −4 + c, so c = 3 and y = −$\frac{2}{3}$x + 3
   −⅔ × 6 = −4, so c = 3. The answer satisfies both conditions: its gradient still matches the given line, and putting x = 6 in gives −4 + 3 = −1, the required point.

Walkthrough 13 — showing a triangle is right-angled Intermediate

1. AB² = (−2 − 1)² + (1 − 5)² = 9 + 16 = 25
   Work with the square of each length, not the length itself. Pythagoras' Theorem is a statement about squares, so stopping at AB² saves taking a root now only to square it again in two lines' time.
2. BC² = (−6 − (−2))² + (4 − 1)² = 16 + 9 = 25
   The same calculation for the second side. −6 − (−2) = −4, and (−4)² = 16 — the minus sign disappears into the square.
3. AC² = (−6 − 1)² + (4 − 5)² = 49 + 1 = 50
   And the third. AC² is the largest of the three, so AC is the longest side — and only the longest side can be a hypotenuse.
4. AB² + BC² = 25 + 25 = 50 = AC²
   This is the converse of Pythagoras' Theorem: if the squares on the two shorter sides add up to the square on the longest, the triangle is right-angled. The two sides that were added are AB and BC.
5. So triangle ABC is right-angled, with the right angle at B. Also AB² = BC² = 25, so AB = BC = 5 units and the triangle is isosceles.

   Fig. 11.7

   The right angle is between AB and BC, the sides whose squares were added, and it is opposite the longest side, AC. Two equal squares mean two equal sides, so it is isosceles. A "show that" ends with the conclusion in words.

Walkthrough 14 — the missing vertex of a parallelogram Intermediate

1. In parallelogram ABCD, AD is parallel to BC and AB is parallel to DC.
   A parallelogram is defined by its two pairs of parallel opposite sides, so those two facts are the only information the shape's name carries. Reading the vertices in order, A to B to C to D, is what identifies which sides are opposite which.
2. Gradient of BC = (1 − 1) ÷ (5 − (−1)) = 0, so BC is horizontal — and therefore so is AD. Hence D has the same y-coordinate as A: D is (d, −2).
   A gradient of 0 means a horizontal line, and a line parallel to a horizontal line is horizontal too. That fixes one of D's two coordinates immediately, leaving a single unknown to chase.
3. Gradient of AB = $\frac{1-(\mathrm{-2})}{\mathrm{-1}-(\mathrm{-5})}$ = $\frac{3}{4}$
   The other pair of parallel sides is AB and DC, and AB is the one with both endpoints known — so it is the one whose gradient can be worked out as a number.
4. Gradient of DC = $\frac{1-(\mathrm{-2})}{5-d}$ = $\frac{3}{5-d}$, and this equals $\frac{3}{4}$
   The same gradient formula, now with the unknown d in it, set equal to the gradient of AB because the two sides are parallel. This is the equation the whole question was built around.
5. 5 − d = 4, so d = 1 and D is (1, −2).

   Fig. 11.8

   Two fractions with equal tops are equal only if their bottoms match, so 5 − d = 4 and d = 1. Check: AD runs from (−5, −2) to (1, −2), horizontal like BC, and DC has gradient 3 ÷ 4 like AB.

Walkthrough 15 — area, and a perpendicular distance from it Intermediate

1. P and R have the same y-coordinate, so PR is horizontal.
   PR = 5 − (−4) = 9 units, and the height of Q above PR is 6 − 2 = 4 units.

   Fig. 11.9

   A horizontal side is worth hunting for: its length is just the difference of the x-coordinates, and the perpendicular height to it is just the difference of the y-coordinates. Neither needs a square root.
2. Area of triangle PQR = ½ × 9 × 4 = 18 units²
   Area of a triangle = ½ × base × height, taking PR as the base and the 4 as the height perpendicular to it. The unit is units², because an area is a length multiplied by a length.
3. PQ = $\sqrt{{(3-(\mathrm{-4}))}^2+{(6-2)}^2}$ = $\sqrt{49+16}$ = $\sqrt{65}$
   The base changes to PQ, which is parallel to neither axis, so this one does need the distance formula. It is left as the exact $\sqrt{65}$: the accuracy the question asks for applies to the final answer only.
4. Taking PQ as the base and the perpendicular from R as the height,
   ½ × $\sqrt{65}$ × height = 18
   A triangle has one area however you measure it, so ½ × base × height gives 18 from this side too. Taking PQ as the base makes the matching height the perpendicular the question asks for, so part (i)'s area is the bridge.
5. height = $\frac{36}{\sqrt{65}}$ = 4.4652… = 4.47 units (to 3 s.f.)
   Multiplying both sides by 2 and dividing by $\sqrt{65}$ gives 36 ÷ $\sqrt{65}$. Evaluate it in one calculator step, so the only rounding is the last one: 4.4652… → 4.47.

Walkthrough 16 — a point equidistant from two others Advanced

1. Let the coordinates of P be (0, k).
   Every point on the y-axis has x-coordinate 0, so half of P is known before any calculation starts. Naming the other half k turns "somewhere on the axis" into something algebra can hold on to.
2. PA² = (−6 − 0)² + (1 − k)² = 36 + (1 − k)²
   PB² = (2 − 0)² + (5 − k)² = 4 + (5 − k)²
   Both distances are written as squares. Two positive lengths are equal exactly when their squares are equal, so working in squares avoids carrying two square roots through the algebra for nothing.
3. 36 + (1 − k)² = 4 + (5 − k)²
   Setting PA² = PB² is the word "equidistant" turned into an equation. This is the line the method marks are attached to; everything after it is routine algebra.
4. 36 + 1 − 2k + k² = 4 + 25 − 10k + k²
   37 − 2k = 29 − 10k
   Expanding both brackets: (1 − k)² = 1 − 2k + k² and (5 − k)² = 25 − 10k + k². The k² appears on both sides and cancels, leaving a linear equation.
5. 8 = −8k, so k = −1 and P is (0, −1).
   Collecting: 37 − 29 = −10k + 2k gives 8 = −8k. Check: PA² and PB² both come to 40. The answer is asked for as coordinates, so it is written as a point.