Coordinate Geometry Problems
You now have three formulas and several forms of the equation of a line. Problems from here on rarely say which to use. Reading the question for the word that names the quantity is the whole method.
Three questions about the same two points: how far apart are they, what is halfway between them, and which direction does the line through them run? Which formula answers each?
Each formula answers exactly one kind of question. How long is distance. Which way is gradient. Halfway is midpoint. Problems combine them, but every single step is one of those three questions asked about two points.
$$\text{length} \to d, \qquad \text{direction} \to m, \qquad \text{halfway} \to M$$
Words in the question map onto formulas. Equal, isosceles, perimeter, length call for distance. Parallel, perpendicular, right-angled, collinear call for gradient. Bisect, halfway, median, centre call for midpoint.
Know
- which of the midpoint, gradient and distance formulas answers each kind of question
- that the perimeter of a figure on the plane is the sum of the distances between successive vertices
- that a triangle with a horizontal or vertical side has a base and height that can be read off directly
Understand
- why exact values rather than rounded ones are needed whenever lengths are being compared
- why a condition involving a squared unknown can produce two valid points
- why a conclusion must name the property being used, not only the numbers computed
Can Do
- classify a triangle from the coordinates of its vertices and find its perimeter
- find the area of a triangle with an axis-parallel side
- find an unknown coordinate from a distance, gradient or midpoint condition
Almost every coordinate geometry problem decomposes into questions about pairs of points, and there are only three of them.
How far apart? Distance formula. Triggered by length, perimeter, equal sides, isosceles, equilateral, radius, and by the converse of Pythagoras theorem.
Which direction? Gradient formula. Triggered by parallel, perpendicular, right-angled, collinear, and by any request for an equation of a line.
What is halfway? Midpoint formula. Triggered by bisect, midpoint, median, centre of a diagonal, and by the perpendicular bisector.
Read the question, underline the word, pick the formula. Time lost on these problems is almost never spent on arithmetic; it is spent deciding what to do, and the vocabulary decides it.
A full classification answers two independent questions: are any sides equal, and is there a right angle?
The first is answered by three applications of the distance formula. Keep the values exact, because the comparison is the point of the calculation and rounding blurs it.
The second can be answered either way. The converse of Pythagoras theorem uses the three lengths you already have, so it costs nothing extra, but you must identify the longest side first. Testing gradients for a product of $-1$ locates the right angle directly, but only tests the pair of sides you chose.
Report both findings. "Isosceles" is an incomplete answer if the triangle is also right-angled, and a question asking you to classify "as fully as possible" is asking for exactly that.
Perimeter is straightforward: find each side with the distance formula and add. Surds rarely combine, so an answer such as $2\sqrt{10} + 2\sqrt{13} + 2\sqrt{5}$ is complete, with a rounded value alongside it if asked.
Area is easy when one side is horizontal or vertical. A horizontal side has a length you can read from the difference in x-coordinates, and the height is the vertical distance from the third vertex to that side, read from the difference in y-coordinates. Then use $A = \tfrac{1}{2}bh$.
For the triangle $P(1, 2)$, $Q(7, 2)$, $R(4, 10)$, the side $PQ$ is horizontal with length $6$, and $R$ is $8$ units above it, so the area is $\tfrac{1}{2}(6)(8) = 24$.
No formula is needed for the height when a side is axis-parallel, which is why exam questions so often arrange one to be.
When a coordinate is unknown, give it a letter and let the stated condition supply the equation.
"The point $(2, k)$ is $5$ units from $(6, 1)$" becomes $\sqrt{(6-2)^2 + (1-k)^2} = 5$. Squaring both sides removes the root and leaves $16 + (1-k)^2 = 25$, so $(1-k)^2 = 9$.
Here is the part worth expecting: $(1-k)^2 = 9$ gives $1 - k = 3$ or $1 - k = -3$, so $k = -2$ or $k = 4$. Two answers, and both are correct.
Geometrically this is obvious once seen. A circle of radius $5$ centred at $(6, 1)$ crosses the vertical line $x = 2$ at two points, and each is a valid answer. Giving only one is giving half the solution.
Gradient and midpoint conditions behave differently: they are linear in the unknown and give exactly one answer. It is specifically the squaring in the distance formula that opens the door to two.
These questions are marked on the reasoning, not on the numbers, and the numbers alone are usually worth less than half.
A complete response has three parts. State what you calculated and what it came to. State what that value means. State the conclusion, naming the property that licenses it.
So not "$AB = 5$, $BC = 5$, so isosceles", but "$AB = 5$ and $BC = 5$, so two sides are equal in length, and a triangle with exactly two equal sides is isosceles."
The difference feels pedantic on an easy question and stops feeling pedantic on a hard one, where the property being invoked is the thing actually in doubt. Build the habit where it is easy so that it is available where it is not.
Watch Me Solve It · 3 examples
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1Find all three sides exactly$AB = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}$$BC = \sqrt{(-4)^2 + (-6)^2} = \sqrt{52} = 2\sqrt{13}$$AC = \sqrt{2^2 + (-4)^2} = \sqrt{20} = 2\sqrt{5}$Exact values, so the comparison in the next step is certain rather than approximate.
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2Compare the lengthsThe three values $2\sqrt{10}$, $2\sqrt{13}$ and $2\sqrt{5}$ are all different, since $10$, $13$ and $5$ are different. A triangle with three unequal sides is scalene.
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3Test the longest side with the converse$BC^2 = 52$$AB^2 + AC^2 = 40 + 20 = 60$The longest side is $BC$. Since $52 \neq 60$, the converse of Pythagoras theorem does not apply and the triangle is not right-angled.
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4Add the sides for the perimeter$P = 2\sqrt{10} + 2\sqrt{13} + 2\sqrt{5} \approx 6.32 + 7.21 + 4.47 = 18.01$The surds have different radicands so they cannot be combined; the exact form is the sum itself.
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1Look for an axis-parallel side$P$ and $Q$ share the y-coordinate $2$, so $PQ$ is horizontal. Choosing it as the base means the height needs no formula.
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2Read the base from the x-coordinates$b = 7 - 1 = 6$The length of a horizontal interval is the difference in x-coordinates.
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3Read the height from the y-coordinates$h = 10 - 2 = 8$The perpendicular distance from $R$ to the horizontal line $y = 2$ is the difference in y-coordinates. The x-coordinate of $R$ plays no part.
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4Apply the area formula$A = \tfrac{1}{2}bh = \tfrac{1}{2}(6)(8) = 24$Square units, since the plane is unitless unless the question supplies units.
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1Write the condition as an equation$\sqrt{(6 - 2)^2 + (1 - k)^2} = 5$The unknown sits inside the distance formula, so the condition becomes an equation in $k$.
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2Square both sides and simplify$16 + (1 - k)^2 = 25$$(1 - k)^2 = 9$Squaring is safe because both sides are lengths and so cannot be negative.
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3Take both roots$1 - k = 3 \quad \text{or} \quad 1 - k = -3$$k = -2 \quad \text{or} \quad k = 4$A squared expression equal to $9$ has two solutions. Taking only the positive root loses half the answer.
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4Check both$(2, -2): \ \sqrt{16 + 9} = 5$$(2, 4): \ \sqrt{16 + 9} = 5$Both points are genuinely $5$ units from $(6, 1)$, one above the given point and one below.
Brain Trainer · 5 problems
Five items. Before calculating, name which of the three formulas the wording calls for.
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1 Is the triangle with vertices $(0,0)$, $(4,0)$ and $(0,3)$ right-angled?
Two sides lie along the axes, so they meet at a right angle at the origin.Yes, at $(0, 0)$ -
2 Find the perimeter of the triangle with vertices $(0,0)$, $(4,0)$ and $(0,3)$.
Sides of $4$, $3$ and $\sqrt{16 + 9} = 5$.$12$ -
3 Find the area of the triangle with vertices $(-3, 1)$, $(5, 1)$ and $(2, 7)$.
The first two share a y-coordinate, so the base is $8$ and the height is $6$.$24$ square units -
4 The point $(a, 3)$ is the midpoint of the interval joining $(-4, 1)$ and $(10, 5)$. Find $a$.
The midpoint's x-coordinate is $\dfrac{-4 + 10}{2}$.$a = 3$ -
5 The point $(0, t)$ is $13$ units from $(5, 0)$. Find all values of $t$.
$25 + t^2 = 169$, so $t^2 = 144$.$t = 12$ or $t = -12$
Multiple Choice · 5 questions
To decide whether a triangle is isosceles from the coordinates of its vertices, you should use:
The triangle with vertices $(0, 0)$, $(6, 0)$ and $(6, 8)$ has perimeter:
For the triangle with vertices $(2, 5)$, $(2, -1)$ and $(9, 3)$, the most convenient base and its height are:
The point $(1, p)$ is $10$ units from $(7, 5)$. The number of possible values of $p$ is:
A student computes three side lengths as $\sqrt{13}$, $\sqrt{13}$ and $\sqrt{26}$ and writes "isosceles". The response is:
Short Answer · 3 questions
The wording names the formula
Length, perimeter and equal sides call for distance. Parallel, perpendicular and collinear call for gradient. Bisect, halfway and median call for midpoint.
Keep values exact while comparing
$\sqrt{32}$ against $\sqrt{32}$ is a decision. $5.66$ against $5.66$ is a coincidence you cannot rely on, since different surds can round the same.
An axis-parallel side gives the height free
Read the base from one coordinate difference and the height from the other, then use $A = \tfrac{1}{2}bh$ with no further calculation.
A distance condition usually has two answers
Squaring the unknown produces a positive and a negative root, and both normally give valid points. Gradient and midpoint conditions stay linear and give one.
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