Points, Lines & Planes

Study Sheet

Points, Lines & Planes

Foundations: undefined terms, segments, midpoints & distance

Undefined Terms: Point, Line, Plane

Concept
The Three Building Blocks

In geometry we start with three ideas so basic that we do not even define them. We describe them instead.

  • Point — a single exact location. It has no size. We draw a dot and name it with a capital letter, like PP.
  • Line — a straight path of points that goes on forever in both directions. Name it by any two of its points with a line symbol, AB↔\overleftrightarrow{AB}, or by a lowercase letter, ℓ\ell.
  • Plane — a perfectly flat surface that goes on forever. Name it with a capital script letter or by three points that are not all on one line, like plane ABCABC.

Collinear points lie on the same line. Coplanar points lie on the same plane.

Geometry starts with three terms it does not define — point, line, plane — and a few axioms about them: two points determine exactly one line, and three points not on one line determine exactly one plane. Nothing is lost by leaving them undefined: every later definition (segment, ray, angle, circle) is built from these words, and every proof ends up leaning on the axioms. That is why vocabulary here is precise — a proof is only as good as the words it is made of.

Example
Naming a line and its points
ABCl
Any two of the three points name the same line.

The points AA, BB, and CC below are collinear because they all lie on one line. That line can be named AB↔\overleftrightarrow{AB}, AC↔\overleftrightarrow{AC}, BC↔\overleftrightarrow{BC}, or simply ℓ\ell.

Side note
Where it goes wrong

Naming a line by two points that are not on it, or treating a ray AB→\overrightarrow{AB} as the same as BA→\overrightarrow{BA}. A ray's first letter is its endpoint.

Example
Collinear or not?
DEFGm
D, E, F share line m; G is off it, but all four lie in the shaded plane.

In the figure, DD, EE, and FF lie on line mm, so they are collinear. Point GG is off the line, so DD, EE, and GG are not collinear. All four points, however, lie on the same flat page, so they are coplanar.

Tip

Tip: Any two points determine exactly one line. Any three non-collinear points determine exactly one plane.

Try it
Quick check

Points PP, QQ, RR lie on one line, and point SS is not on that line. How many planes contain PP, QQ and SS? How many contain PP, QQ and RR?

Show the answer

PP, QQ, SS are three points not on one line, so exactly one plane contains them. PP, QQ, RR are collinear, so they pin down only a line — and infinitely many planes contain a line (think of pages of a book turning about the spine).

Segments, Rays, and Opposite Rays

Concept
Segments and Rays
ABCDEFsegment ABray CDline EF
Two endpoints, one endpoint, none.
  • A segment AB‾\overline{AB} is the part of a line between two endpoints AA and BB, including both endpoints. It has a definite length.
  • A ray AB→\overrightarrow{AB} starts at endpoint AA and goes forever through BB. Order matters: AB→\overrightarrow{AB} and BA→\overrightarrow{BA} point opposite ways.
  • Opposite rays are two rays with the same endpoint that together form a straight line.

Notation: AB‾\overline{AB} names the segment (a figure); ABAB with no bar means its length (a number).

The notation carries the picture: the bar over AB‾\overline{AB} has no arrows because the segment stops at both ends; CD→\overrightarrow{CD} has one arrow because the ray stops at CC and runs on forever through DD; EF↔\overleftrightarrow{EF} has two. Because a ray starts at its first letter, CD→\overrightarrow{CD} and DC→\overrightarrow{DC} are different rays — they point in opposite directions and share only the segment between them.

Example
Reading a figure of rays
ABC
Both rays start at B and head in opposite directions.

Here BA→\overrightarrow{BA} and BC→\overrightarrow{BC} share endpoint BB and form a straight line, so they are opposite rays. The segment AC‾\overline{AC} is the piece between AA and CC.

Side note
Where it goes wrong

Calling AB→\overrightarrow{AB} and BA→\overrightarrow{BA} opposite rays. Opposite rays must share their endpoint; these two start at different points (and overlap along AB‾\overline{AB}). The opposite ray of BA→\overrightarrow{BA} here is BC→\overrightarrow{BC}.

Tip

Remember: A bar means a figure (AB‾\overline{AB}, AB→\overrightarrow{AB}, AB↔\overleftrightarrow{AB}); no bar means a number (AB=AB= the length).

Try it
Quick check

Points PP, QQ, RR lie on a line in that order. Name a pair of opposite rays, and decide whether QP→\overrightarrow{QP} and PQ→\overrightarrow{PQ} are the same ray.

Show the answer

QP→\overrightarrow{QP} and QR→\overrightarrow{QR} are opposite rays: both start at QQ and together make the line. QP→\overrightarrow{QP} and PQ→\overrightarrow{PQ} are not the same: the first starts at QQ and runs through PP to the left forever; the second starts at PP and runs through QQ and RR to the right.

Measuring Segments & the Segment Addition Postulate

Concept
Segment Addition Postulate
ABC64AC = 10
The two pieces laid end to end make the whole.

If point BB lies on AC‾\overline{AC} (between AA and CC), then the two shorter lengths add to the whole:

AB+BC=AC.AB + BC = AC.

This lets you find any one of the three lengths when you know the other two.

This is a postulate, not a theorem: it is part of what we mean by length on a line. Put a ruler along the line with AA at 00; if BB is between, it sits at ABAB and CC sits at ACAC, and the gap from BB to CC is the difference. The word between is doing real work — if BB is off the segment, the two pieces overshoot, and for a point off the line, AB+BC>ACAB + BC > AC (a triangle's two sides are longer than the third).

Example
Adding segment lengths

Point BB is between AA and CC with AB=6AB = 6 and BC=4BC = 4. Then

AC=AB+BC=6+4=10.AC = AB + BC = 6 + 4 = 10.
Example
Solving for a missing length

If AC=15AC = 15 and AB=9AB = 9 with BB between AA and CC, then

BC=AC−AB=15−9=6.BC = AC - AB = 15 - 9 = 6.
Example
Segment addition with algebra

BB is between AA and CC, with AB=2x+1AB = 2x + 1, BC=3x−4BC = 3x - 4 and AC=22AC = 22. Find xx and both pieces.

The postulate turns the picture into one equation:

(2x+1)+(3x−4)=225x−3=22x=5.\begin{aligned} (2x + 1) + (3x - 4) &= 22 \\ 5x - 3 &= 22 \\ x &= 5 . \end{aligned}

So AB=2(5)+1=11AB = 2(5) + 1 = 11 and BC=3(5)−4=11BC = 3(5) - 4 = 11. Check: 11+11=2211 + 11 = 22. (BB happens to be the midpoint.)

Side note
Where it goes wrong

Stopping at xx. The question asks for lengths; x=5x = 5 is not a length here, AB=11AB = 11 is. A second slip: setting AB=BCAB = BC when nothing says BB is the midpoint — "between" only means the pieces add up.

Tip

Tip: The point named in the middle of the sentence (“BB is between AA and CC”) is the one that splits the segment. It appears on both short pieces.

Try it
Quick check

QQ is between PP and RR. PQ=3xPQ = 3x, QR=x+5QR = x + 5 and PR=25PR = 25. Find QRQR.

Show the answer

Segment addition: 3x+(x+5)=253x + (x + 5) = 25, so 4x=204x = 20 and x=5x = 5. Then QR=5+5=10QR = 5 + 5 = \mathbf{10} (and PQ=15PQ = 15; check 15+10=2515 + 10 = 25).

Congruent Segments & Midpoints

Concept
Congruence and Midpoint

Two segments are congruent (AB‾≅CD‾\overline{AB} \cong \overline{CD}) when they have equal length (AB=CDAB = CD). We mark congruent segments with matching tick marks.

The midpoint MM of AB‾\overline{AB} is the point exactly halfway between AA and BB. It splits the segment into two congruent halves:

AM=MB=12 AB.AM = MB = \tfrac{1}{2}\,AB.

Congruent segments are simply segments with the same length, so every congruence statement about segments is really an equation between numbers. A midpoint exists and is unique for every segment — on a ruler it is the average of the two end readings — and it is exactly the point where AM=MBAM = MB and AM+MB=ABAM + MB = AB hold together, which is why each half is 12AB\tfrac12 AB.

Example
Using the midpoint
AMB77AB = 14
Matching tick marks: the two halves are congruent.

MM is the midpoint of AB‾\overline{AB} and AB=14AB = 14. Then each half is

AM=MB=12(14)=7.AM = MB = \tfrac{1}{2}(14) = 7.

The tick marks show the two halves are congruent.

Example
A midpoint with algebra

MM is the midpoint of AB‾\overline{AB}, with AM=4x−1AM = 4x - 1 and MB=2x+9MB = 2x + 9. Find ABAB.

A midpoint makes the halves equal, so set them equal (not either one equal to the whole):

4x−1=2x+9  ⇒  2x=10  ⇒  x=5.4x - 1 = 2x + 9 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5 .

Each half is 4(5)−1=194(5) - 1 = 19, so AB=2⋅19=38AB = 2 \cdot 19 = 38.

Side note
Where it goes wrong

Setting a half equal to the whole: writing 4x−1=384x - 1 = 38 when 3838 is ABAB. Decide first whether each expression names a half or the whole segment, then use half == half or whole =2×= 2 \times half.

Tip

Remember: A midpoint makes two equal halves, so if you know the whole, halve it; if you know a half, double it.

Try it
Quick check

MM is the midpoint of PQ‾\overline{PQ}. PM=3x+2PM = 3x + 2 and PQ=40PQ = 40. Find xx.

Show the answer

PMPM is half of PQPQ, so PM=20PM = 20: 3x+2=203x + 2 = 20, giving x=6x = \mathbf{6}. (Check: MQ=20MQ = 20 too, and 20+20=4020 + 20 = 40.)

Midpoint Formula & Distance Formula

Concept
Formulas on the Coordinate Plane

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

Midpoint M=(x1+x22, y1+y22)\textbf{Midpoint } M = \left( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2} \right)
Distance AB=(x2−x1)2+(y2−y1)2\textbf{Distance } AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

The midpoint averages the coordinates. The distance formula is the Pythagorean Theorem in disguise.

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2

Both formulas come from looking along each axis separately. Halfway from x1x_1 to x2x_2 is x1+12(x2−x1)=x1+x22x_1 + \tfrac12(x_2 - x_1) = \tfrac{x_1 + x_2}{2}, and the same for yy — so the midpoint averages. For distance, the horizontal and vertical gaps are the legs of a right triangle whose hypotenuse is the segment, so the Pythagorean Theorem gives (Δx)2+(Δy)2\sqrt{(\Delta x)^2 + (\Delta y)^2}.

Example
Midpoint of two points
xy123451234ABM
M is halfway across (2.5) and halfway up (2).

Find the midpoint of A(1,1)A(1,1) and B(4,3)B(4,3).

M=(1+42, 1+32)=(52, 2)=(2.5, 2).M = \left(\frac{1+4}{2},\ \frac{1+3}{2}\right) = \left(\frac{5}{2},\ 2\right) = (2.5,\ 2).
Example
Distance between two points
xy1234512345AB345
The segment is the hypotenuse of a 3-4-5 right triangle.

Find the distance from A(1,1)A(1,1) to B(4,5)B(4,5).

AB=(4−1)2+(5−1)2=32+42=9+16=25=5.AB = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5.

This is the famous 33-44-55 right triangle: the horizontal leg is 33, the vertical leg is 44, and the segment is the hypotenuse.

Side note
Where it goes wrong

Subtracting in the midpoint formula or adding in the distance formula. For (1,1)(1,1) and (4,5)(4,5): (4−12,5−12)=(1.5,2)\left(\tfrac{4-1}{2}, \tfrac{5-1}{2}\right) = (1.5, 2) is half the change, not the midpoint (2.5,3)(2.5, 3). And 32+42=7\sqrt{3^2} + \sqrt{4^2} = 7 is the length of the two legs, not the distance 55 — square, add, then take one root.

Reminder — Distance and midpoint:  d=(x2−x1)2+(y2−y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)
Tip

Tip: In the distance formula the subtractions are squared, so it does not matter which point you call first — a negative squared is positive. Just be careful and neat with the arithmetic.

Try it
Quick check

Find the midpoint and the length of the segment from (−2,3)(-2, 3) to (4,−5)(4, -5).

Show the answer

Midpoint: (−2+42,3+(−5)2)=(1,−1)\left(\tfrac{-2+4}{2}, \tfrac{3+(-5)}{2}\right) = \mathbf{(1, -1)}. Length: Δx=6\Delta x = 6, Δy=−8\Delta y = -8, so 36+64=100=10\sqrt{36 + 64} = \sqrt{100} = \mathbf{10} (a 66-88-1010 triangle).

Segment Bisectors & Basic Constructions

Concept
Segment Bisector

A segment bisector is any line, ray, or segment that passes through the midpoint of a segment, cutting it into two congruent halves. A line that bisects a segment and meets it at a right angle is a perpendicular bisector.

Example
A perpendicular bisector

Line kk passes through MM, the midpoint of PQ‾\overline{PQ}, and meets it at 90∘90^\circ. So PM=MQPM = MQ (equal tick marks) and kk is the perpendicular bisector of PQ‾\overline{PQ}.

The same picture proves the Perpendicular Bisector Theorem: every point XX on kk is equidistant from PP and QQ. Triangles XMPXMP and XMQXMQ have PM=MQPM = MQ, a shared side XMXM, and right angles at MM, so they are congruent by SAS, and XP=XQXP = XQ.

Side note
Where it goes wrong

Changing the compass opening between the arc from PP and the arc from QQ. With radii 44 and 55, the crossing points are 44 from PP but 55 from QQ — no longer equidistant — and the line through them misses the midpoint.

Concept
Two Classic Compass Constructions
PQMk
Equal arcs from P and Q cross at two points; the line through them is k.

Copying a segment AB‾\overline{AB}: Draw a ray with endpoint CC. Open your compass to the length ABAB. Without changing that opening, place the point on CC and swing a small arc across the ray; label the crossing DD. Then CD‾≅AB‾\overline{CD} \cong \overline{AB}.

Bisecting a segment AB‾\overline{AB}: Open the compass wider than half of ABAB. From AA, draw arcs above and below the segment; from BB with the same opening, draw two more arcs. Connect the two crossing points. That line is the perpendicular bisector, and where it meets AB‾\overline{AB} is the midpoint.

Why the construction works: each crossing point was drawn with the same compass opening from PP and from QQ, so it is the same distance from both. The points equidistant from PP and QQ are exactly the points of the perpendicular bisector (a point XX with XP=XQXP = XQ makes two congruent triangles XMPXMP and XMQXMQ by SSS, which forces equal right angles at MM). Two such points determine the line, so the ruler only has to join them.

Tip

Remember: Good constructions never measure with a ruler — they use only a compass and straightedge. Keep the compass opening unchanged for each step.

Try it
Quick check

Line kk is the perpendicular bisector of PQ‾\overline{PQ}, and XX is a point on kk with XP=2y+3XP = 2y + 3 and XQ=5y−9XQ = 5y - 9. Find XPXP.

Show the answer

Every point on the perpendicular bisector is equidistant from PP and QQ, so 2y+3=5y−92y + 3 = 5y - 9, giving 3y=123y = 12 and y=4y = 4. Then XP=2(4)+3=11XP = 2(4) + 3 = \mathbf{11} (and XQ=20−9=11XQ = 20 - 9 = 11).

Going Deeper: Advanced Ideas

Concept
The Section (Partition) Formula

The midpoint splits a segment in the ratio 1:11:1. To find a point PP that divides AB‾\overline{AB} in any ratio m:nm:n (so that AP:PB=m:nAP:PB = m:n), with A(x1,y1)A(x_1,y_1) and B(x2,y2)B(x_2,y_2), use the section formula:

P=(m x2+n x1m+n,  m y2+n y1m+n).P = \left( \frac{m\,x_2 + n\,x_1}{m+n},\ \ \frac{m\,y_2 + n\,y_1}{m+n} \right).

Notice the “cross” pattern: the weight mm (nearest to BB) multiplies the BB-coordinates, and nn multiplies the AA-coordinates. Setting m=n=1m=n=1 recovers the midpoint formula, since each coordinate becomes a plain average.

Example
Dividing a segment in the ratio 2:32:3
xy1234561234567ABP
Cut AB into 5 equal steps: P is 2 steps from A and 3 from B.

Find the point PP on AB‾\overline{AB} with A(1,2)A(1,2) and B(6,7)B(6,7) such that AP:PB=2:3AP:PB = 2:3. Here m=2m=2 and n=3n=3, so m+n=5m+n = 5:

P=(2(6)+3(1)5,  2(7)+3(2)5)=(155, 205)=(3, 4).P = \left( \frac{2(6) + 3(1)}{5},\ \ \frac{2(7) + 3(2)}{5} \right) = \left( \frac{15}{5},\ \frac{20}{5} \right) = (3,\ 4).

Check: PP is closer to AA than to BB, as expected when the first part of the ratio (22) is the smaller one.

Side note
Where it goes wrong

Attaching each weight to the wrong endpoint. For AP:PB=2:3AP : PB = 2 : 3 with A(1,2)A(1,2), B(6,7)B(6,7), the "obvious" (2⋅1+3⋅65,2⋅2+3⋅75)=(4,5)\left(\tfrac{2\cdot1 + 3\cdot6}{5}, \tfrac{2\cdot2 + 3\cdot7}{5}\right) = (4, 5) is 33 steps from AA — the ratio 3:23 : 2. The weight next to BB in the ratio multiplies BB's coordinates: (3,4)(3, 4).

Concept
Distance and Midpoint in Three Dimensions

A point in space needs three coordinates, A(x1,y1,z1)A(x_1,y_1,z_1). The plane formulas simply grow a zz-term:

Distance AB=(x2−x1)2+(y2−y1)2+(z2−z1)2\textbf{Distance } AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}
Midpoint M=(x1+x22, y1+y22, z1+z22)\textbf{Midpoint } M = \left( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2},\ \frac{z_1+z_2}{2} \right)

This is still just the Pythagorean Theorem, applied twice: once in the base plane and once up the vertical direction.

Example
Distance between two points in space
AB2312
A and B are opposite corners of a 2 by 3 by 12 box.

Find the distance from A(1,2,3)A(1,2,3) to B(3,5,15)B(3,5,15).

AB=(3−1)2+(5−2)2+(15−3)2=4+9+144=157.AB = \sqrt{(3-1)^2 + (5-2)^2 + (15-3)^2} = \sqrt{4 + 9 + 144} = \sqrt{157}.

Because 157157 is not a perfect square, we leave the answer as the exact radical 157\sqrt{157} (about 12.5312.53).

Concept
Counting Lines, Planes, and Regions
1234567
Three lines in general position: 1 + 3 + 3 = 7 regions.

When points and lines are in “general position” (no needless overlaps), the counts follow neat patterns. For nn points with no three collinear, the number of distinct lines through pairs of them is

(n2)=n(n−1)2.\binom{n}{2} = \frac{n(n-1)}{2}.

For nn points with no four coplanar, the number of distinct planes through triples of them is

(n3)=n(n−1)(n−2)6.\binom{n}{3} = \frac{n(n-1)(n-2)}{6}.

And nn lines drawn in a plane, no two parallel and no three meeting at one point, cut the plane into

1+n+(n2)1 + n + \binom{n}{2}

regions — each new line adds one more region than the line before it.

The region count grows because each new line is cut by the earlier ones. The kk-th line crosses the k−1k - 1 lines already drawn at k−1k - 1 different points, which chop it into kk pieces, and each piece splits one old region into two — so line kk adds kk regions. Starting from the 11 region of the empty plane, nn lines give 1+(1+2+⋯+n)=1+n+(n2)1 + (1 + 2 + \cdots + n) = 1 + n + \binom{n}{2}.

Example
How many lines and planes from 55 points?

Take 55 points, no three collinear and no four coplanar. The number of lines is

(52)=5⋅42=10,\binom{5}{2} = \frac{5 \cdot 4}{2} = 10,

and the number of planes is

(53)=5⋅4⋅36=10.\binom{5}{3} = \frac{5 \cdot 4 \cdot 3}{6} = 10.
Concept
Reflection for Shortest Paths
lABA′X
A to X to B has the same length as the straight segment from A' to B.

To find the shortest path from point AA to a line ℓ\ell and then on to point BB (both AA and BB on the same side of ℓ\ell), reflect one point across the line. If A′A' is the mirror image of AA across ℓ\ell, then for any point XX on the line AX=A′XAX = A'X. So the bent path A→X→BA \to X \to B has the same length as A′→X→BA' \to X \to B, which is shortest when A′A', XX, and BB are collinear. Draw the straight segment A′B‾\overline{A'B}; where it crosses ℓ\ell is the best point XX.

A concrete case: A=(0,3)A = (0, 3), B=(6,1)B = (6, 1), and ℓ\ell is the xx-axis. Reflecting gives A′=(0,−3)A' = (0, -3). The segment A′B‾\overline{A'B} rises 44 over a run of 66, so it crosses the axis where −3+46x=0-3 + \tfrac46 x = 0, at X=(4.5,0)X = (4.5, 0). The shortest path has length A′B=62+42=52=213≈7.21A'B = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13} \approx 7.21; any other point YY on the line gives A′Y+YB>A′BA'Y + YB > A'B by the triangle inequality.

Concept
The Idea of a Locus

A locus is the set of all points that satisfy a given condition, and nothing else. Two rules of thumb turn everyday shapes into loci:

  • The locus of points a fixed distance rr from a point OO is a circle of radius rr centered at OO.
  • The locus of points equidistant from two fixed points AA and BB is the perpendicular bisector of AB‾\overline{AB}.

To describe a locus, ask: what shape do all the qualifying points trace out?

Example
A coordinate proof using the locus idea
xy−1123456−1123ABP
Every point on the dashed line x = 3 is equally far from A and B; on the axis it is P.

Prove that the point equidistant from A(0,0)A(0,0) and B(6,0)B(6,0) with yy-coordinate 00 is the midpoint. A point P(x,0)P(x,0) is equidistant when AP=BPAP = BP, so AP2=BP2AP^2 = BP^2:

x2=(x−6)2=x2−12x+36.x^2 = (x-6)^2 = x^2 - 12x + 36.

Subtracting x2x^2 from both sides gives 0=−12x+360 = -12x + 36, so x=3x = 3. Thus P(3,0)P(3,0), which is exactly the midpoint (0+62,0+02)\left(\frac{0+6}{2},\frac{0+0}{2}\right). This is a coordinate proof: by placing the figure on convenient axes, an algebra step settles a geometric claim.

Tip

Remember: Coordinate proofs work best when you place a key point at the origin and a key segment along an axis — the zeros make the distance and midpoint arithmetic short and clean.

Try it
Quick check

Find the point one quarter of the way from A(−2,1)A(-2, 1) to B(6,9)B(6, 9).

Show the answer

One quarter of the way means AP:PB=1:3AP : PB = 1 : 3. With m=1m = 1, n=3n = 3:

P=(1⋅6+3⋅(−2)4, 1⋅9+3⋅14)=(04,124)=(0,3).P = \left(\frac{1\cdot 6 + 3\cdot(-2)}{4},\ \frac{1\cdot 9 + 3\cdot 1}{4}\right) = \left(\frac{0}{4}, \frac{12}{4}\right) = \mathbf{(0, 3)}.

Check by steps: the whole change is (8,8)(8, 8); a quarter of it is (2,2)(2, 2), and (−2,1)+(2,2)=(0,3)(-2,1) + (2,2) = (0,3).

Why It Works

The midpoint formula averages coordinates because the midpoint is equally far from both ends along each axis: halfway from x1x_1 to x2x_2 is x1+x2−x12=x1+x22x_1 + \tfrac{x_2 - x_1}{2} = \tfrac{x_1 + x_2}{2}. Segment addition (AB+BC=ACAB + BC = AC when BB is between) is an axiom about lengths on a line, and it is what lets algebra enter geometry: an unknown length becomes a variable in an equation.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Distance and midpoint
d=(x2−x1)2+(y2−y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)

What it means. The straight-line distance between two points, and the point exactly halfway between them.

Example. (1,2)(1,2) to (4,6)(4,6): d=32+42=5d=\sqrt{3^2+4^2}=5, midpoint =(2.5, 4)=(2.5,\,4).

Why it works. The two points are opposite corners of a right triangle with legs ∣x2−x1∣|x_2-x_1| and ∣y2−y1∣|y_2-y_1|; the distance is the hypotenuse, so Pythagoras gives the formula. The midpoint is just the average of the coordinates, since averaging lands halfway along each axis.

Tip. Distance is the Pythagorean theorem in disguise. Squaring removes any sign worry, so you never need absolute values here.

Tip
The Triangle Inequality
a+b>cfor every pair of sidesa+b>c \quad\text{for every pair of sides}

What it means. Any two sides of a triangle must together exceed the third.

Example. Sides 2,3,62,3,6 cannot form a triangle since 2+3<62+3<6.

Why it works. The straight path between two vertices is the shortest one, so going via the third vertex can only be longer.

Tip. To test whether three lengths form a triangle, just check the two shortest against the longest.