Parallel & Perpendicular Lines

Study Sheet

Parallel & Perpendicular Lines

Everything you need to name angles, find them, and work with slopes

Kinds of Lines: Parallel, Perpendicular, and Skew

Concept
The three relationships
parallelperpendicularpqskew (p, q)
Edge p (front top) and edge q (back right) never meet and are not parallel.

Parallel lines lie in the same flat surface (plane) and never meet, no matter how far you stretch them. We write ℓ∥m\ell \parallel m. Perpendicular lines cross to form a perfect square corner, a 90∘90^\circ angle. We write ℓ⊥m\ell \perp m. Skew lines do not lie in the same plane. They never meet, but they are not parallel either — think of an airplane's path crossing high above a straight road.

When a transversal crosses two parallel lines it creates equal angles (corresponding, alternate interior, alternate exterior) and supplementary ones (same-side interior); the converses give ways to prove lines parallel; and in coordinates parallel lines share a slope while perpendicular slopes multiply to −1-1. Skew lines need three dimensions: in a single plane, two lines that never meet are parallel by definition, but in space a line can pass over another without ever meeting it or running alongside it.

Concept
Transversal

A transversal is a line that crosses two (or more) other lines. When a transversal cuts two lines it creates eight angles — four at each crossing — and those angles come in special matching pairs.

Tip

Quick check: Parallel lines stay the exact same distance apart forever. Perpendicular is a special case of intersecting — the angle just happens to be exactly 90∘90^\circ.

Side note
Where it goes wrong

Using the parallel-line angle facts before knowing the lines are parallel. The facts are IF-THEN: parallel implies equal corresponding angles, and the converse needs its own justification.

Try it
Quick check

In a rectangular box, two edges have no vertex in common. Must they be parallel?

Show the answer

No. They can be parallel (opposite edges of one face, like the top-front and bottom-front edges) or skew (like the top-front edge and a back vertical edge: they are not in one plane, never meet, and point in different directions). Edges with no common vertex are never perpendicular intersecting lines, because they do not meet.

The Eight Angles and Their Pairs

When a transversal tt crosses lines mm and nn, number the angles like this.

Concept
Name the four pairs
mnt12345678
At each crossing: 1 (or 5) upper left, 2 upper right, 3 lower left, 4 lower right.

Corresponding angles sit in the same corner at each crossing (one up, one down). Pairs: (1,5), (2,6), (3,7), (4,8)(1,5),\ (2,6),\ (3,7),\ (4,8). Alternate interior angles are between the two lines and on opposite sides of the transversal. Pairs: (3,6), (4,5)(3,6),\ (4,5). Alternate exterior angles are outside the two lines and on opposite sides of the transversal. Pairs: (1,8), (2,7)(1,8),\ (2,7). Same-side interior angles (also called co-interior or consecutive interior) are between the lines and on the same side of the transversal. Pairs: (3,5), (4,6)(3,5),\ (4,6).

The names are just coordinates. "Interior" and "exterior" say whether the angle lies between the two lines or outside them; "alternate" and "same-side" say which side of the transversal. Corresponding angles occupy the same corner at the two crossings — slide the top crossing down the transversal and each angle lands on its corresponding partner. None of the names assumes the lines are parallel; that assumption only enters when we claim the angles are equal.

Tip

Memory helper: Interior means inside (between the two lines: angles 3, 4, 5, 6). Exterior means outside (angles 1, 2, 7, 8). Alternate always means opposite sides of the transversal; same-side means the same side.

Example
Naming a pair

Look at the diagram above. What is the relationship between angle 44 and angle 55? Both angles are between lines mm and nn (interior), and angle 44 is on the right of tt while angle 55 is on the left (opposite sides). They are alternate interior angles.

Side note
Where it goes wrong

Calling angles 33 and 55 alternate interior because they are both "inside". They are interior, but both lie to the left of tt, so they are same-side interior angles — and with parallel lines they add to 180∘180^\circ rather than being equal.

Try it
Quick check

Using the numbering in the figure, name the relationship: (a) ∠1\angle 1 and ∠5\angle 5 (b) ∠3\angle 3 and ∠6\angle 6 (c) ∠1\angle 1 and ∠8\angle 8 (d) ∠4\angle 4 and ∠6\angle 6.

Show the answer

(a) Same corner at each crossing: corresponding. (b) Between the lines, opposite sides of tt: alternate interior. (c) Outside the lines, opposite sides of tt: alternate exterior. (d) Between the lines, both right of tt: same-side interior.

Angle Relationships When the Lines Are Parallel

Everything above works for any two lines. The magic happens when the two lines are parallel.

Concept
The parallel-line angle rules

If two parallel lines are cut by a transversal, then:

  • Corresponding angles are congruent (equal).
  • Alternate interior angles are congruent.
  • Alternate exterior angles are congruent.
  • Same-side interior angles are supplementary (they add to 180∘180^\circ).

Also, vertical angles (the ones directly across an X) are always congruent, and a straight line gives a linear pair that adds to 180∘180^\circ.

Only the first rule is assumed (the Corresponding Angles Postulate): sliding along the transversal from one parallel line to the other, the crossing looks exactly the same. The rest follow in one step each. Alternate interior: ∠3\angle 3 equals its corresponding angle ∠7\angle 7, which equals its vertical angle ∠6\angle 6. Same-side interior: ∠3=∠7\angle 3 = \angle 7, and ∠7\angle 7 forms a linear pair with ∠5\angle 5, so ∠3+∠5=180∘\angle 3 + \angle 5 = 180^\circ.

Example
One angle unlocks them all
mnt70°110°110°70°70°110°110°70°
m∥nm \parallel n and ∠1=70∘\angle 1 = 70^\circ: only two different measures appear.

In the figure, m∥nm \parallel n and angle 1=70∘1 = 70^\circ. Find every other angle.

Angle 1=70∘1=70^\circ. Its vertical angle is 44, so angle 4=70∘4 = 70^\circ. Angles 22 and 33 form linear pairs with angle 11, so each is 180∘−70∘=110∘180^\circ - 70^\circ = 110^\circ. Now jump to the bottom crossing using the parallel rules: angle 55 corresponds to angle 11, so angle 5=70∘5 = 70^\circ. Then angle 8=70∘8 = 70^\circ (vertical to 55), and angles 66 and 77 are 110∘110^\circ each. Result: angles 1,4,5,8=70∘1,4,5,8 = 70^\circ and angles 2,3,6,7=110∘2,3,6,7 = 110^\circ.

Example
Same-side interior are supplementary
mntx115°
Angles 3 and 5: both between the lines, both left of t.

Suppose m∥nm \parallel n and one same-side interior angle measures 115∘115^\circ. Its partner satisfies 115∘+x=180∘115^\circ + x = 180^\circ, so x=65∘x = 65^\circ.

Side note
Where it goes wrong

Setting same-side interior angles equal. With 115∘115^\circ given, x=115∘x = 115^\circ is wrong; the figure shows one acute and one obtuse angle between the lines, and they add to 180∘180^\circ, so x=65∘x = 65^\circ.

Tip

Shortcut: With parallel lines, every one of the eight angles is either equal to the angle you know or is its supplement (180∘180^\circ minus it). There are really only two different numbers in the whole picture!

Try it
Quick check

m∥nm \parallel n and ∠2=128∘\angle 2 = 128^\circ (numbering as in the figures). Find ∠4\angle 4, ∠5\angle 5 and ∠6\angle 6.

Show the answer

∠4\angle 4 forms a linear pair with ∠2\angle 2: 180−128=52∘180 - 128 = \mathbf{52^\circ}. ∠5\angle 5 is alternate interior with ∠4\angle 4: 52∘\mathbf{52^\circ}. ∠6\angle 6 corresponds to ∠2\angle 2: 128∘\mathbf{128^\circ}.

Finding Unknown Angles with Algebra

Concept
Turn a relationship into an equation

Decide whether the two labeled angles are congruent (equal) or supplementary (add to 180∘180^\circ). Then write the matching equation and solve for the variable. Finally, plug back in to find the actual angle measures.

Look at the picture before writing anything: with parallel lines, two angles are equal exactly when both are acute or both are obtuse, and supplementary when one of each. That quick check catches the most common mistake — setting a pair equal that should add to 180∘180^\circ — before any algebra is done.

Example
Congruent angles (equal)
mnt(3x+15)°(5x−25)°
Corresponding angles, drawn at their solved size of 75∘^\circ.

Lines m∥nm \parallel n. Two corresponding angles measure (3x+15)∘(3x+15)^\circ and (5x−25)∘(5x-25)^\circ. Find xx and each angle.

Corresponding angles are congruent, so set them equal:

3x+15=5x−2540=2xx=20\begin{aligned} 3x+15 &= 5x-25\\ 40 &= 2x\\ x &= 20 \end{aligned}

Each angle is 3(20)+15=75∘3(20)+15 = 75^\circ (and 5(20)−25=75∘5(20)-25 = 75^\circ as a check). x=20x=20, both angles =75∘=75^\circ.

Example
Supplementary angles (add to 180)
mnt(2x+20)°(3x−10)°
Same-side interior angles, drawn at 88∘^\circ and 92∘^\circ.

Lines m∥nm \parallel n. Two same-side interior angles measure (2x+20)∘(2x+20)^\circ and (3x−10)∘(3x-10)^\circ. Find xx and each angle.

(2x+20)+(3x−10)=1805x+10=1805x=170x=34\begin{aligned} (2x+20)+(3x-10) &= 180\\ 5x+10 &= 180\\ 5x &= 170\\ x &= 34 \end{aligned}

The angles are 2(34)+20=88∘2(34)+20 = 88^\circ and 3(34)−10=92∘3(34)-10 = 92^\circ. Check: 88+92=18088+92 = 180. ✓

Side note
Where it goes wrong

Solving correctly but for the wrong relationship. If you set 2x+20=3x−102x + 20 = 3x - 10 you get x=30x = 30 and two 80∘80^\circ angles — but same-side interior angles of 80∘80^\circ each add to 160∘160^\circ, so the lines would not be parallel. The answer contradicts the given, which is the signal to recheck the relationship.

Tip

Watch the word: “congruent” ⇒\Rightarrow set the expressions equal. “Supplementary” ⇒\Rightarrow set their sum equal to 180180.

Try it
Quick check

m∥nm \parallel n. A pair of alternate interior angles measure (7x−9)∘(7x - 9)^\circ and (4x+24)∘(4x + 24)^\circ. Find xx and the angle.

Show the answer

Alternate interior angles are equal: 7x−9=4x+247x - 9 = 4x + 24, so 3x=333x = 33 and x=11x = 11. Each angle is 7(11)−9=68∘7(11) - 9 = \mathbf{68^\circ} (and 4(11)+24=684(11) + 24 = 68).

Proving Lines Are Parallel (the Converses)

The rules above run backwards too. If you can find one of the special angle relationships, you can conclude the lines are parallel.

Concept
The converse statements

Two lines cut by a transversal are parallel if any one of these is true:

  • a pair of corresponding angles is congruent, or
  • a pair of alternate interior angles is congruent, or
  • a pair of alternate exterior angles is congruent, or
  • a pair of same-side interior angles is supplementary.

Why the converse holds: suppose alternate interior angles are equal but the lines meet at a point PP on one side. Then the transversal and the two lines form a triangle, and one of the equal angles is an exterior angle of that triangle while the other is a remote interior angle. An exterior angle equals the sum of the two remote interior angles, so it is strictly larger than either — it cannot equal one of them. The lines therefore never meet: they are parallel.

Example
Are these lines parallel?
mnt63°63°
Equal alternate interior angles force m∥nm \parallel n.

A transversal crosses two lines. A pair of alternate interior angles measure 63∘63^\circ and 63∘63^\circ. Because these alternate interior angles are congruent, the converse rule tells us the two lines are parallel. If instead they had measured 63∘63^\circ and 70∘70^\circ, they would not be congruent, so the lines would not be parallel.

Side note
Where it goes wrong

Arguing backwards from the conclusion. "The lines look parallel, so the alternate interior angles are equal, so the lines are parallel" is circular. To prove lines parallel you must get the angle fact from somewhere else (given measures, vertical angles, a triangle), then apply the converse.

Tip

Careful: Same-side interior angles do the opposite — the lines are parallel when those two angles add to 180∘180^\circ, not when they are equal.

Try it
Quick check

A transversal cuts lines mm and nn. A pair of same-side interior angles measure 72∘72^\circ and 118∘118^\circ. Are mm and nn parallel?

Show the answer

Same-side interior angles of parallel lines add to 180∘180^\circ. Here 72+118=190≠18072 + 118 = 190 \neq 180, so the lines are not parallel. (Because the sum is more than 180∘180^\circ, the lines lean apart on that side and meet on the other side of the transversal.)

Perpendicular Lines and Right Angles

Concept
Perpendicular means right angles

When two lines are perpendicular (ℓ⊥m\ell \perp m), they cross to make four 90∘90^\circ angles. A small square drawn in the corner is the symbol for a right angle. If one of the four angles is 90∘90^\circ, all four are.

Why one right angle gives four: the angle next to it forms a linear pair with it, so it is 180∘−90∘=90∘180^\circ - 90^\circ = 90^\circ; the other two are vertical to these. A perpendicular also gives the shortest route from a point to a line — any slanted segment is the hypotenuse of a right triangle whose leg is the perpendicular — which is why "distance to a line" always means the perpendicular distance.

Example
Splitting a right angle
52°x
The two pieces fill a 90∘^\circ corner, so x = 38∘^\circ.

A right angle is split by a ray into two smaller angles of x∘x^\circ and 52∘52^\circ. Since the whole angle is 90∘90^\circ:

x+52=90⇒x=38∘.x + 52 = 90 \quad\Rightarrow\quad x = 38^\circ.
Side note
Where it goes wrong

Assuming a corner is a right angle because it looks square. Without a right-angle mark, a "⊥\perp" or a given 90∘90^\circ, an 88∘88^\circ angle and a 90∘90^\circ angle look the same on paper.

Tip

Remember: Perpendicular is the same idea whether you are looking at drawn lines (right angles) or graphed lines (slopes that are negative reciprocals — next section).

Try it
Quick check

Lines ℓ⊥m\ell \perp m meet at OO. A ray from OO lies inside one of the right angles and makes angles of 3y∘3y^\circ with ℓ\ell and (2y+15)∘(2y + 15)^\circ with mm. Find both angles.

Show the answer

The two pieces fill a right angle: 3y+(2y+15)=903y + (2y + 15) = 90, so 5y=755y = 75 and y=15y = 15. The angles are 3(15)=45∘3(15) = \mathbf{45^\circ} and 2(15)+15=45∘2(15) + 15 = \mathbf{45^\circ} — the ray bisects the right angle.

Slopes of Parallel and Perpendicular Lines

On the coordinate plane, slope measures steepness: slope=riserun=change in ychange in x\text{slope}=\dfrac{\text{rise}}{\text{run}}=\dfrac{\text{change in }y}{\text{change in }x}.

Concept
The slope rules

Parallel lines have equal slopes. If line 1 has slope mm, then any line parallel to it also has slope mm. Perpendicular lines have slopes that are negative reciprocals. Flip the fraction and change the sign: if one slope is mm, the perpendicular slope is −1m-\dfrac{1}{m}. A quick test: two slopes are perpendicular exactly when their product is −1-1.

Parallel lines rise at the same rate, so equal slopes is just "the same steepness". For perpendicular lines, turn the slope triangle of the first line a quarter turn: a step of run 44, rise 33 becomes run −3-3, rise 44 (the legs swap and one changes sign). The new slope is 4−3=−43\tfrac{4}{-3} = -\tfrac43 — the negative reciprocal of 34\tfrac34 — and the product is −1-1. Horizontal and vertical lines are the one exception, because a vertical line has no slope to take a reciprocal of.

Example
Read off the slopes
xy4334
Turn the run-4, rise-3 triangle a quarter turn: run −-3, rise 4, slope −-4/3.

Line AA has slope 34\dfrac{3}{4}.

  • A line parallel to AA has slope 34\dfrac{3}{4} (same).
  • A line perpendicular to AA has slope −43-\dfrac{4}{3} (flip 34\tfrac34 to 43\tfrac43, change the sign). Check: 34⋅(−43)=−1\dfrac34 \cdot \left(-\dfrac43\right) = -1. ✓
Example
Parallel, perpendicular, or neither?

Line AA has slope −2-2 and line BB has slope 12\dfrac{1}{2}. Their product is −2⋅12=−1-2 \cdot \dfrac12 = -1, so the lines are perpendicular.

Side note
Where it goes wrong

Taking the reciprocal but forgetting the sign (or the reverse). Slopes 22 and 12\tfrac12 multiply to 11, not −1-1: those lines are neither parallel nor perpendicular; both rise to the right, so they cannot meet at a right angle.

Tip

Special cases: A horizontal line has slope 00; a vertical line has an undefined slope. A horizontal line and a vertical line are perpendicular to each other.

Try it
Quick check

Are the lines y=25x+1y = \tfrac25 x + 1 and 5x+2y=85x + 2y = 8 parallel, perpendicular, or neither?

Show the answer

Solve the second for yy: 2y=−5x+82y = -5x + 8, so y=−52x+4y = -\tfrac52 x + 4 with slope −52-\tfrac52. The product of slopes is 25⋅(−52)=−1\tfrac25 \cdot \left(-\tfrac52\right) = -1, so the lines are perpendicular.

Writing Equations of Parallel and Perpendicular Lines

We use slope-intercept form y=mx+by = mx + b and point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1).

Concept
The three-step recipe
  • Find the slope of the given line (read mm from y=mx+by=mx+b).
  • Choose the new slope: same mm for parallel, negative reciprocal for perpendicular.
  • Plug that slope and the given point into point-slope form, then simplify to y=mx+by=mx+b.

A line is fixed by one point and one slope, and the problem hands you both: the point directly, the slope through the parallel or perpendicular condition. Point-slope form is the natural tool because it is literally the slope definition, y−y1x−x1=m\tfrac{y - y_1}{x - x_1} = m, cleared of fractions: every point (x,y)(x, y) on the line makes the same slope with (x1,y1)(x_1, y_1).

Example
A parallel line through a point
xy−11234−1123456(1, 4)
Solid: y = 2x + 1. Dashed: the parallel through (1, 4), y = 2x + 2.

Write the equation of the line through (1,4)(1,4) that is parallel to y=2x+1y = 2x + 1. The given slope is 22; parallel means the new slope is also 22. Point-slope:

y−4=2(x−1)  ⇒  y−4=2x−2  ⇒  y=2x+2.y - 4 = 2(x - 1) \;\Rightarrow\; y - 4 = 2x - 2 \;\Rightarrow\; y = 2x + 2.

Check the point: 2(1)+2=42(1)+2 = 4. ✓

Example
A perpendicular line through a point
xy−1123456−1123456(4, 3)
The dashed line y = −-x/2 + 5 passes through (4, 3) and meets y = 2x + 1 at a right angle.

Write the equation of the line through (4,3)(4,3) that is perpendicular to y=2x+1y = 2x + 1. The given slope is 22; the perpendicular slope is −12-\dfrac12. Point-slope:

y−3=−12(x−4)  ⇒  y−3=−12x+2  ⇒  y=−12x+5.y - 3 = -\tfrac12(x - 4) \;\Rightarrow\; y - 3 = -\tfrac12 x + 2 \;\Rightarrow\; y = -\tfrac12 x + 5.

Check the point: −12(4)+5=3-\tfrac12(4)+5 = 3. ✓

Side note
Where it goes wrong

Keeping the old intercept. The line through (4,3)(4, 3) perpendicular to y=2x+1y = 2x + 1 is not y=−12x+1y = -\tfrac12 x + 1: test it — at x=4x = 4 it gives y=−1y = -1, not 33. The intercept must be recomputed from the new point.

Tip

Final tip: Always test your equation by plugging the given point back in. If it makes the equation true, your line really does pass through that point — and the slope guarantees it is parallel or perpendicular as required.

Try it
Quick check

Write the equation of the line through (−2,5)(-2, 5) perpendicular to y=−23x+4y = -\tfrac23 x + 4.

Show the answer

The given slope is −23-\tfrac23, so the perpendicular slope is 32\tfrac32. Point-slope: y−5=32(x+2)=32x+3y - 5 = \tfrac32(x + 2) = \tfrac32 x + 3, so y=32x+8\mathbf{y = \tfrac32 x + 8}. Check: 32(−2)+8=5\tfrac32(-2) + 8 = 5. ✓

Going Deeper: Advanced Ideas

Up to now we have used the parallel-line rules. Here we see why they are true, chase angles through several transversals at once, and push the slope ideas into real coordinate computations: distances, feet of perpendiculars, and equations of lines.

Concept
Why the transversal theorems are true

The whole system rests on one assumed fact, the Corresponding Angles Postulate: if two parallel lines are cut by a transversal, corresponding angles are congruent. Everything else is a short deduction from it, using two facts you already know:

  • Vertical angles (across an X) are congruent.
  • A linear pair (two angles on a straight line) is supplementary.

Chain them together: a corresponding angle is congruent to the one directly across from it (vertical), which is the alternate interior angle — so alternate interior angles are congruent. Swap the vertical step for a linear pair and you get same-side interior angles are supplementary. No new assumptions are needed.

Example
Worked proof: alternate interior angles are congruent
mnt367
Angle 3 = angle 7 (corresponding) and angle 7 = angle 6 (vertical).

Using the eight-angle numbering from earlier, suppose m∥nm \parallel n. Prove ∠3≅∠6\angle 3 \cong \angle 6.

The angle you start with (∠3\angle 3) hops to its corresponding partner (∠7\angle 7), then across the X to its vertical partner (∠6\angle 6). Two congruences in a row give the result. The same two-step hop, run backwards, proves the converse used to show lines are parallel.

Concept
The auxiliary parallel line trick (zigzag chases)

When a path bends between two parallel lines, there is no single transversal to use. The fix: draw a new line through the bend point, parallel to the two given lines. This splits the bend angle into two pieces, and each piece is now an alternate-interior angle with one of the original lines. Add (or subtract) the pieces to finish. This “build your own parallel line” move turns an impossible-looking figure into two ordinary transversal problems.

Example
Worked zigzag angle chase
35°25°ABPrs
The dashed parallel through P splits the bend into 35∘^\circ and 25∘^\circ.

Lines rr and ss are parallel. A path goes from AA on rr down to a corner PP, then to BB on ss. The path makes a 35∘35^\circ angle with rr at AA and a 25∘25^\circ angle with ss at BB. Find ∠APB\angle APB.

Draw the dashed line through PP parallel to rr and ss. It splits ∠APB\angle APB into an upper piece and a lower piece.

  • Upper piece =35∘=35^\circ: alternate interior angles with AA's angle across transversal APAP.
  • Lower piece =25∘=25^\circ: alternate interior angles with BB's angle across transversal BPBP.

So ∠APB=35∘+25∘=60∘\angle APB = 35^\circ + 25^\circ = \mathbf{60^\circ}. (Whenever the bend points away from the region between the parallels you add the pieces; if the path stays on one side you subtract them instead.)

Concept
Distance between two parallel lines

Parallel lines never meet, so they stay a fixed distance apart — but that distance is measured perpendicular to the lines, not vertically. For two lines written as

y=mx+b1andy=mx+b2(same slope m),y = mx + b_1 \qquad\text{and}\qquad y = mx + b_2 \quad(\text{same slope } m),

the perpendicular distance between them is

d=∣b2−b1∣1+m2.d = \frac{\lvert b_2 - b_1 \rvert}{\sqrt{1 + m^2}}.

The numerator is the vertical gap between the intercepts; dividing by 1+m2\sqrt{1+m^2} tilts that vertical gap into the true perpendicular gap. (If both lines are given in the form ax+by+c=0ax+by+c=0 with matching a,ba,b, the distance is ∣c2−c1∣/a2+b2\lvert c_2 - c_1\rvert / \sqrt{a^2+b^2}.)

Example
Worked distance between parallels
xy−3−2−1123−1123456√55
The perpendicular gap (5\sqrt{5}) is shorter than the vertical gap (5).

Find the distance between y=2x+1y = 2x + 1 and y=2x+6y = 2x + 6. Same slope m=2m = 2, so they are parallel. Here b1=1b_1 = 1 and b2=6b_2 = 6:

d=∣6−1∣1+22=55=5≈2.24.d = \frac{\lvert 6 - 1 \rvert}{\sqrt{1 + 2^2}} = \frac{5}{\sqrt{5}} = \sqrt{5} \approx 2.24.

Notice this is less than the vertical gap of 55 — the perpendicular route between two slanted lines is always the shortest one.

Side note
Where it goes wrong

Reporting the vertical gap between intercepts as the distance. For y=2x+1y = 2x + 1 and y=2x+6y = 2x + 6 the intercepts differ by 55, but the lines are slanted, and the shortest (perpendicular) gap is 5≈2.24\sqrt5 \approx 2.24.

Concept
The foot of a perpendicular

The foot of the perpendicular from a point PP to a line ℓ\ell is the single point FF on ℓ\ell where the perpendicular dropped from PP lands. It is the closest point on ℓ\ell to PP, and the distance PFPF is the true distance from the point to the line. To find it:

  • Take the slope of ℓ\ell and use its negative reciprocal as the slope of the perpendicular through PP.
  • Write that perpendicular line in point-slope form.
  • Solve the two equations together — the solution is the foot FF.
Example
Worked foot of a perpendicular
xy−112345−112345PF
F = (2, 3) is the point of y = x + 1 closest to P = (4, 1).

Find the foot of the perpendicular from P=(4,1)P = (4,1) to the line ℓ: y=x+1\ell:\ y = x + 1, and the distance from PP to ℓ\ell. Line ℓ\ell has slope 11, so the perpendicular has slope −1-1. Through P=(4,1)P=(4,1):

y−1=−1(x−4)  ⇒  y=−x+5.y - 1 = -1(x - 4) \;\Rightarrow\; y = -x + 5.

Set the two lines equal to find where they cross:

x+1=−x+52x=4x=2,y=2+1=3.\begin{aligned} x + 1 &= -x + 5\\ 2x &= 4\\ x &= 2, \qquad y = 2 + 1 = 3. \end{aligned}

The foot is F=(2,3)F = (2,3). The distance is

PF=(4−2)2+(1−3)2=4+4=8=22≈2.83.PF = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83.

No other point on ℓ\ell is closer to PP than FF.

Try it
Quick check

Find the distance from the point (5,0)(5, 0) to the line y=x+1y = x + 1.

Show the answer

The perpendicular through (5,0)(5, 0) has slope −1-1: y=−x+5y = -x + 5. It meets y=x+1y = x + 1 where x+1=−x+5x + 1 = -x + 5, so x=2x = 2 and the foot is (2,3)(2, 3). The distance is (5−2)2+(0−3)2=18=32≈4.24\sqrt{(5-2)^2 + (0-3)^2} = \sqrt{18} = \mathbf{3\sqrt2} \approx 4.24.

Tip

Why the slope criteria are consistent: a line with slope ba\tfrac{b}{a} points in direction (a,b)(a,b). Rotating that direction a quarter turn (90∘90^\circ) sends (a,b)↦(−b,a)(a,b) \mapsto (-b,a), whose slope is a−b=−ab\dfrac{a}{-b} = -\dfrac{a}{b} — exactly the negative reciprocal, and their product is −1-1. So the algebra (negative reciprocals) and the geometry (a right angle) are the same statement. Likewise, two lines have equal slopes precisely when they point the same direction, which is precisely when they never cross — the definition of parallel.

Why It Works

Take the parallel postulate as the axiom (through a point not on a line there is exactly one parallel). Corresponding angles equal follows: if they were unequal, the second line could be rotated slightly to make them equal, producing a second parallel through the same point — impossible. Alternate interior angles equal then comes from combining corresponding angles with vertical angles, and same-side interior angles supplementary from combining with a linear pair. One axiom, three theorems.

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
Slope of a line
m=y2−y1x2−x1m = \frac{y_2-y_1}{x_2-x_1}

What it means. Rise over run: how much yy changes for each 11 that xx increases.

Example. Through (1,2)(1,2) and (4,8)(4,8): m=8−24−1=2m=\dfrac{8-2}{4-1}=2.

Why it works. Between two points on a line the vertical change is y2−y1y_2-y_1 and the horizontal change is x2−x1x_2-x_1. Similar triangles guarantee this ratio is the same wherever you measure it, so it is a property of the line itself.

Tip. Keep the points in the same order top and bottom. Reversing both gives the same slope; reversing only one flips the sign.

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Angle pair relationships
vertical angles are equal;linear pair sums to 180∘;complements sum to 90∘\text{vertical angles are equal};\quad \text{linear pair sums to }180^\circ;\quad \text{complements sum to }90^\circ

What it means. Angles formed by crossing lines come in predictable pairs.

Example. If two angles are supplementary and one is 110∘110^\circ, the other is 70∘70^\circ.

Why it works. Two angles on a straight line total 180∘180^\circ. If ∠1\angle 1 and ∠2\angle 2 form a line, and ∠2\angle 2 and ∠3\angle 3 also form a line, then both equal 180∘−∠2180^\circ-\angle 2 — so the vertical angles ∠1\angle 1 and ∠3\angle 3 are equal.

Tip. With parallel lines cut by a transversal, corresponding and alternate angles are equal, while same-side interior angles are supplementary.

Tip
Parallel and perpendicular slopes
m1=m2 (parallel),m1m2=−1 (perpendicular)m_1=m_2 \ (\text{parallel}), \qquad m_1m_2=-1 \ (\text{perpendicular})

What it means. Parallel lines share a slope; perpendicular slopes are opposite reciprocals.

Example. A line perpendicular to slope 22 has slope −12-\tfrac12 (their product is −1-1).

Why it works. Rotating a direction by 90∘90^\circ turns a run of aa and rise of bb into a run of −b-b and rise of aa, changing slope ba\tfrac{b}{a} into −ab-\tfrac{a}{b} — whose product with the original is −1-1.

Tip. A horizontal line (m=0m=0) is perpendicular to a vertical line, whose slope is undefined — the product rule does not apply to that pair.