Distance in the Plane

The absolute value function is defined as

|x| =

{x if x ≥ 0; and

−x if x < 0.

If the number a is positive or zero, then |a| = a. If a is negative, then |a|is the number you’d get by “erasing” the minus sign in front of it. Thus,|5| = 5, |17| = 17, |0| = 0, | − 2| = 2, and | − 9| = 9.

Because it’s the job of the absolute value function to erase the minus signin front of those numbers that have them, the absolute value of a numberx is the same as the absolute value of its negative, −x. Written with mathsymbols

|x| = | − x|As examples, |4| = 4 = | − 4| and | − 10| = 10 = | − (−10)|.

Whether you square a number or its negative, you’ll get the same result.That is, x2 = (−x)2. And since |x| equals either x or −x, depending onwhether x is negative, we know that

|x|2 = x2

As examples, |3|2 = 32 and | − 7|2 = 72 = (−7)2.178

fills up the little circle of the second piece at the point (0, 0). The result isthat there is no little circle in the graph of the absolute value function

32

Distance in RThe distance between the two numbers a, b ∈ R is |a− b|.

Examples.

• The distance between the numbers 5 and 3 is |5− 3| = |2| = 2.

• The distance between the numbers 3 and 5 is |3− 5| = | − 2| = 2.

• The distance between −4 and 12 is | − 4− 12| = | − 16| = 16.

We saw an illustration in the first two examples above that

|a− b| = |b− a|

That makes sense. It means that the distance between a and b is the same asthe distance between b and a, as it should be. We can check this fact usingalgebra: |a− b| = | − (b− a)| = |b− a|.

As was discussed above, |x|2 = x2. Thus, substituting a− b for x, we have

|a− b|2 = (a− b)2

Lengths in right trianglesA triangle is three points in the plane, with each pair of points joined by

the straight line segment between them. A triangle is a figure with threesides, or a trigon. We usually place so much emphasis on the angles of threesided figures that we usually call them triangles instead of trigons.

A right triangle is a triangle one of whose angles is a right angle. We’ll havemore to say about right angles and angles in general soon. Most people arecomfortable with what a right angle is, although there are different names

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Distance in RThe distance between the two numbers a, b E R is a — bI.

Examples.

• The distance between the numbers 5 and 3 is 5—3) = 121 = 2.

• The distance between the numbers 3 and 5 is 3 —

5) = I — 2) = 2.

• The distance between —4 and 12 is — 4 — 12) = I — 16) = 16.

We saw an illustration in the first two examples above that

la-b) = lb-alThat makes sense. It means that the distance between a and b is the same asthe distance between b and a, as it should be. We can check this fact usingalgebra: a — bl = I — (b — a)l = )b

— al.As was discussed above, 1x12 = x2. Thus, substituting a — b for x, we have

a — b)2 = (a —

Lengths in right trianglesA triangle is three points in the plane, with each pair of points joined by

the straight line segment between them. A triangle is a figure with threesides, or a trigon. We usually place so much emphasis on the angles of threesided figures that we usually call them triangles instead of trigons.

A right triangle is a triangle one of whose angles is a right angle. We’ll havemore to say about right angles and angles in general soon. Most people arecomfortable with what a right angle is, although there are different names

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for them. The three most common are an angle of 90◦, an angle of π2 , or one

of four equal angles resulting from the intersecting of perpendicular lines.

The side lengths of right triangles satisfy a well known equation. It’s famousenough and useful enough that instead of calling it a “claim” we call it a“theorem”.

The Pythagorean Theorem (1). If a, b, and c are the three lengths ofthe sides of a right triangle, and if c is the length of the side opposite theright angle in the triangle, then

a2 + b2 = c2

Proof: We can draw the same triangle four times to create a big square,each of whose sides have length a+b. This picture is drawn on the next page.

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for them. The three most common are an angle of 900, an angle of , or oneof four equal angles resulting from the intersecting of perpendicular lines.

rit

The side lengths of right triangles satisfy a well known equation. It’s famousenough and useful enough that instead of calling it a “claim” we call it a“theorem”.

The Pythagorean Theorem (1). If a, b, and c are the three lengths ofthe sides of a right triangle, and if c is the length of the side opposite theright angle in the triangle, then

Proof: We can draw the same triangle four times to create a giant square,each of whose sides have length a + b. This picture is drawn on the next page.

rL J

9 9 2a+b=c

bC.

0

164

Notice that the area of the big square is

(a+ b)2

The big square can be dissected into 4 triangles and a smaller square. Eachof the 4 triangles has a base of a and a height of b, so they each have area12ab. The smaller square in the middle of the picture has sides whose lengthsequal c. Thus, the area of the square in the middle of the picture is c2. Thearea of the entire picture is the sum of the areas of the 4 triangles and thesmaller square in the middle

4(1

2ab) + c2

We’ve calculated the area of the entire picture in two different ways, andthey must be equal

(a+ b)2 = 4(1

2ab) + c2

We can multiply out the left side of the equation and simplify the right side:

a2 + 2ab+ b2 = 2ab+ c2

Now subtract 2ab from both sides:

a2 + b2 = c2

�181

a. 6

Notice that the area of the giant square is(a+b)2

The giant square can be dissected into 4 triangles and a smaller square.Each of the 4 triangles has a base of a and a height of b, so they each havearea -ab. The smaller square in the middle of the picture has sides whoselengths equal c. Thus, the area of the square in the middle of the picture isc2. The area of the entire picture is the sum of the areas of the 4 trianglesand the smaller square in the middle

4(ab) + c2

We’ve calculated the area of the entire picture in two different ways, andthey must be equal

(a+b)2 =4(ab)+c2

We can multiply out the left side of the equation and simplify the right side:a2 + 2ab + b2 = 2ab + c2

Now subtract 2ab from both sides:a2 + b2 =

6

b

b

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Distance in R2

The Pythagorean Theorem allows us to determine the distance betweenany pair of points in the plane.

Proposition (2). The distance between two points (x1, y1) and (x2, y2) is√(x1 − x2)2 + (y1 − y2)2

Proof: The distance between (x1, y1) and (x2, y2) is the length, c, of a sideof a right triangle. Notice that c is a length, so c ≥ 0.

The other two sides of the triangle have length |x1 − x2| and |y1 − y2| sothe Pythagorean theorem tells us that the distance, c, between (x1, y1) and(x2, y2) satisfies the equation

c2 = |x1 − x2|2 + |y1 − y2|2

Therefore, either

c =√|x1 − x2|2 + |y1 − y2|2 or c = −

√|x1 − x2|2 + |y1 − y2|2

Recall that c ≥ 0, so the only solution for c is

c =√|x1 − x2|2 + |y1 − y2|2

We discussed earlier in this chapter that |a− b|2 = (a− b)2 for any numbersa and b, so we can write that c, the distance between (x1, y1) and (x2, y2),equals √

(x1 − x2)2 + (y1 − y2)2�

182

I x-x1

(x)

C

a

Example.

• The distance between the points (2,−3) and (−5, 8) is√(2− (−5))2 + (−3− 8)2 =

√(7)2 + (−11)2 =

√49 + 121 =

√170

Norms of vectorsThe norm of a vector (a, b) is a number, usually written as ||(a, b)||, that

is the distance between the point (a, b) and the point (0, 0). Thus, ||(a, b)||equals

√(a− 0)2 + (b− 0)2 which simplifies as

||(a, b)|| =√a2 + b2

If you think of a vector as an arrow, then its norm is the length of the arrow.

Example.

• The norm of the vector (−1, 3) is

||(−1, 3)|| =√

(−1)2 + 32 =√

1 + 9 =√

10

Norms via dot productsRecall from the exercises in the chapter “The Plane of Vectors” that

(a, b)

(ab

)= a2 + b2

is sometimes called the “dot product” of the vectors (a, b) and (a, b). Thus,the norm of (a, b) is the square root of the dot product of the vectors (a, b)and (a, b). That is how some prefer to think about norms.

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(-1,3) 3

b

0.

(-1,3) 3

b

0.

Exercises

For #1-14, provide the value asked for.

1.) |4|

2.) |0|

3.) | − 4|

4.) |3− 4|

5.) |4− 7|

6.) |10− (−4)|

7.) the distance between 7 and −4

8.) the distance between −5 and 9

9.) the distance between (8,−2) and (2, 5)

10.) the distance between (3, 4) and (1, 7)

11.) the distance between (10, 4) and (−2, 4)

12.) the distance between (−3,−7) and (−1,−5)

13.) the distance between (−9, 3) and (0, 0)

14.) | − (−6)|

For #15-18, give the norms of the vectors.

15.) ||(1, 3)|| 16.) ||(−4, 6)|| 17.) ||(−1,−5)|| 18.) ||(7,−3)||

For #19-21, find the solutions of the equations in one variable.

19.) (2x2 − 3x− 2)2 = 1 20.) (2− ex)2 = 9 21.)√

loge(x) + 27 = −2

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For #22-27, use the Pythagorean Theorem to determine the value of x, thelength of the specified side of the right triangle.

22.) 25.)

23.) 26.)

24.) 27.)

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Distance in R2The Pythagorean Theorem allows us to determine the distance between

any pair of points in the plane.

Proposition (2). The distance between two points (x1,y) and (x2, y2) is

— x2)2 + (Yi — Y2)2Proof: The distance between (x1,y) and (x2, y2) is the length, c, of a side

of a right triangle.

3

The other two sides of the triangle have length lxi — x21 and lYi — Y2l °the Pythagorean theorem tells us that the distance, c, between (x1, Yi) and(x2, Y2) satisfies the equation

Therefore,= jx1

— x2l + IYi Y21

C = lxi — x212 + lYi — Y212We discussed earlier in this chapter that ja

—bj2 = (a — b)2 for any numbers

a and b, so we can write that c, the distance between (xi, Yi) and (x2, y2),equals

____________________

— x2)2 + (Yi — Y2)2

i66

• 2Distance in RThe Pythagorean Theorem allows us to determine the distance between

any pair of points in the plane.

Proposition (2). The distance between two points (x1, yj) and (x2, y2) is

(x1 —x2)2 + (Yi — Y2)2Proof: The distance between (x1,Yl) and (x2, y2) is the length, c, of a side

of a right triangle.

The other two sides of the triangle have length x1— x21 and li — Y2l °

the Pythagorean theorem tells us that the distance, c, between (x1,Yl) and(x2, y2) satisfies the equation

Therefore,= lxi — x2l + lYi — Y2l

C = lxi — x212 + li — Y212We discussed earlier in this chapter that a — bl2 = (a — b)2 for any numbersa and b, so we can write that c, the distance between (Xi, y) and (x2, y2),equals

____________________

—x2)2+ (Yi — Y2)2

I0

i66

Distance in R2The Pythagorean Theorem allows us to determine the distance between

any pair of points in the plane.

Proposition (2). The distance between two points (x1, y’) and (x2, Y2) is

— x2)2 + (yl — Y2)Proof: The distance between (x1,Yi) and (x2, Y2) is the length, c, of a side

of a right triangle.

x

The other two sides of the triangle have length x1— x21 and Ii — Y2i °

the Pythagorean theorem tells us that the distance, c, between (x1,y) and(x2, Y2) satisfies the equation

Therefore,= lxi — xj2 + IYi — Y21

C = — x212 + lYi — Y212We discussed earlier in this chapter that Ia — bj2 = (a — b)2 for any numbersa and b, so we can write that c, the distance between (x1, yj) and (x2, y2),equals

_____________________

— x2)2 + (yi — Y2)2

12

i66

Distance in R2The Pythagorean Theorem allows us to determine the distance between

any pair of points in the plane.

Proposition (2). The distance between two points (x1, Yi) and (x2, Y2) is

— x + (y’ — Y2)

Proof: The distance between (x1,Yi) and (x2, y2) is the length, c, of a sideof a right triangle.

9

The other two sides of the triangle have length x1— x2 and ly’ — Y21 °

the Pythagorean theorem tells us that the distance, c, between (x1,y) and(x2,Y2) satisfies the equation

Therefore,C2 = jx1 — x2j2 + Ii — Y212

C = VIxi — X22 + ly’ — Y22We discussed earlier in this chapter that a —

= (a — b)2 for any numbersa and b, so we can write that c, the distance between (x1, yi) and (x2, y2),equals

_____________________

— x2)2 + (Y1 — Y2)2

15

166

Distance in R2The Pythagorean Theorem allows us to determine the distance between

any pair of points in the plane.

Proposition (2). The distance between two points (x1,Yi) and (x2, Y2) is

— x2)2 + (Yi — Y2)2Proof: The distance between (x1, Yi) and (x2, y2) is the length, c, of a side

of a right triangle.

The other two sides of the triangle have length lxi — x21 and lYi - Y21 °the Pythagorean theorem tells us that the distance, c, between (x1,Yi) and(x2, Y2) satisfies the equation

Therefore,= lxi — x21 + lYi — Y21

C = \,/JXi — x212 + Yi — Y212We discussed earlier in this chapter that a — b12 = (a — b)2 for any numbersa and b, so we can write that c, the distance between (xi, Yi) and (x2, y2),equals

____________________

— x2)2 + (Yi — Y2)2

x

7

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Distance in R2The Pythagorean Theorem allows us to determine the distance between

any pair of points in the plane.

Proposition (2). The distance between two points (xi, yl) and (12, y2) is

(x1 — 12)2 + (y— Y2)2

Proof: The distance between (xi, Yi) and (12, Y2) is the length, c, of a sideof a right triangle.

x

The other two sides of the triangle have length x— 121 and lyi — Y21 °

the Pythagorean theorem tells us that the distance, c, between (Xj, Yl) and(12, y2) satisfies the equation

Therefore,C2 = 121 + yi

— Y21

C = Ixi 1212 + Iyl — Y2lWe discussed earlier in this chapter that Ia — bI2 = (a — b)2 for any numbersa and b, so we can write that c, the distance between (11, y) and (12, y2),equals

____________________

(x1 12)2 + (yi— Y2)2

I0

9

166