Unit 3. Analytic Geometry

IES VIRREY MORCILLO. VILLARROBLEDO (ALBACETE)

UNIT 3. ANALYTIC GEOMETRY

1 APM

UNIT 3.

ANALYTIC GEOMETRY



2 APM

INDEX

1. FIXED VECTORS IN THE PLANE.

2. FREE VECTORS IN THE PLANE.

3. COORDINATES OF A VECTOR. POSITION VECTOR.

MAGNITUDE OF A VECTOR. DISTANCE BETWEEN TWO

POINTS.

4. OPERATIONS WITH VECTORS.

4.1. ADDING VECTORS.

4.2. SUBTRACTING VECTORS.

4.3. PRODUCT OF A VECTOR AND A NUMBER.

5. LINEAR COMBINATION OF VECTORS. LINEARLY

DEPENDENT/INDEPENDENT VECTORS.

6. SCALAR PRODUCT OF VECTORS.

7. ANGLE BETWEEN TWO VECTORS.

8. MIDPOINT IN A LINE SEGMENT. SYMMETRIC POINT.

9. EQUATIONS OF A STRAIGHT LINE.

9.1. VECTOR EQUATION.

9.2. PARAMETRIC EQUATIONS.

9.3. CONTINUOUS FORM.

9.4. STANDARD FORM (GENERAL FORM OF THE STRAIGHT

LINE).

9.5. SLOPE-INTERCEPT FORM OR EXPLICIT EQUATION OF THE

STRAIGHT LINE.

9.6. POINT-SLOPE FORM.

10. RELATIVE POSITIONS BETWEEN TWO STRAIGHT LINES IN

THE PLANE.

11. ANGLE BETWEEN TWO LINES.

12. EXERCISES AND PROBLEMS.

13. SELF EVALUATION.



3 APM

1. FIXED VECTORS IN THE PLANE.

A scalar is a quantity that only has one magnitude (for example, the temperature).

2. FREE VECTORS IN THE PLANE.

Two vectors are equivalent when they have the same magnitude, the same or

parallel direction and the same sense.

It is called that they are the same free vector.

Observe this figure:

3. COORDINATES OF A VECTOR. POSITION VECTOR.

MAGNITUDE OF A VECTOR. DISTANCE BETWEEN TWO

POINTS.

A vector, , has magnitude and direction.

We represent vectors as arrows:

The length of the segment is the magnitude (quantity).

That is the distance from A to B.

The arrow head is the sense.

= ( , )

If the coordinate of A and B are:

A(x1, y1), B(x2, y2)



4 APM

Example 1: Find the coordinates of the vector , if we know that A(2,2) and

B(5,7)

= 5 2, 7 2 = (3,5)

Example 2: The vector has the coordinates (5, -2). Find the coordinates of A if

the terminal point is B(12, -3).

(12 x1, -3 y1) = (5, -2)

12 x1 = 5 x1 = 7

-3 y1 = -2 y1 = -1

So, A (7, -1)

The magnitude of a vector, , is the length of the segment.

A vector with magnitude 1 is called a unit vector.

Example: = 1,3 = 12 + (3)2 = 10 3.162

The vector that joins the coordinates

origin, O, with a point, P, is the position vector

of the point P.

To calculate the magnitude we use

Pythagoras theorem:

If = , = +



5 APM

Example: Calculate the distance between these points: A(3, -1) and B(-4,2)

= 4 3, 2 1 = 7,3 = (7)2 + 32 = 58 7.616

4. OPERATIONS WITH VECTORS.

4.1. ADDITION OF TWO VECTORS.

To add two vectors, add their coordinates:

= , , = , + = ( + , + )

Example: = 3,5 , = 4, 1 + = 3 + 4 ,5 + 1 = (1,4)

To add two vectors we put the vectors head to

tail one after the other.

The result is a triangle. The third side is the

sum of the two vectors.

Parallelogram Rule: If there are two vectors

with a common origin and parallel lines are drawn to

the vectors, a parallelogram is obtained whose

diagonal coincides with the sum of the vectors.

, = = ( ) + ( )

The distance between two points P(x1, y1)

and Q(x2, y2), is the magnitude of the vector .

Using Pythagoras theorem:

http://4.bp.blogspot.com/-dzCdoyMJ4rE/UCK7FaE4xrI/AAAAAAAAAE4/Pokr7QzKV8I/s1600/20070926klpmatgeo_285.Ges.SCO.png



6 APM

4.2. SUBTRACTING VECTORS.

Example: = 4, 5 and = (3,2), then:

= + = 4, 5 + 3,2 = (7, 7)

4.3. PRODUCT OF A VECTOR AND A NUMBER.

Example: = 4,5 . Calculate 3 .

3 = 3 4 , 3 5 = (12,15)

= + ( )

To subtract two vectors and , add

with the opposite . So,

= (1,2)

To multiply a vector by a

number, we multiply each coordinate

of the vector by the number. So:

http://4.bp.blogspot.com/_isAkcamREJQ/S8ZM8_C0u5I/AAAAAAAAAAU/Wl7-Znm6C5c/s1600/20070926klpmatgeo_286.Ges.SCO[1].pnghttp://www.youbioit.com/es/article/17277/como-se-multiplica-un-vector-por-un-escalar?size=_original



7 APM

5. LINEAR COMBINATION OF VECTORS. LINEARLY

DEPENDENT/INDEPENDENT VECTORS.

A linear combination of two or more vectors is the vector obtained by adding

two or more vectors (with different directions) which are multiplied by scalar values.

= 11 + 22 + +

Example 1: Given the vectors = 1,2 and = (3,1). Calculate the linear

combination vector = 2 + 3 .

= 2 1,2 + 3 3,1 = 2,4 + 9,3 = (11,1)

Example 2: Can the vector = 2,1 be expressed as a linear combination of the

vectors = 3,2 and = (1, 4)?

(2, 1) = a(3, -2) + b(1,4)

(2, 1) = (3a, -2a) + (b, 4b)

(2, 1) = (3a + b, -2a + 4b)

3 + = 2

2 + 4 = 1

You must solve the system of equations. The solution is a = and b =

So, =1

2 +

1

2

Vectors are linearly dependent if there is a linear combination of them that equals

the zero vector, without the coefficients of the linear combination being zero.

11 + 22 + + = 0



8 APM

IMPORTANT PROPERTIES

1. Two vectors in the plane are linearly dependent if, and only if they are parallel.

2. Two vectors in the plane = (u1, u2) and = (v1, v2) are linearly dependent if

their coordinates are proportional. So:

= u1, u2 = k v1, v2 k =u1v1

=u2v2

Several vectors are linearly independent if none of them can be expressed as a

linear combination of the others.

11 + 22 + + = 0 1 = 2 = = = 0

Example 1: Determinate if the vectors are linearly dependent or independent:

= (3, 1) and = (2, 3).

3

2

1

3, because 33 = 9 and 21 = 2

Example 2: Determine x for the vectors are linearly dependent or independent:

= (x 1, 3) and = (x + 1, 5).

1

+ 1=

3

5 5 1 = 3 + 1 5 5 = 3 + 3 2 = 8 = 4

Then, these vectors are linearly dependent for x = 4.

If x4, these vectors are linearly independent.

6. SCALAR PRODUCT OF VECTORS.

The scalar product or dot product of two vectors and is equal to:

=

Example: If = (3, 0) and = (5,5) and , = 45, calculate the scalar product.

= cos 45 = 32 + 02 52 + (5)2 cos 45 = 3 50 2

2= 15



9 APM

It can also be expressed as:

= +

Example: If = (3, 0) and = (5,5), calculate the scalar product.

= 1 1 + 2 2 = 3 5 + 0 5 = 15 + 0 = 15

When two vectors are orthogonal (90) then their scalar product is zero, regardless

of their lengths.

The sign of the scalar product of two vectors and depends on whether the

angle between them is acute or obtuse. If the value of the scalar product is zero, then the

vectors are perpendicular:

7. ANGLE BETWEEN TWO VECTORS.

The angle between two vectors and is given by the formula:

= +

+

+

Example 1: Calculate the angle between these vectors: = (3, 0) and

= (5,5).

cos = 3 5 + 0 (5)

32 + 02 52 + (5)2=

15

3 50=

15

3 5 2=

1

2= 2

2 = 45



10 APM

Example 2: Calculate the angles of the triangle with vertices A(6, 0), B(3, 5) and

C(-1, -1).

= 3,5 ; = 3,5 ; = 7,1 ; = (4,6)

cos =

=

3 7 + 5 1

3 2 + 52 7 2 + 1 2=

16

34 50= 0.388057

A = 67 9 58.84

cos =

=

3 4 + (5) 6

32 + (5)2 4 2 + 6 2=

18

34 52= 0.428

B = 64 39 13.77

C = 180 (67 9 58.84 + 64 39 13.77) = 48 10 47.39

C = 48 10 47.39



11 APM

8. MIDPOINT IN A LINE SEGMENT. SYMMETRIC POINT.

Example 1: What is the midpoint here?

The midpoint is halfway between the two end

points:

Its x value is halfway between the two x values.

Its y value is halfway between the two y values.

= +

2, +

2

To calculate the midpoint in a line segment:

Add both x coordinates, divide by 2.

Add both y coordinates, divide by 2.

We can use this formula:

= +

2, +

2

= 3 + 8

2,5 + (1)

2

= 5

2,4

2

= 2.5, 2

Use the formula:



12 APM

Example 2: Calculate the coordinates of point C in the line segment AC,

knowing that the midpoint is M(2, -2) and the endpoint is A(-3, 1).

If C(x1, y1)

2 =3 + 1

2 4 = 3 + 1 1 = 7

2 =1 + 2

2 4 = 1 + 2 2 = 5

So, C(7, 5)

9. EQUATIONS OF A STRAIGHT LINE.

9.1. VECTOR EQUATION.

Any point X(x, y) of the line can be calculated from a point P(p1, p2) plus a

multiple of the direction vector of the line.

To define a line we need:

A point P(p1, p2) and

A direction vector, = u1, u2 .

=

+ =

= +

, = 1,1 + u1, u2

The expression of point (x, y) is

the vector equation of the line:



13 APM

9.2. PARAMETRIC EQUATIONS.

If in the vector equation we separate x and y, we get the parametric equations:

= 1 + 1 = 2 + 2

9.3. CONTINUOUS FORM.

If we isolate in each equation:

= 11

= 22

The continuous form is:

11

= 22

9.4. STANDARD FORM (GENERAL FORM OF THE STRAIGHT

LINE).

We cross-multiply:

2 21 = 1 12

Now, we transfer everything to the first side and order the coefficients:

2 21 1 + 12 = 0

If = 2 , = 1, = 21 + 12

The General form of the straight line or Implicit equation is:

+ + = 0



14 APM

OBSERVATION: If the line is the general form, ie, + + = 0, the

direction vector of the line is: = (,)

9.5. SLOPE-INTERCEPT FORM OR EXPLICIT EQUATION OF THE

STRAIGHT LINE.

But we can isolate in different ways. The first way is the following:

2 21 = 1 12

2 21 + 12 = 1

=21

+12 21

1

The slope-intercept form or explicit equation is:

= +

9.6. POINT-SLOPE FORM.

The second way is the following:

2 21 = 1 12

2( 1) = 1( 2)

2 =21

( 1)

The point-slope form is:

2 = ( 1)

Example 1: Draw the line that passes though point A(2,2) and has (4,2) as a

direction vector. Calculate the equations of this straight line.

, = 2,2 + 4,2

= 2 4 = 2 + 2

2

4= 2

2

2 4 = 4 + 8

2 + 4 = 12

=2 + 12

4 =

+



15 APM

Example 2: Write the equation of the line that passes through points A(5,3) and

B(-2,-1). Draw this line.

= 2,1 5,3 = (7,4)

Now, we calculate the line that passes though point A(5,3) and has (7,4) as

a direction vector.

, = 5,3 + 7,4

= 5 7 = 3 4

5

7= 3

4

4 + 20 = 7 + 21

4 + 7 = 1

=4 + 1

7 =

+



16 APM

Two lines are parallel if their slopes are equal. So, =

Two lines are perpendicular if their slopes are the inverse of each other and their

signs are opposite. So, =1

Example 1: The slope of the line through the points A = (2, 1) and B = (4, 7) is:

=

=

Example 2: The line passes through points A = (1, 2) and B = (1, 7) and has no

slope since division by 0 is undefined.

=7 2

1 1=

5

0

Example 3: The slope of the line 2 3 4 = 0 is =2

3=

2

3

=2 12 1

=21

+ + = 0 =

REMEMBER: SLOPE

The slope is the inclination of a line with respect

to the x-axis. It is denoted by the letter m.

Slope of two given points:

Slope of a given angle: = tan

Slope of a given vector of a line:

= (1, 2):

Slope of a straight line in general form:



17 APM

10. RELATIVE POSITIONS BETWEEN TWO STRAIGHT LINES IN

THE PLANE.

Parallel lines have the same slope. So, = . They dont have any point in

common.

Intersecting lines have a different slope. They have one point in common.

Perpendicular lines have the inverse slope. So, =1

. Perpendicular lines

are a special case of intersecting lines.

Coincident lines have the same slope and all the points on the lines are

common. They are the same line!

If we want to find the relative position of two lines we can follow two ways:

First way: Solve the simultaneous equations and:

- If the system has one solution, the lines intersect.

- If the system has more than one solution, we have coincident lines.

- If the system has no solution the lines have no point in common, so the lines

are parallel.

Second way: Compare the coefficients of the equations. If + + = 0 and

+ + = 0 are the equations of the lines, we have:

- If

the lines intersect.

- If

=

=

the lines coincide.

- If

=

the lines are parallel.

Example 1: Calculate k so that the lines r x + 2y 3 = 0 and s x ky + 4 = 0,

are parallel.

=1

2, =

1

the lines r and s are parallel if

1

2=

1

= 2



18 APM

Example 2: Determine the equation for the line parallel to r x + 2 y + 3 = 0 that

passes through the point A = (3, 5).

The line r and s are parallel if = =1

2

We use the point-slope form: 2 = ( 1)

5 =1

2( 3)

2 10 = + 3

+ 2 13 = 0

Example 3: The line r 3x + ny 7 = 0 passes through the point A(3, 2) and is

parallel to the line s mx + 2y 13 = 0. Calculate the values of m and n.

3 3 + 2 7 = 0 9 + 2 7 = 0 2 + 2 = 0 = 1

The lines r and s are parallel. So: = 3

1=

2 = 6

11. ANGLE BETWEEN TWO LINES.

= +

+

+

=

+

The angle between two lines is the smaller

of the angles formed by the intersection of the two

lines. The angle can be obtained from:

Formula 1: Their direction vectors:

Formula 2: Their slopes:



19 APM

Example 1: Find the angle between the lines r and s, if their directional vectors are

2,2 and 2,3

cos = 2 2 + 2 (3)

(2)2 + 22 22 + (3)2=

10

8 13=

10

2 2 13=

5

26=

5 26

26= 0.981

= 111876

Example 2: Given the lines r 3x + y - 1 = 0 and s 2x + by - 8 = 0, calculate the

value of b so that the two lines form an angle of 45.

= 1,3 and = , 2

cos 45 = 2

2= 1 + 3 2

10 2 + 4 2

2=

+ 6

10 2 + 4

2

2

2

= + 6

10 2 + 4

2

1

2=

+ 6 2

10 2 + 4

10 2 + 4 = 2 + 6 2

5 2 + 4 = 2 + 12 + 36

52 + 20 = 2 + 12 + 36

42 12 16 = 0

2 3 4 = 0 1 = 4 , 2 = 1



20 APM

12. EXERCISES AND PROBLEMS.

1. Calculate the coordinates of the vector with A(3,4) and B(-2,1).

2. The coordinates of fix vector are (2,3) and B coordinates are B(0,1).

Calculate the point A geometrically and analytically.

3. Are the vectors and equipollent if A(5,3), B(0,-3), C(1,1) and D(-4,-5)?

4. Lets the points P(1,0), Q(5,7) and M(0,1), calculate the coordinates to a point N

if . Do it geometrically, too.

5. Lets the points be A(2,2), B(1,3) and C(-1,-2). Represent in the Cartesian axis.

Calculate the length of the sides of the triangle.

6. Calculate the following operations with vectors 1,1 , 0,5 (3,1):

a) 4 1

2

b) + 2 3

c) + + 2 +

d) + 2

7. Calculate, if it is possible, a and b to (3a, 5)=(-2b, 3).

8. Calculate to = 3 1

2 , if (-1,3) and (7,-2).

9. Write the vector (1,5) like linear combination of the vectors

3,2 4,12 .

10. A(-2,0), B(0,0) and C(-3,2) are three points. Represent these points and calculate

the vectors , . Calculate the vectors , . What do you

observe? Calculate the magnitude of the vectors , . Calculate the

angles of the triangle.

11. Calculate the distance between these points:

a) A(1,2), B(5,-3)

b) A(-17,-2), B(0,1)

c) A(-1,-1), B(1,1)

d) A(0,-6), B(-1,8)



21 APM

12. A line passes though A(0,3) and B(4,-3).

a) Draw the line.

b) Find the direction vector.

c) Calculate the slope of the line.

d) Calculate the equation of this line.

13. The equation of a line is 3 + 4 = 2.

a) What type of equation is it?

b) What is the slope?

c) Write two points of this line.

d) Draw the line.

14. Find an equation of the line that satisfies the given conditions:

a) It passes through (2,-3) and its slope is 7.

b) It passes through (2,-1) and (1,7).

c) It passes through (-4,5), parallel to the x-axis.

d) It passes through (4,-5), parallel to the y-axis.

15. Find an equation of the line that it passes through (1,-6) and it is parallel to the

line + 2 = 6.

16. Study the position of these lines:

a) + 3 1 = 0; 2 + 6 + 5 = 0.

b) , = 3,1 + 2,1 ; + 2 = 3.

c) =1

2;

3

2=

1

1.

d) + 3 = 2; = 2

= 3 + 2 .

17. Find the parametric equations of the line that passes through point A(3,-2) and

that it is parallel to vector (3,-1).

18. The segments between A(-4,-3); B(-3,2); C(4,4) and D form a trapezium. Find

the coordinates of point D.

19. If the straight line pass by the point A(2,1) and director vector is (2,4).

Calculate its equations in all the forms you know.



22 APM

20. The implicit equation of a straight line is 2x-3y+1=0. Write the continuous

equation, slope-point, explicit form, vectorial and parametrics.

21. Find the equation of a line through the points (1,2) and (3,-1). What is its slope?

22. What is the equation of the line that passes through the point (1,1) and is parallel

to the line = 2 + 2 ?

23. What is the equation of the line that passes through the point (1,1) and is

perpendicular to the line = 2 + 2 ? Where do the two lines intersect?

24. Calculate the parallelogram area ABCD, knowing the side equations AB is

x 2y = 0, the side equation AD is 3x + y = 0 and the coordinates of C are (3,5).

25. The points B(-1,3) and C(3,-3) are the vertex in an isosceles triangle which third

vertex is in the straight line x + 2y = 15, lets AB and AC the equal sides.

Calculate the A coordinates and the length of the sides.

26. Calculate a for the lines + 1 2( + 2) and

3 3 + 1 5 + 4 = 0 are:

a) Parallel.

b) Perpendicular.

27. Calculate m for the lines mx + y = 12 and 4x 3y = m+1 are parallel. Calculate

the distance between these lines.

28. A rhombus has the following vertices A(1,3), B(4,6), C(4,y). Calculate y and the

vertex D. Calculate the length of the diagonals and angles of this rhombus.

29. What is the position of these lines? If these lines are parallel calculate the

distance between them. If these lines are intersecting lines, calculate the point of

intersection and the angle between them.

a) 3 = 8; 3 = 2

b) 2 3 = 4; 4 + 6 = 4

30. Find an equation of the line that is parallel to the line 2x 3y 5 = 0 through the

point (1, -2).



23 APM

13. SELF EVALUATION.

1. Calculate the value of k knowing the magnitude of the vector = , 3 is 5.

2. Calculate the scalar product and the angle formed by the following vectors:

= 3,4 and = 8,6

3. Given the vectors = 2, and = 3,2 , calculate the value of k so that

the vector and are:

a) Perpendicular.

b) Parallel.

c) Make an angle of 60.

4. Prove that the points A(1,7), B(4,6) and C(1, -3) belong to a circumference of a

circle of center (1,2). Note: If O is the center of the circle, the distances from O

to A, B and C should be equal.

5. Identify the type of triangle determined by the points: A(4, -3), B(3,0) and

C(0,1).

Important note:

If (,)2 = , 2 + (,)2, it is right triangle.

If (,)2 < , 2 + (,)2, it is acute triangle.

If (,)2 > , 2 + (,)2, it is obtuse triangle.

6. Calculate the coordinates of point C in the line segment AC, knowing that the

midpoint is B(2, -2) and A(3, -1).

7. Calculate the equation of the line that passes through the point P(-3,2) and it is

perpendicular to the line 8 1 = 0.

8. The vertices of a parallelogram are A(3,0), B(1,4), C(-3,2) and D(-1,-2).

Calculate the area.

9. A parallelogram has a vertex A(8,0) and the point of intersection of its two

diagonals is M(6,2): If the other vertex is at the origin, calculate:

a) The other two vertices.

b) The equations of the diagonals.

c) The length of the diagonal.

10. Calculate the angle between the following lines 2 3 = 8;

3 = 2.

Documents

Unit 3. Analytic Geometry