CALCULUS I( with Analytic Geometry)MATH 21-1

CO2Discuss comprehensively the fundamental

concepts in Analytic Geometry and use them to solve application problems and problems

involving lines.

FUNDAMENTAL CONCEPTS OF ANALYTIC GEOMETRY

Lesson 1: Rectangular Coordinate System, Directed Distance, Distance Formula

OBJECTIVE:

At the end of the lesson, the students should be able to illustrate properly and solve application problems involving distance formula.

• Analytic Geometry – is the branch of mathematics, which deals with the properties, behaviours, and solution of points, lines, curves, angles, surfaces and solids by means of algebraic methods in relation to a coordinate system(Quirino and Mijares) .

• It is a unified algebra and geometry dealing with the study of relationships between different geometric figures and equations by means of the geometric properties and processes of algebra in relation to a coordinate system ( Marquez, et al).

DEFINITION:

FUNDAMENTAL CONCEPTS

Two Parts of Analytic Geometry

1. Plane Analytic Geometry – deals with figures on a plane surface (two-dimensional geometry, 2D).

2. Solid Analytic Geometry – deals with solid figures ( three-dimensional geometry, 3D).

Directed Line – a line in which one direction is chosen as positive and the opposite direction as negative.

Directed Line Segment – portion of a line from one point to another.

Directed Distance – the distance from one point to another; may be positive or negative depending upon which direction is denoted positive.

DEFINITION:

RECTANGULAR COORDINATES

A pair of number (x, y) in which x is the first and y the second number is called an ordered pair. It defines the position of a point on a plane by defining the directed distances of the point from a vertical line and from a horizontal line that meet at a point called the origin, O. The x-coordinate of a point , known also as its abscissa, is the directed distance of the point from the vertical axis, y-axis; while the y-coordinate, also known as the ordinate, is its directed distance from the horizontal axis, the x-axis.

DISTANCE BETWEEN TWO POINTS

The horizontal distance between any two points is the difference between the abscissa (x-coordinate) of the point on the right minus the abscissa (x-coordinate) of the point on the left; that is,

Horizontal Distance Between Points

Vertical Distance Between Any Two Points

The vertical distance between any two points is the difference between the ordinate (y-coordinate) of the upper point minus the ordinate (y-coordinate) of the lower point; that is,

Distance Between Any Two Points on a Plane

The distance between any two points on a plane is the square root of the sum of the squares of the difference of the abscissas and of the difference of the ordinates of the points. That is, if

SAMPLE PROBLEMS

• By addition of line segments verify whether the points A ( - 3, 0 ) , B(-1, -1) and C(5, -4) lie on a straight line.

• The vertices of the base of an isosceles triangle are at (1, 2) and (4, -1). Find the ordinate of the third vertex if its abscissa is 6.

3. Find the radius of a circle with center at (4, 1), if a chord of length 4 is bisected at (7, 4).

1. Show that the points A(-2, 6), B(5, 3), C(-1, -11) and D(-8, -8) are the vertices of a rectangle.

2. The ordinate of a point P is twice the abscissa. This point is equidistant from (-3, 1) and (8, -2). Find the coordinates of P.

6. Find the point on the y-axis that is equidistant from (6, 1) and (-2, -3).

Lesson 2: DIVISION OF A LINE SEGMENT

OBJECTIVE:

At the end of the lesson, the students should be able to illustrate properly and solve problems involving division of line segments.

Let us consider a line segment bounded by the points . This line segment can be subdivided

in some ratio and the point of division can be determined. It is also possible to determine terminal point(s) whenever the given line segment is extended beyond any of the given endpoints or beyond both endpoints . If we consider the point of division/ terminal point to be P (x, y ) and define the ratio, r, to be

then the coordinates of point P are given by:

If the line segment is divided into two equal parts, then the point of division is called the midpoint. The ratio, r, is equal to ½ and the coordinates of point P are given by:

or simply by:

SAMPLE PROBLEMS•Find the midpoint of the segment joining (7, -2) and (-3, 5).•The line segment joining (-5, -3) and (3, 4) is to be divided into five equal parts. Find all points of division.•The line segment from (1, 4) to (2, 1) is extended a distance equal to twice its length. Find the terminal point.•On the line joining (4, -5) to (-4, -2), find the point which is three-seventh the distance from the first to the second point. •Find the trisection points of the line joining (-6, 2) and (3, 8). •What are the lengths of the segments into which the y-axis divided the segment joining ( -6, -6) and (3, 6)?•The line segment joining a vertex of a triangle and the midpoint of the opposite side is called the median of the triangle. Given a triangle whose vertices are A(4,-4), B(10, 4) and C(2, 6), find the point on each median that is two-thirds of the distance from the vertex to the midpoint of the opposite side.

Lesson 3: INCLINATION AND SLOPE A LINE

OBJECTIVES:

At the end of the lesson, the students should be able to use the concept of angle of inclination and slope of a line to solve application problems.

INCLINATION AND SLOPE OF A LINE

The angle of inclination of the line L or simply inclination , denoted by , is defined as the smallest positive angle measured from the positive direction of the x-axis to the line.

The slope of the line, denoted by m , is defined as the tangent of the angle of inclination; that is, And if two points are points on the line L then the slope m can be defined as

PARALLEL AND PERPENDICULAR LINES

If two lines are parallel their slope are equal. If two lines are perpendicular, the slope of one line is the negative reciprocal of the slope of the other line.

If m1 is the slope of L1 and m2 is the slope of L2 then , or

Sign Conventions:Slope is positive (+), if the line is leaning to the right.Slope is negative (-), if the line is leaning to the left.Slope is zero (0), if the line is horizontal.Slope is undefined , if the line is vertical.

x

y y

x

SAMPLE PROBLEMS1. Find the slope, m, and the angle of inclination of the line through the points (8, -4) and (5, 9).2. The line segment drawn from (x, 3) to (4, 1) is perpendicular to the segment drawn from (-5, -6) to (4, 1). Find the value of x.3. Find y if the slope of the line segment joining (3, -2) to (4, y) is -3.

ANGLE BETWEEN TWO INTERSECTING LINES

L1

L2

ti

it

mm1mmtan

Where: mi = slope of the initial side mt = slope of the terminal side

The angle between two intersecting lines is the positive angle measured from one line (L1) to the other ( L2).

0180:note

Sample Problems

1.Find the angle from the line through the points (-1, 6) and (5, -2) to the line through (4, -4) and (1, 7). 2.The angle from the line through (x, -1) and (-3, -5) to the line through (2, -5) and (4, 1) is 450 . Find x3.Two lines passing through (2, 3) make an angle of 450 with one another. If the slope of one of the lines is 2, find the slope of the other.

AREA OF A POLYGON BY COORDINATESConsider the triangle whose vertices are P1(x1, y1), P2(x2, y2) and P3(x3, y3) as shown below. The area of the triangle can be determined on the basis of the coordinates of its vertices.

o

y

x

111 y,xP

222 y,xP

333 y,xP

Label the vertices counterclockwise and evaluate the area of the triangle by:

1yx1yx1yx

21A

33

22

11

The area is a directed area. Obtaining a negative value will simply mean that the vertices were not named counterclockwise. In general, the area of an n-sided polygon can be determined by the formula :

1n54321

1n54321

yy..yyyyyxx..xxxxx

21A

SAMPLE PROBLEMS

1. Find the area of the triangle whose vertices are (-6, -4), (-1, 3) and (5, -3).2.Find the area of a polygon whose vertices are (6, -3), (3, 4), (-6, -2), (0, 5) and (-8, 1).

Lesson 4: EQUATION OF A LOCUS

OBJECTIVE:

At the end of the lesson, the students should be able to determine the equation of a locus defining line, circle and conics and other geometries defined by the given condition.

EQUATION OF A LOCUS

An equation involving the variables x and y is usually satisfied by an infinite number of pairs of values of x and y, and each pair of values corresponds to a point. These points follow a pattern according to the given equation and form a geometric figure called the locus of the equation. Since an equation of a curve is a relationship satisfied by the x and y coordinates of each point on the curve (but by no other point), we need merely to consider an arbitrary point (x,y) on the curve and give a description of the curve in terms of x and y satisfying a given condition.

Sample Problems

Find an equation for the set of all points (x, y) satisfying the given conditions.1. It is equidistant from (5, 8) and (-2, 4).2.The sum of its distances from (0, 4) and (0, -4) is 10.3.It is equidistant from (-2, 4) and the y-axis

Lesson 5: STRAIGHT LINES / FIRST DEGREE EQUATIONS

OBJECTIVE:

At the end of the lesson, the students is should be able to write the equation of a line in the general form or in any of the standard forms; as well as, illustrate properly and solve application problems concerning the normal form of the line.

STRAIGHT LINEA straight line is the locus of a point that

moves in a plane in a constant slope.

Equation of Vertical/ Horizontal Line If a straight line is parallel to the y-axis ( vertical line ), its equation is x = k, where k is the directed distance of the line from the y-axis.

Similarly, if a line is parallel to the x-axis ( horizontal line ), its equation is y = k, where k is the directed distance of the line from the x-axis.

General Equation of a Line

A line which is neither vertical nor horizontal is defined by the general linear equation Ax + By + C=0, where A and B are nonzeroes.The line has y-intercept of and slope of .

DIFFERENT STANDARD FORMS OF THE EQUATION OF A STRAIGHT LINE

A. POINT-SLOPE FORM:If the line passes through the points ( x , y) and (x1, y1), then the slope of the line is .

Rewriting the equation we have

which is the standard equation of the point-slope form.

1

1

xxyym

11 xxmyy

B. TWO-POINT FORM:If a line passes through the points (x1, y1) and (x2, y2), then the

slope of the line is . Substituting it in the point-slope formula will result to

the standard equation of the two-point form.

12

12

xxyym

112

121 xx

xxyyyy

C. SLOPE-INTERCEPT FORM:

Consider a line containing the point P( x, y) and not parallel to either of the coordinate axes. Let the slope of the line be m and the y-intercept ( the intersection point with the y-axis) at point (0, b), then the slope of the line is .

Rewriting the equation, we obtain

the standard equation of the slope-intercept form.

0xbym

bmxy

D. INTERCEPT FORM:Let the intercepts of a line be the points (a, 0), the x-

intercept, and (0, b), the y-intercept. Then the slope of the line is defined by .

Using the Point-slope form, the equation is written as

or simply as

the standard equation of the Intercept Form.

abm

0xabby

1by

ax

E. NORMAL FORM:

Suppose a line L, whose equation is to be found, has its distance from the origin to be equal to p. Let the angle of inclination of p be . Since p is perpendicular to L, then the slope of p is equal to the negative reciprocal of the slope of L,

Substituting in the slope-intercept form y = mx + b , we obtain

orthe normal form of the straight line

sincosm or ,cot

tan1m

sinpx

sincosy

p y sin cos x

Reduction of the General Form to the Normal Form

The slope of the line Ax+By+C=0 is . The slope of p which is perpendicular to the line is therefore ; thus, . From Trigonometry, we obtain the values and .

If we divide the general equation of the straight line by , we have or This form is comparable to the normal form .Note: The radical takes on the sign of B.

BA

AB

ABtan

22 BA

Bsin

22 BA

Acos

22 BA 0 BA

C yBA

B xBA

A222222

BA

C yBA

B xBA

A222222

p y sin cos x

PARALLEL AND PERPENDICULAR LINES

Given a line L whose equation is Ax + By + C = 0. The line Ax + By + K = 0 , for any constant K not

equal to C, is parallel to L; and the Bx – Ay + K = 0 is perpendicular to L.

DIRECTED DISTANCE FROM A LINE TO A POINT

The directed distance of the point P(x1, y1) from the

line Ax + By + C = 0 is , where the radical takes on the sign of B.

2211

BACByAxd

y

x

111 y,xP

222 y,xP

0CByAx 11

0d1

0d2

line the below is point the 0,d if

line the above is point the 0,d if

:note

Sample Problems1. Determine the equation of the line passing through (2, -3) and parallel to the line through (4,1) and (-2,2).2. Find the equation of the line passing through point (-2,3) and perpendicular to the line 2x – 3y + 6 = 03. Find the equation of the line, which is the perpendicular bisector of the segment connecting points (-1,-2) and (7,4).4. Find the equation of the line whose slope is 4 and passing through the point of intersection of lines x + 6y – 4 = 0 and 3x – 4y + 2 = 0.5. The points A(0, 0), B(6, 0) and C(4, 4) are vertices of triangles. Find: a. the equations of the medians and their intersection point

6. Find the distance from the line 5x = 2y + 6 to the point (3, -5).7. Find the equation of the bisector of the acute angles and also the bisector of the obtuse angles formed by the lines x + 2y – 3 = 0 and 2x + y – 4 = 0. 8. Determine the distance between the lines: 2x + 5y -10 =0 ; 4x + 10y + 25 = 0.9. Write the equation of the line a) parallel to b) perpendicular to 4x + 3y -10 = 0 and is 3 units from the point ( 2, -1).

CLASSWORK 11. The abscissa and ordinate of a point units from (3, 3) are numerically equal but of opposite signs. Find the point.2.Given two points A(8, 6) and B(–7, 9), determine a third point P(x, y) such that the slopes of AP and BP are ½ and –2/3 respectively. 3.A line through (–6,–7) and (x, 7) is perpendicular to a line through (1,–4) and (–5, 2). Find x.4.A line passes through (6,–4) and makes an angle of 1350 with the x-axis. Find the equation of the line.5.The angle from the line through (–1, y) and (4,–7) to the line through (4, 2) and (–1,–9) is 1350. Find y.6.Find the equation of the bisector of the obtuse angle between the lines x + 2y – 3 = 0 and 2x + y – 4 = 0.

52

REFERENCES

Analytic Geometry, 6th Edition, by Douglas F. RiddleAnalytic Geometry, 7th Edition, by Gordon Fuller/Dalton TarwaterAnalytic Geometry, by Quirino and MijaresFundamentals of Analytic Geometry by Marquez, et al.Algebra and Trigonometry, 7th ed by Aufmann, et al.