COORDINATE GEOMETRY

THE STRAIGHT LINES

THE CARTESIAN PLANE: We can represent real numbers by points on a real line, we can also represent ordered pairs by points in a plane. This plane is called Rectangular coordinate system or the Cartesian plane after the mathematician Rene Descartes (1596 – 1650). On the Cartesian plane, the horizontal number line is called x – axis and the vertical number line is usually y – axis. The point of intersection of the two axes (plural of axis) separate the plane into four regions called quadrants. Each point in the plane corresponds to an ordered pair (x,y) of real number x and y called coordinates of the point.

Consider a point B can be located on a plane by noting its distance along the x – axis and it distance along the y – axis from the origin 0.

B(x,y)

FIG 1

The point B is said to be located by the coordinate (x,y)

Example

Plot the points (–1,2), (2,3), (1,–4), (0,0), (–2,3),(4,–2) in the Cartesian plane.

DISTANCE BETWEEN TWO POINTS

Distance between two points A (x,y), and B (x₂y₂)

y B (x₂y₂)

d y₂–y₁ y₂

A C

A(z,y) x₂–x y₁ x

O x₁ p Q

x₂ FIG 3

Consider the figure above BQ and AP are the ordinate of A and B and AC is perpendicular to BC

From the diagram

AC² = x₂–x₁, BC = y₂ – y₁

Then from Pythagoras theorem

AB² = AC² + BC²

AB =

d =

Example

Find the distance between the following pairs of points

1. A(5,7) B(7,10)

2. C(3,–2) D(–5,–7)

3. E(½a, 3b) F(3a, ½b)

Solution

1. Let d₁ be the distance between A and B

2. Let d₂ be the distance between C and D

3. Let d₂ be the distance between E and F

FINDING POINTS AT A SPECIFIED DISTANCE FROM A GIVEN POINT.

Example

Find x so that distance between (1,2), (x, –10) is 13

Solution

13²= (x – 2)² + (–12)² Squaring both sides

169 – 144 = x² – 2x + 1

25 = x² – 2x + 1

0 = x² – 2x – 24

0 = (x – 6) (x – 4)

x = 6 or x = –4

There are two possible coordinates (6, –10) and (–4, –10) both of which lies 13 units from (1,2)

APPLICATION OF DISTANCE FORMULA

Example: Show that the points A (4,0) B(2,1) and C (–1, –5) are vertices of a right angle

Solution

d₃d₁

Fig 3 d₂

Using distance formula, one can find the of the sides

Since

One can conclude from Pythagorean theorem that the triangle is a right triangle.

Example: Show that point A(2,4), B(5,1), C (2,–2), D (–1,1) are the vertices of a square.

A(2,4) d₁ B(5,1)

d₄ d₂

D(–1,1) d₃ C(2,–2)

Solution

Using the distance formula gives the following

Since all the sides are equal, we can conclude it is square

THE MIDPOINT OF A LINE SEGMENT

The coordinates of the midpoint of a line segment are the average values of the corresponding coordinates of the two endpoints. The coordinates of the mid–point of the joining (x,y) and (x₂y₂) are

Let A and C have coordinates (x, y) and (x₂y₂) and point B be the midpoint with coordinate

(x, y) ABC is similar to BCD

Hence, the midpoint of the line segment joining the point (x₁,y₁) and (x₂,y₂) in the coordinate plane is

Example

Find the midpoint of the line joining the following line segment.

1. A(6,–3) , B(6,5)

2. X(1,1) , Y(9,7)

3. P(7,–4) , , Q(2,8)

Solution

EXAMPLE : A line segment has (4,10)as one endpoint and (6,8)as its midpoint. Find the other end point (x_2
,y₂)

Solution

Fig 7

(6,8)=

SLOPE OR GRADIENT OF A STRAIGHT LINE

The slope (m) of a non- vertical straight line passing through point (x_{1 ,}y₁) and (x_{2 ,}y₂) is .Whenever this formula is used, the order of subtraction is vital. If given two coordinate of a straight line .You are to choice either one of them as (x_{1 ,}y₁) and the other as (x_{2 ,}y₂). When this is done the numerate and denominator formed must be of the same order of subtraction.

Example

Find the slope of the line through the pairs of points

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

2, C(1,3) D(5,4)

3, E(-7,-2) F(4,-3)

Solution

y ANGLES OF SLOPE

Every non-horizontal line must intersect with x-axis .The angle formed by such an intersection determines the ANGLES OF THE SLOPE OF THE LINE.

The angle of slope of non-horizontal line is positive θ (less than 180^o) anticlockwise from the x-axis

Fig 9

The slope o the line AB

Example

Find the slope of the line passing through the pairs of point and the angle of slope of the line.

1. A(7,2) B(11,9)

2. C(–3,19) D(10,4)

3. E(–2,–15) F(7,12)

Solution

Let m₁ be the slope o the line

EQUATION OF STRAIGHT LINE

Consider the figure below

y

C

A

O E

Let B(x,y) be any point on the straight line and A the point in which line cuts OY which is C. The slope m =

Y – C = mx

y = mx + C

The equation mx + C = y is know as the Equation of a straight line of slope m.

Note: y = mx + C is the Equation of a line in slope form

Also ax + by = C equation of line

This equation is known as slope – Intercept form of equation of a line. Where is the intercept on the x –axis and is the intercept on y -axis

Example

Find the intercept that the line 7x – 14y – 28 = 0 makes on the axes what is the slope of this line.

Solution

Which is the intercept form; hence, the intercept on the axes is 4 and –2

THE POINT SLOPE FORM OF THE EQUATION OF A LINE

Once the slope of a line and coordinate of one point on a line can be determined, then one can find the equation for the line consider the figure below.

(x,y)

y – y₁

(x₁y₁)

x – x₁ x Fig 11

The slope of the line

m (x – x₁) = y – y₁

This is the point slope form of the equation of a line

Example

Find the equation of the line that passes through the (–1,–4) and has a slope of 4

Solution

y – y₁ = m(x – x₁) {Substitute y₁ = 4, x₁ = 1}

y – (–4) = 4(x–(–1))

y – 4 = 4(x+1)

y – 4 = 4x + 4

y = 4x {The equation of the line}

Two Point Form

The one point – slope form can be used to find the equation of a line passing through two (x₁y₁) and (y₂y₂)

C (x₂y₁)

B(x,y) D

F E

x FIG 13

This is called the two – point form of the equation of a line.

Example

Find the equation of the equation of a lines joining the following pairs of points.

1. A(–3,4) and B(8,1)

2. P(3,2) and q(–7,–3)

Solution

A (–3,4) and B(8,1)

Let (x₁y₁) = (–3,4), (x₂y₂) = (8,11)

APPLICATION OF LINES TO DAILY BUSINESS

Example

During the first two quarter of the year, Cadbury Nigeria Plc. Had a total sales of N6.4million and N10.6 million respectively.

1. Write a linear equation to predict the total sales y in term of the quarter

2. Use the equation to predict the total sales during the fourth quarter

Solution

The coordinate that will be obtained will be in the form (quarter, sales). Therefore (1.6, 4) and (2, 10.6) will be two point on the line respectively. the slope of the line passing through these two point is

Using the point slope form, the equation of the line is

y – y₁ = m (x – x₁)

y – 6.4 = 4.2 (x – 1)

y – 6.4 = 4.2x–4.2

y = 4.2x + 2.2

Using the equation y = 4.2x + 2.2

We can find the fourth = quarter sales

y = 4.2(4) + 2.2

y = N19 million

ANGLE BETWEEN TWO STRAIGHT LINE

C B

m₂ m₁

₂ ₂

A D x FIG 13

Consider the figure above, such that line has slope m₁ and CD has slope m₂ each making ₂ and ₂ respectively with the x – axis is

Then tan ₁ = m₁

tan ₂ = m₂

From the diagram

₂ + = ₁ {Sum of the two opposite angle of a is equal to opposite

exterior angle}

₂ = ₁ = ₂ where ₁ > ₂

tan = tan ( ₁– ₂)

If two non perpendicular lines have slopes m₁ and m₂, then the angle between the two lines is given by

Example : Find the angle , between the two lines

5x + 3y – 18 = 0

2x + 5y – 13 = 0

Solution

Line 1:

5x + 3y – 18 = 0

3y = –5x + 18

Line 2 : 2x + 5y – 13 = 0

5y = –2x + 13

The two lines have slopes of

Thus, the angle between the two lines is given by

CONDITION FOR PERPENDICULARITY WHEN TWO LINES ARE PERPENDICULAR

When two lines are parallel the angle between is 90^o

= 90 and hence tan =

1+m₁m₂ = 0

m₁ =

Therefore two lines are perpendicular if and only if their slopes are negative reciprocal of each other

CONDITION FOR PARALLELISM

When the lines are parallel, the angle between them is zero

tan = 0

= 0

That is

m₁– m₂ = 0

m₁ = m₂

Two distinct times are parallel if and only if their slopes are equal

Example

Find the equation of the straight line which passes through the point (5,8) and is perpendicular to the line 2y – 3x – 4 = 0

SOLUTION

Let the slope of the line required be m₂

From the equation of line 2y – 3x – 4 = 0

2y = 3x + 4

Since the line required is perpendicular to 2y – 3x – 4 = 0, m₂=

Using one – point form equation of a line

Hence the equation of the line required is

Example

Find the equation of the straight line which passes through (4, 5) and parallel to the line

5y –10x +7 = 0

Solution

Let the slope of the line required be m₂

From 5y = 10x + 7 = 0

5y = 10x – 7

Since the line required is parallel to 2x + 5y – 50 = 0, m₂ = 2

Using the one point form to find the equation which passes through the point (4,5)

y = y = (x – x₁)

y – 5 = 2 (x – 4)

y – 5 = 2x – 8

2x – y – 3 = 0

Hence the equation of the line required is 2x – y – 3 = 0

INTERSECTING LINE

To find the coordinates of the points of intersection of two lines

Let the two lines be

m₁x + C₁ = y – – – – – – – –(1)

m₂x + C₂ = y – – – – – – – –(2)

The lines y = m₁x + C₁ and y = m₂x + C₂

When m₂x + C₁ = m₂x + C₂, where m₁ m₂

m₂x + m₁x = C₁ – C₂

(m₂ + m₁)x = C₁ – C₂

The value of y can be gotten by substituting back the value of x into (i) or (ii)

It should be noted, that the value point of intersection can be obtained by solving simultaneously

Example

Find the point of intersection of the lines 2x + 5y – 4 = 0 and 3x – 2y + 2 = 0

Solution

2x + 5y – 4 = 0 – – – – – – – – (1)

3x – 2y + 2 = 0 – – – – – – – – (2)

Multiply (1) by 3

6x + 15y – 12 = 0 – – – – – (3)

Multiply (2) by 2

6x – 4y +4 = 0 – – – – – – – (4)

Subtract (4) from (3)

19y – 16 = 0

y =

Substitute y = into (1)

2x + 5 – 4 = 0

x = –

The coordinate of the point of intersection is

Example

Find the coordinate of the foot of the perpendicular from the point (3,4) on the line

2x + 3y – 4 = 0

Solution

First, we find the equation of the line which passing through (3,4) and is perpendicular to

2x + 3y – 4 = 0

3y = –2x +4

y =

m₁ =

The slope of the line which is perpendicular to 2x + 3y –4 is m₂ = (m₁m₂ = –1)

Using the one point. Form equation we can find the equation of this perpendicular that passes through (3,4)

y – y₁ = m (x – x₁)

y – 4 = (x – 3)

3x – 2y + 1 = 0

To find the point of intersection, we solve the two equation

2x + 3y – 4 = 0

3x – 2y + 1 = 0

We get x =

Thus the required foot of perpendicular is

LINE THROUGH THE INTERSECTION OF TWO GIVEN LINES

Consider the equations of the straight lines be

a₁x + b₁y + C₁ = 0 ––––––––––––––––––––––––––––(1)

a₂x + b₂y + C₂ = 0 ––––––––––––––––––––––––––––(2)

and let them meet the point (x₁y₁). Since this point lies both (i) and (ii)

Consider the equation (a₁x + b₁y + C₁) + (a₂x + b₂y + C₂) = 0 ––––––––––––––––(3)

Where is any constant

This is the equation of a straight lines since it is of the first degree in x and y. since point (x₁y₁) lies both (1) and (2), hence it must satisfy the equation (3). Thus it is the required line.

Example

Find the equation of the line draw through the point of intersection of the lines 3x + 2y – 1 = 0

And 5x + 1 = 0 which passes through the point (0,0)

Solution

Any line through the point of intersection is given by (3x+2y–1)+ (5x+6y+1)=0

If this passes through the point (0,0), its co–ordinates satisfy the equation and hence

(3(0) – 2(0)–1) + (5(0) + 6(0) + 6(0) +1) = 0

– 1 + = 0

= 1

Hence the required equation is

(3x + 2y – 1) + 1 (5x+6y+1) = 0

8x + 8y = 0

x + y = 0

Example

Find the equation of two lines through the point of intersection of the lines 3x + 2y – 1 = 0 and 2x – y + 7 = 0

(a) Perpendicular to 3x + 2y – 1 = 0

(b) Perpendicular to 2x – y + 7 = 0

Solution

Ay line through the point of intersection is given by (3x+2y–1)+s (2x–y+7)=0

(3+2 )x + (2 – ) y – 1 + 7 = 0

(2 – )y = –(3+2 )x + 1 – 7

Hence the slope is =

(a) Hence it is perpendicular to 3x + 2y – 1 = 0

Whose slope is

THE PERPENDICULAR FORM

· To find the equation of the straight line the perpendicular on which form the origin O makes an angle with x –axis and the length of the perpendicular is given as P.

P

O

E

D

C

FIG 15

P(x,y) is any point on the line and OB = P is the perpendicular on the is line and let is make and angle with the x –axis. Drop CE perpendicular to OX

Then < EDC = 90 –

<ECD =

OB = OD Cos ( OBD)

= (OE + ED) Cos

= (x + CE tan ) Cos

= x Cos + y sin

P = x cos + y sin

Example

Find the equation of the straight line whose perpendicular distance from the origin is of length 4 units and is at angle of 60 to the x–axis.

Solution

x FIG 16

Let the equation of the line be

P = x cos + y Sin

P = 4, = 60

4 = x Cos 60 + y Sin 60

8 = x +

Example

Find the equation of the straight line whose perpendicular distance from the origin O is of length 7 unit ad is at angle of 380 to the positive y – axis

38⁰

x Fig 1

Equation of the line

P = x Cos + y Sin

P = 90 – 38 = 52⁰

7 = x Cos 52 + y Sin 52

7= 0.6157x + 0.7880y

If the equation

ax + by + c = 0

x Cos + y Sin – p = 0 represent the same line

Then

LENGTH OF PERPENDICULAR

If the equation of line lien ax + by + c = 0 it may be rewritten

which is in he perpendicular from because

are the sine and cosine of an angle since the sum of the squares is unit

Then the length of the perpendicular is

If the denominator i.e is always positive than the length of perpendicular form any point of the positive side of the line will be positive, and from a negative side of the line, it will be negative.

EXAMPLE

Find the length of the perpendicular from the point P(2, 3) to the line 4x + 3y – 10 = 0 and state which side of the line P is on of the line P is on

Solution

The general form of the given equation is

4x+ 3y – 10 = 0

Hence, the length of the perpendicular is

The length of the perpendicular is 1.4 and it is on positive side of the line. Subustituting the coordinate (0,0) of the origin in the equation of the line 4x + 3x – 10 , we obtain – 2 . Thus the origin is on the negative of the line . Hence P in on opposite side ( positive side to the origin )

Example

Find the length of the perpendicular from the (–2,1) to the line x – y – 2 = 0 and state side of the line P is on

Solution

The general from of the given equation is

x – y – 2 = 0

Hence , the length of the perpendicular is

The length of the perpendicular is and it is on the negative side of the line substituting the coordinate (0,0) of the origin in the equation of the line x – y –2 = 0 , we obtain –2 , showing that origin is also on the negative side of the line . Therefore P is on the same side of line as the origin.