EQUATIONS OF STRAIGHT LINES(PERSAMAAN GARIS LURUS)

Standard Competence : 1. Understanding an algebraic form, relation, function, and linear equation

Basic Competence : . Understanding an algebraic form, relation, function, and linear equation

Standard Competence :

- 1.6 Determine the gradien, the equation, and the graph of a straight line.

After learning this topic, the students are expected to be able to :

Identify Linear Equation in various form and variables

Draw the graph of Linear Equation in cartessian coordinate

Determine the gradient of Linear Equation in various form

Determine the Linear Equation passing through two points

-

• Determine the Linear Equation passing through a point and gradient• Determine a point of itersection between two lines in various positions•And apply a linear equation concept in problem solving

The Straight Line

y

x

y

x

All straight lines have an equation of the form y mx c

m = gradient y axis intercept

C C

+ ve gradient

- ve gradient

1 1( , )A x y

2 2( , )B x y

Vertical HeightGradient =

Horizontal Distancem

2 1y y

2 1x x

2 1

2 1

y y

x x

Undefined and zero gradient

Gradient is a measure of slope. If a line has zero gradient it has zero slope.

A line with zero slope is horizontal.

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

Consider two points on this graph.

( 4,4) (6,4)2 1

2 1

y ym

x x

4 4

6 4

0

The equation of the line is 4y

All horizontal lines have an equation of the form y c

y

x

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

2

2

4

4

6

6

8

8

10

10

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

(4, 6)

(4,8)

Consider two points on this graph.

2 1

2 1

y ym

x x

8 6

4 4

(undefined)

The equation of the line is 4x

All vertical lines have an equation of the form x a

x

y

yx

From the diagram we can see that

tany

x

m

tanm

Note that is the angle the line makes with the positive direction of the x axis.

Collinearity

Two lines can either be:

A

B C

At an angle Parallel and Distinct

A

B

C

D

A

B

C

Parallel and form a straight line

Points that lie on the same straight line are said to be collinear.

To prove points are collinear:

1. Show that two pairs of points have the same gradient. (parallel)

2. If the pairs of points have a point in common they MUST be collinear.

1. Prove that the points P(-6 , -5), Q(0 , -3) and R(12 , 1) are collinear.

3 5

0 6pqm

2

6

1

3

1 3

12 0qrm

4

12

1

3

As the gradients of PQ and QR are equal PQ is parallel to QR.

Since Q is a point in common to PQ and QR, the points P, Q and R

are collinear.

Page 3 Exercise 1B

Perpendicular Lines

x

y

( , )a b

If we rotate the line through 900.

Perpendicular Lines

x

y

( , )a b

If we rotate the line through 900.

Perpendicular Lines

x

y

( , )a b

If we rotate the line through 900.

Perpendicular Lines

x

y

( , )a b

B( , )b a

O

A

This is true for all perpendicular lines.

If two lines with gradients m1 and m2 are perpendicular then

0

0OA

bm

a

b

a

0

0OB

am

b

a

b

1OA OB

b am m

a b

1 2 1m m Conversely, if then the lines with gradients m1 and m2

are perpendicular.

1 2 1m m

1. If P is the point (2,-3) and Q is the point (-1,6), find the gradient of the line perpendicular to PQ.

6 3

1 2PQm

9

3 3

To find the gradient of the line perpendicular to PQ we require thenegative reciprocal of –3.

Remember: 1a b

b a

3Since 3

1 1

The negative reciprocal would be 3

1The gradient of the line perpendicular to PQ is .

3

2. Triangle RST has coordinates R(1,2), S(3,7) and T(6,0). Show that the triangle is right angled at R.

S

T

R

7 2

3 1RSm

5

2

0 2

6 1RTm

2

5

5 21

2 5RS RTm m

Since 1, RS is perpendicular to RT.RS RTm m

Hence the triangle is right angled at R.

Equation of a Straight Line

All straight lines have an equation of the form y mx c

x

y

( , )P x y

A(0,C)

m

P(x,y) is any point on the line except A.

For every position P the gradient of AP is

0

y cm

x

y c

x

y c mx

y mx c

1. What is the equation of the line with gradient 2 passing through the point (0,-5)?

2 and 5m c

the equation of the line is 2 5y x

2. Find the gradient and the y intercept of the line with equation 4 3 2x y

Rearranging gives: 3 4 2y x 4 2

3 3y x

4 2Gradient is - and the axis intercept is .

3 3y

3. Show that the point (2,7) lies on the line 4 1.y x

When 2,x 4 2 1y 7

Because (2,7) satisfies the equation y = 4x – 1, the point must lie on the line.

General Equation of a Straight Line

0 is used as an alternative to Ax By C y mx c

0 is the GENERAL EQUATION of a straight line.Ax By C

1. Rearrange 2 5 into the form 0

and identify the values of A, B and C.

y x Ax By C

2 5 0x y 2, 1, 5A B C

42. Rearrange into the form 0

3and identify the values of A, B and C.

xy Ax By C

3 4y x 4, 3, 0A B C 4 3 0x y

3. Rearrange 7 into the form 0

and identify the values of A, B and C.

x Ax By C

7 0x 1, 0, 7A B C

Finding the equation of a Straight Line

To find the equation of a straight we need•A Gradient•A Point on the line

The equation of a straight line with gradient m passing through (a,b) is

( )y b m x a

x

y

( , )P x y

A(a,b)

m

P(x,y) is any point on the line except A.

For every position P the gradient of AP

y bm

x a

1

m y b

x a

( )y b m x a

11. Find the equation of the straight line passing through (5, 2) with gradient .

2

1(5, 2),

2P m

1( 2) ( 5)

2y x

1 52

2 2y x

1 9

2 2y x or 2 9y x

Equating with ( )y b m x a

2. Find the equation of the line passing through P(-2,0) and Q(1,6).

6 0

1 2PQm

2

Using point P

0 2( 2)y x

2 4y x

But what if we used point Q?

6 0

1 2PQm

2

Using point Q

6 2( 1)y x

2 4y x

6 2 2y x

Regardless of the point you use the equation of the straight line will ALWAYS be the same as both points lie on the line.

Lines in a Triangle

1. The Perpendicular Bisector.

A perpendicular bisector will bisect a line at 900 at the mid point.

The point of intersection is called the Circumcentre.

1. A is the point (1,3) and B is the point (5,-7). Find the equation of the perpendicular bisector of AB.

To find the equation of any straight line we need a point and a gradient.

Mid-point of AB = 1 5 3 ( 7)

,2 2

3, 2

7 3

5 1ABm

10 5

4 2

2

5m 1 2since 1m m

22 3

5y x

2 16

5 5y x

2. The Altitude.

An altitude of a triangle is a line from a vertex perpendicular to the opposite side. A triangle has 3 altitudes.

The point of intersection is called the Orthocentre.

3. The Median of a Triangle.

The median of a triangle is a line from a vertex to the mid point of the opposite side. A triangle has 3 medians.

The point of intersection is called the centroid

A further point of information regarding the centroid.

2

1

The centroid is a point of TRISECTION of the medians. It divides each median in the ratio 2:1.

1. F, G and H are the points (1,0), (-4,3) and (0,-1) respectively. FJ is a median of triangle FGH and HR is an altitude. Find the coordinates of the point of intersection D, of FJ and HR.

(Draw a sketch – It HELPS!!)

F

G

H

MEDIAN

J

4 0 3 1,

2 2J

2,1

1 0

2 1FJm

1

3

10 1

3y x

1 1

3 3y x

ALTITUDE

F

G

H

JR

H(0,-1)

3 0

4 1FGm

3

5

5

3HRm

51

3y x

51

3y x

The point D occurs when;

D

1 1 51

3 3 3x x

42

3x

2

3x

5 21

3 3y

1

9

2 1,

3 9D