Presentation Math Coordinate Geometry

1) Nasirah Binti Che Daud D20081032296 2) Wan Masturah Binti Wan Mad Mohtar D20081032356 3) Ayuni Amalina Binti Mukhtar D20081032369 4) Nur Ain Bt Ahmad Fikri D20081032311  5) Nur Syazwani Bt Wan Aziz

Coordinate Plane • A basic concept for coordinate geometry. • It describes a two-dimensional plane in terms of two perpendicular axes: x and y. • The x-axis-horizontal direction • the y-axis-vertical direction of the plane.

y-axis

X-axis

• Points are indicated by their positions along the x and y-axes in the form (a,b) • L coordinates is (–3, 1.5)

Equation Of A Line • An equation of a line can be written y = mx + b where m is the slope b is the y-intercept

• Slant of a line is called the slope/gradient. • Slope is the ratio of the change in the yvalue over the change in the x-value. Slopes = Change in y value Change in x value

- The rate at which line rises (or falls) vertically for every unit across to the right. Gradient line y

) :P(x1,y1) ,Q (x2,y2 2 1

y -y m= x2 - x1

Q(x2,y2) m (y2-y1) P(x1, y1)

(x2-x1) 0

x

Where..

x 2 ≠ x1

y

Gradient line of PQ : Q(6,5)

mPQ

5 −2 = 6 −2

θ P(2,2) x

0

If θ < 90, m is positive

y2 - y1 = x2 - x1

∴ mPQ

3 = 4

Example 1: • Given two points, P = (0, –1) and Q = (4,1), on the line we can calculate the slope of the line. • Slopes= Change in y value Change in x value = 1-(-1) 4-0 =2 4 =1 2

Gradient line of MN :

y

mMN

M (2,7)

5 = −3

θ N (5,2) x

0

If θ > 90, m is negative

7 −2 = 2 −5

5 ∴ mMN = − 3

Example 2: • Consider the two points, R(–2, 3) and S(0, –1) on the line. What would be the slope of the line? Slopes = Change in y value Change in x value = -1-3 0-(-2) = -4 2 = -2

y

P

0

m=0 Q

X

-If the line PQ is parallel with the x-axis, θ = 0° OR θ = 180° ….Hence, m = 0

How about parallel & perpendicular line ?? What are their gradient ??.... 1.) Two lines are parallel if and only if both have the same gradient ; m1=m2 2.) Two lines with m1 and m2 gradient perpendicular if and only if m2m1= -1

Slopes Of Parallel Lines

y

Q(6,5)

P(2,2)

R (4,2) x

0

Gradient line of PQ :

y2 - y1 mPQ = x2 - x1

5 −2 = 6 −2

∴mPQ

3 = 4

Gradient line of RS : y2 - y1 mRS = x2 - x1 5 −2 = 8 −4

∴mRS ∴ mPQ // mRS

S (8,5)

3 = 4

Question… 1.) Does the straight-line AB and CD below parallel??....

A(1,9) , B(5,8) , C(5,2) ,D(1,3) y 2 − y1 mAB = x 2 − x1 8 −9 = 5 −1 =−

∴

1 4

y 2 − y1 mCD = x 2 − x1 3−2 = 1−5 =−

mAB = mCD…….hence, AB // CD

1 4

Slopes Of Perpendicular Lines • Two lines are perpendicular if the product of their slopes (m) is –1 • The line y= ½ x-1 perpendicular to y= -2x-1 Because: ½ x (-2) =-1

y

P(2,5)

R (9,5)

Q (6,3) x

0

Gradient line of PQ :

Gradient line of RS : y2 - y1 mQR = x2 - x1 5 −3 = 9 −6

y2 - y1 mPQ = x2 - x1 5−2 = 2−4

∴mPQ

3 = −2

∴mRS ∴ mPQ

mRS

2 = 3

Y-intercept

• The y-intercept is where the line intercepts (meets) the y-axis.

• The midpoint of a segment divides the segment into two segments of equal length. • The midpoint between the two points (x1,y1) and (x2,y2) is

 x1 + x 2 y 1 + y 2  ,   2   2

Example: • The midpoint of the points A(1,4) and B(5,6) is

 1 + 5 4 + 6   6 10  ,   =  ,  = (3,5) 2  2 2   2

Distance Formula • The distance between the two points (x1,y1) and (x2,y2) is

Dividing Point with Ratio Formula for inside point :

mx 2 + nx1 x= m+n

my2 + ny1 y= m+ n

Inside dividing ratio A( x1, y1)

P ( x, y ) B ( x 2, y 2 )

Given points A(1,2) and B(19,23). If P (x,y) dividing inside AB with ratio 1:2, find the value of x and y.

y n

B(19,23)

m A(1, 2) A(1,2 )

P(x,y )

x

Solution: y= my2 + ny1 m+ n

(1)(23) + (2)(2) = 1+ 2 27 = 3 =

9

x = mx 2 + nx1 m+n

(1)(19) + (2)(1) 19 + 2 = = 1+ 2 3 21 = 3

=7

P (7,9)

Outside dividing ratio y

P ( x, y ) B ( x 2, y 2 )

A( x1, y1)

x

Outside dividing point Formula : mx 2 − nx1 x= m−n

my 2 − ny 1 y= m −n

Given that points A(-5,-6) and B(1,-2). Get the coordinates that dividing outside AB with the ratio of 5:3 x1 = −5 x 2 = −1

y1 = −6 y 2 = −2

m =5 n =3

mx 2 −nx 1 my 2 −ny 1 ) P =( , m −n m −n  5( −1) − 3( −5) 5( −2) − 3( −6)  P = ,  5 −3 5 −3    − 5 + 15 − 10 + 18  P= ,  2  2 

= (5,4)

STRAIGHT-LINE EQUATION Equation Type

EQUATION TYPE • Gradient Type • Interception Type • General Type

GRADIENT TYPE y

y = mx + c P(x,y)

m = gradient

c

c = y-interception x

INTERCEPTION TYPE x

x+y =1 a b

P(0,b)

Q(a,0)

a = x-interception b = y-interception

y

GENERAL TYPE ax + by + c = 0

a, b, c are constant

METHOD TO FIND STRAIGHT-LINE EQUATION •Give Two Points or •One Point, One Gradient

GIVEN TWO POINTS First, find gradient (m): m = y2 – y1 x2 – x1

Q(3,4)

∴ m = 4-(-4) = 8 3-1 2 m=4 Then, find c Substitute q(3,4) in equation y = mx + c (4) = (4)(3) + c

P(1,-4)

4 = 12 + c C= -8

y = 4x - 8

∴ the equation y = 4x-8

ONE POINT, ONE GRADIENT Substitute m and point p(2,8) In equation y = mx + c m=3

(8) = (3)(2) + C 8 =6+C Hence c = 2

P(2,8)

So the equation= y = 3x + 2

y = 3x + 2

CHANGING EQUATION TYPE

y = 2x + 10 Change this equation into General Type y = 2x + 10 ∴ y – 2x – 10 = 0 or 2x – y + 10 = 0

Then, change 2x – y + 10 = 0 into Interception type 2x – y + 10 = 0 2x – y = -10 2x – y = -10 -10 -10

-10

2x + y = 1 10 10

or

x + y=1 5

10

Brain-Storming Corner

Given a line with two given point. Find the equation of the straight line y (2, 7)

(-3, 2) x

Gradient ; 7– 2 =5 2 – (-3)

5

=1 Find c by substitute (2,7) in equation y = mx + c 7 = (1)(2) + c ∴c = 5 Equation y = mx + c

∴ y=x+5

Change the equation into general type and interception type y=x+5 ∴ y – x – 5 = 0 or x – y + 5 = 0

Then, change x – y + 5 = 0 into Interception type x–y+5=0 x – y = -5 x – y = -5 -5 -5

-5

x+ y=1 5

5

Subtopics…. • The nearest point to the straight line • The distance between two straight lines that parallel to each other • Intersection of the straight line • Area of rectangle • Area of triangle

The nearest point to straight line y Perpendicular distance :

P

-The shortest distance between them / the length of a perpendicular line segment from the line to the point

ax + by +c = 0

d

x

(h,k) Q

The nearest point to straight line The nearest distance from point (h,k) to straight line ax +by +c =0 is ;

P

ax + by +c = 0 d

d=

ah + bk + c a +b 2

2

(h,k) Q

if

ah + bk + c a2 + b2

<0

Hence, the point is on the other side.

•

Example Find the point distance and location of:(2,1) and (-3,2) towards straight line 2y-3x-1=0 Solution : From straight line 2y-3x-1= 0 , a = -3 b = 2 c = -1

Point (2,1), hence h=2, k=1

Point (-3,2), hence, h=-3, k=2

d1

=

− 3(2) + 2(1) −1 (−3 ) + (2) 2

−5 = 13

2

d2 =

− 3(−3) + 2(2) − 1 (−3) 2 + (2) 2

12 = 13

d1

−5 = 13

d2

12 = 13

y

2y-3x-1=0

(-3,2)

(2,1) d2

d1 0

x

The distance between 2 straight-lines that parallel to each other L1 L2

METHO D 1) Find the coordinate on one of the line 2) Find the point perpendicular distance from the other line

Example : •

Find the distance between the parallel lines 5x+12y+1=0 and 5x+12y+8=0 The distance 

Solution: Take 5x+12y +1=0

1 When x=0, y= − 12 1 Coordinate is (0, − ) 12

1  to line 5x+12y+8=0 is :  0,−   12 

a= 5

b= 12 c=8

d=

h= 0

1 − k= 12

ah +bk +c a 2 +b 2

 1  0 +12  −  + 8 7 12   d= = 13 25 +144

Intersection of straightline The coordinate for two straight lines intersection can be found by solving both equation stimultaneously Solution :

y

2x-3y=6 …………(1)

P

4x+y =16 …………(2)

S 2x-3y=6

(1) x 2 4x+y =16 …………(2) 4x-6y=12 …………(3) (2) - (3) 7y= 4

4x+y=16 0

x R

Q

4 ∴y = 7

27 ∴x= 7

Choose A Quizzes SELECT THE DIFFICULTY LEVEL

EASY EXPERT

y A

QUESTION 1

5

QUESTION 2

4

B3

QUESTION 3 QUESTION 4 QUESTION 5

C

D

E F

QUESTION 6

y A

QUESTION 1

5 4

B3

3 UNITS LEFT, 5 UNITS UP

C

D QUESTION 2 QUESTION 3

E

QUESTION 4 F

QUESTION 5 QUESTION 6

y A

QUESTION 1

5

QUESTION 2

4

B3

2 UNITS RIGHT, C

D

4 UNITS DOWN

QUESTION 3 E

QUESTION 4 F

QUESTION 5 QUESTION 6

y A

QUESTION 1 5

QUESTION 2

4

B3

QUESTION 3

5 UNITS RIGHT, C

D

0 UNITS UP/DOWN E

QUESTION 4 F

QUESTION 5 QUESTION 6

y A

QUESTION 1 5

QUESTION 2

4

B3

QUESTION 3 QUESTION 4

C

D

5 UNITS LEFT, 3 UNITS DOWN

E F

QUESTION 5 QUESTION 6

y A

QUESTION 1

5

QUESTION 2

4

B3

QUESTION 3 QUESTION 4 QUESTION 5

C

D

0 UNITS LEFT/RIGHT 4 UNITS UP

E F

QUESTION 6

y A

QUESTION 1

5

QUESTION 2

4

B3

QUESTION 3 QUESTION 4 QUESTION 5

C

D

QUESTION 6

0 UNITS RIGHT/LEFT

E F

0 UNITS UP/DOWN

Positive slope

For example: Given two points, P = (0, –1) and Q = (4,1), on the line we can calculate the slope of the line.

• y-intercept -1

2

• Gradient 1/2

4

• Equation Y = 2x + 4 Y = 1/2x - 1

Q1 Q2 Q3 Q4 Q5 Q6

Negative slope

For example: Consider the two points, R(–2, 3) and S(0, –1) on the line. What would be the slope of the line?

• y-intercept -1

-2

• Gradient -2

3

• Equation Y = -2x - 1 Y = 3x - 2

Q1 Q2 Q3 Q4 Q5 Q6

Slopes Of Parallel Lines • In coordinate geometry, two lines are parallel if their slopes (m) are equal. • For example: The line y=1/2x+1 is parallel to the line y=1/2x-1. Their slopes are both the same. Q1 Q2 Q3 Q4 Q5 Q6

Slopes Of Perpendicular Lines • In the coordinate plane, two lines are perpendicular if the product of their slopes (m) is –1. • For example: The line Y=1/2X-1 is perpendicular to the line y = –2x– 1. The product of the two slopes is 1/2 x (-2) = -1

Q1 Q2 Q3 Q4 Q5 Q6

Mid Point Formula • To find a point that is halfway between two given points, get the average of the xvalues and the average of the y-values. • The midpoint between the two points (x1,y1) and (x2,y2) is:    

• For example: The midpoint of the points A(1,4) and B(5,6) is

Q1 Q2 Q3 Q4 Q5 Q6

Distance Formula • For example: To find the distance between A(1,1) and B(3,4), we form a right angled triangle with AB as the hypotenuse. The length of AC = 3 – 1 = 2. The length of BC = 4 – 1 = 3. Applying Pythagorean Theorem: •

AB2 = 22 + 32 AB2 = 13 AB = /13

Q1 Q2 Q3 Q4 Q5 Q6

REALLY QUIT? YES

NO

THAT ALL FROM US THANK YOU