Same direction opposite direction different direction

Parallel or carried

Parallel or carried

Not carried , not parallel

Scalars , vectors and directed line

segment

1] scalar quantity :

It is a quantity determined completely by a real number which

magnitude of this quantity.

For example: length , mass , time ,……

2] vector quantity :

It is a quantity determined completely by a real number which is

the magnitude and the direction

For example : force – displacement – velocity

# Distance : it is the length of the actual path covered during

movement from the position to anther and it is a scalar quantity

# Displacement: it is the smallest distance between the starting point

and the ending point and it is vector quantity.

#Directed line segment :

It is a line segment with : 1) starting point

2) ending point

3) direction from staring point to ending point

#the norm of directed line segment :it is the length of it

#Equivalent directed line segments if :

1) they have same direction.

2) they have same length.

Exercise

1] Complete :

1) to define scalar quantity you should know …….

2) to define vector quantity you should know …….

3) the directed line segment is a line segment which has ………. , …… , …….

4) two directed line segment are equivalent if they have ……

5) in the opposite figure :

ABCDEF is a regular hexagon , then

a) 𝐴𝐵 is equivalent to ……. And equivalent to …..

b) 𝑀𝐷 is equivalent to ……. And equivalent to …..

c) 𝑀𝐷 is equivalent to ……. And equivalent to …..

2] On the lattice , if : A(3,-2) , B(6,2) , C(1,3) , D(4,7)

a) Find : 𝐴𝐵 and 𝐶𝐷

b) prove that : 𝐴𝐵 equivalent to 𝐶𝐷

1

1

2

3

4

5

2

the position vector :

the position vector of a given point A with respect

to the origin point O is the directed line segment 𝑂𝐴 which its starting point is the origin point O and the

given point A is its ending point

the norm of the vector :

it is the length of the line segment that represent the vector.

If 𝐴 =(x,y) , then 𝐴 = the length of 𝑂𝐴

The length of 𝑂𝐴 = (𝑥 − 0)2 + (𝑦 − 0)2

Then 𝐴 = 𝑥2 + 𝑦2

The unit vector : it is a vector whose norm is unity.

A=( 3

5 ,

4

5 ) is unit vector because : 𝑨 = (

3

5)𝟐

+ (4

5 )

𝟐 = 1

Zero vector : it is a vector whose norm equals zero and denoted by 𝑂

=(0,0)

Cartesian form : 𝑨 = (x,y)

Polar form : 𝑨 = ( 𝑨 , 𝜽 )

Fundamental form : 𝑨 = x𝒊 + y𝒋

( 𝑨 , 𝜽 )polar form = ( 𝑨 cos 𝜽 , 𝑨 sin 𝜽 ) Cartesian form

(x,y) Cartesian form = ( 𝒙𝟐 + 𝒚𝟐 , tan-1 𝒚

𝒙 ) polar from

Vectors

( , )A x y

( , )A x y

For every non zero vectors 𝐴 = ( x1, y1 ) and 𝐵 =( x2 , y2 )

1) if A // B Then tan θ1 = tan θ2

And y1

x1 =

y2

x2

And x1y2 – x2y1 = 0

2) if A ⊥ B Then tan θ1 × tan θ2 = -1

And y1

x1 ×

y2

x2 = -1

And x1x2+y1y2 = 0

Parallel and perpendicular vector

Exercise

1] Complete :

1) if : 𝐴 = (4,5) and 𝐵 =(3,-2) then 2𝐴 + 𝐵 = …..

2) if : 𝐴 = 2i + 3j and 𝐵 = 3i – j , then 2𝐴 – 𝐵 = ….

3) if : (6,4) and (3,m) are two perpendicular vectors , then m = …..

4) if : (-2,1) and (-3,k) are two parallel vectors , then k = …..

5) if 3𝑘𝐴 = −15𝐴 , then k = …..

2] if 𝑀 = 2i + 3j , 𝑁 = -8i – 12j , 𝐿 = ai + 15j and 𝐹 = 6i + bj :

1) prove that : 𝑀 // 𝑁

2) Find : a ∈ ℝ if 𝑁 // 𝐿

3) Find the value of : 4 𝑀 + 𝑁

4) Find : b ∈ ℝ , 𝐹 ⊥ 𝑁

5) Is 𝐹 ⊥ 𝑀 ? explain your answer .

3] Convert :

1) ( 2 , -1 ) to Fundamental form

2) 3i + 5j to Cartesian form

3) ( 5 , 5 𝟑 ) to polar form

4) ( 12 , 1500 ) to fundamental form

5) (6 , 𝟓𝝅

𝟑 ) to Cartesian form

1

2

2

3

4

5

1

1

2

3

4

5

3

1

2

3

4

5

Subtracting two vectors geometrically

1] the triangle rule :

𝑨𝑩 + 𝑩𝑪 = 𝑨𝑪

2] the parallelogram rule :

𝑨𝑩 + 𝑨𝑫 = 𝑨𝑪

3] the median rule:

𝑨𝑩 + 𝑨𝑪 = 𝟐𝑨𝑫

𝑨𝑩 - 𝑨𝑪 = 𝑪𝑩

𝑨𝑩 = 𝑩 - 𝑨

Operation On Vectors

A B

C

A B

CD

A

BC D

Second

Adding vectors geometrically First

A B

C

Exercise

1] Complete :

a) if : 𝐴 = (-1,5) , 𝐵 = (2,1) ,then 𝐴𝐵 = ……

b) if : 𝐴 = (4,-2) , 𝐴𝐵 = (3,5) , then 𝐵 = …..

c) if : M is a midpoint of 𝑋𝑌 ,then 𝑋𝑀 + 𝑌𝑀 = ……

d) if : ABC is a triangle ,then 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴 = ….

e) if : ABC is a triangle ,then 𝐴𝐵 - 𝐶𝐵 = ……. , 𝐵𝐴 - 𝐵𝐶 = …….

2] ABCD is a trapezium in which in which 𝐴𝐷 //𝐵𝐶 , E is the midpoint of

𝐴𝐵

F is the midpoint of 𝐷𝐶 .

prove that : 𝐴𝐷 + 𝐵𝐶 = 2 𝐸𝐹

3] ABCD is a quadrilateral in which : 𝐵𝐶 = 3 𝐴𝐷 .prove that :

a) ABCD is a trapezium b) 𝐴𝐶 + 𝐵𝐷 = 4 𝐸𝐹

4] ABCDEF is regular hexagon prove that :

𝐴𝐵 + 𝐴𝐶 + 𝐴𝐸 + 𝐴𝐹 = 2 𝐴𝐷

A B

C

DE

F

1

1

2

3

4

5

2

3

4

Application On Vectors

Exercise

1] ABCD is a parallelogram ,E is a midpoint of AB

F is a midpoint of DC

Prove that : DEBF is a parallelogram

2] ABCD is a quadrilateral , if 𝐴𝐶 + 𝐵𝐷 = 2 𝐷𝐶 prove that :

ABCD is a parallelogram

3] using vectors prove that : A(3,4) , B(1,-1) , C(-4,-3) , D(-2,2)

are vertices of a rhombus

4] using vectors prove that : A(1,3) , B(6, 1) , C(4,-4) , D(-1,-2)

are vertices of a square and find its area.

5] ABCD is a trapezium , AD//BC

AD = 1

2 BC , 𝐴𝐵 = 𝑁 , 𝐴𝐷 = 𝑀

a) Express in term of 𝑀 and 𝑁 each of the following :

𝐵𝐶 , 𝐴𝐶 , 𝐷𝐶 , 𝐷𝐵

b)if : X ∈ 𝐴𝐶 where AX = 1

3 ×AC

prove that : the point D , X and B are collinear .

M

N

Geometry First

1

2

3

4

5

Physical application

1] Complete:

1) If: 𝐹1 = i - 3j , 𝐹2

= 3i – j act on a particle, then the norm of the

resultant = ….N

2) If: 𝐹1 =(a,b) , 𝐹2

= -3i + 4j act on a particle and the system is in equilibrium

,then a = …… , b = …..

3) If: 𝑉𝐴 = 12 𝑒 , 𝑉𝐵

= 8 𝑒 , then 𝑉 AB = ……

4) If : : 𝑉𝐴 = 120 𝑒 , 𝑉𝐵

= -80 𝑒 , then 𝑉 BA = …… , 𝑉 AB = ……

5) If: 𝑉 AB = 75 𝑒 , 𝑉𝐴 = -60 𝑒 , then 𝑉 BA = …… , 𝑉𝐵

= ……

2] Find the resultant force 𝐹 acting in each of the following:

Second

1 2

3 4

1

2

1

2

3

4

5

3] In each of the following, the two forces 𝐹1 and 𝐹2

act at a particle. Show the

magnitude and the direction of the resultant of each two forces:

4] Forces 𝐹1 = 7i – 5j , 𝐹2

= ai + 3j , 𝐹3 = -4i +(b-3)j ,find the values of a and b

if:

(1) The system of forces are in equilibrium.

(2) The resultant of the forces = -5j

1

2

3

4

3

4

Unit 2

Lesson (1)

If C 𝐴𝐵 , then C divides 𝐴𝐵

Internally , where 𝑚2

𝑚1> 0

Let A = (X1, y1), B (X2, y2) and

The ratio of division is m1: m2

X =m X m X

1 1 2 2m m

1 2

Y = m Y m Y

1 1 2 2m m

1 2

If C = (X,Y) is the midpoint of 𝐴𝐵 , then X = 2

2X

1X

,

Y =2

2y

1y

If H = (X1 , y1), B = (X2, y2) and C = (X3, y3) vertices of ∆ ABC,

M is the point of intersection of its medians

m = (3

3X

2X

1X

, 3

3y

2y

1y

)

y

y\

X\ X o

(X1y1)

(X , y)

(X2 , y2)

A

C

B

M2

M1

The division of a line segment

The internal division First

If C 𝐴𝐵 , C 𝐴𝐵 , then C divides 𝐴𝐵 , where 𝑚2

𝑚1< 0

Externally. Let A = (X1 , y1), B = (X2 , y2)

And C = (X , Y)

X = 2

m X m X11 2

m + m1 2

Y = m Y m Y

1 21 2m + m

1 2

If A = (– 3 , –7) and B = (4 , 0), find the coordinates of the

point C which divides 𝐴𝐵 by the ratio 5 : 2 internally.

Answer

C = (X , Y)

X =

m X m X1 21 2

m m1 2

= 2741

7026

52)4(53) (2

Y = m y m y

(2 7) (5 0) 141 1 2 2 2m m 72 5

1 2

C = (2 , –2)

y

y\

X\ X o

(X1 , y1)

(X , y)

(X2 , y2)

A

B

C M2

M1

𝐴𝐵 A 2

5 : 2 B 5

internally

A(-3,-7) B(4,0)

2 5

The External division Second

Ex 1

Ex2: If A=(6 , 3) and B = (–7, 2), find the coordinates of the

point C which divides 𝐴𝐵 by the ratio 3 : 2 externally.

solution

Let C = (X , y)

X =

m X m X1 21 2

m m1 2

= (-2) 6 3 7 ( 12) ( 21)

331( 2) 3

Y =2

m Y m Y3 2 ( 2) 31 1 2 0

m m 3 ( 2)1 2

Ex3: If A = (3 , -2) and B = (-1, 5) find:

1) The coordinates of the point C which divides 𝐴𝐵 with the

ratio 2:3 internally.

2) The coordinates of the point D which divides 𝐴𝐵 by the

ratio 4:3 externally.

………………………………………………………………………..

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

A(6,3) B(-7,2) m2

m1< 0

Put –ve sign

- 2 3 For any term

Of ratio

𝐴𝐵 A 2

3 : 2 B 3

Externally

Ex 2

Ex 3

Ex4: If A = (4 , 3) and B = (-3, 5) find C 𝐴𝐵 such that

3AC= 5CB

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

Ex5: If the point C = (4 , 4) divides 𝐴𝐵 by the ratio 1:2 internally

and A = (7 , 8) find the coordinates of B.

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

Ex6: Find the ratio by which the Y – axis divides the line

segment 𝐴𝐵 where A = (2 , 3), B = (-3 , 7) showing the

kind of division and find the point of division.

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

Ex 4

Ex 5

Ex 6

Ex7: If A = (4 , 12) , B = (-2 , 10), C = (1 , 3) , D = (2 , 7) E is

the midpoint of 𝐴𝐵 and M divides 𝐶𝐷 externally by the

ratio 3:2, find the length of 𝐸𝑀

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

Ex8: ABCD is parallelogram. If A= (7,-2), B= (15,4) and C= (9,6),

Find the coordinates of the point of intersection of its

diagonals 𝐴𝐶 and 𝐵𝐷 then find the coordinates of D.

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

Ex 8

Ex 7

Lesson (2)

1) If the straight line passing through two points

A = (X1, y1) and B = (X2 , y2)

m = 2 1

2 1

y y

X X

2) The slope of the straight line whose equations is ax+by+c=0 is ba

3) The slope of the straight line which makes the positive direction of the

X-axis a positive angle of measure θ is m = tan θ.

4) If 𝑈 = (x,y) directed vector ,so the slope will be 𝑦

𝑥

1)If L1 and L2 are two straight lines of slopes m1 and m2 respectively

then:

a) L1 // L 2 m1 = m2

b) L1 ⊥ L2 m1m2 = -1

2)The slope of any vertical straight line(parallel to Y– axis) is undefined.

3)The slope of any horizontal straight line (parallel to X – axis) = Zero.

4)If the slope of 𝐴𝐵 = the slope of 𝐵𝐶 , then the point A , B and C are

collinear.

The straight line

Notes:

The slope of the straight line:

1

2

3

4

1

2

3

4

1) [ vector form ]

(x,y) point slope

(𝑥, 𝑦) = (𝑎, 𝑏) + 𝑘(𝑙, 𝑚)

(𝑥, 𝑦) = (𝑎, 𝑏) + (𝑘𝑙, 𝑘𝑚)

(𝑥, 𝑦) = (𝑎 + 𝑘𝑙 , 𝑏 + 𝑘𝑚)

So , [ Parametric form ]

Then 𝑋 – 𝑎 = 𝐾𝑙 , 𝑌 – 𝑏 = 𝑘𝑚

𝑋−𝑎

𝑙 = k ,

𝑦−𝑏

𝑚 = k

𝑋−𝑎

𝑙 =

𝑦−𝑏

𝑚

So, [ General form ]

2)The equation of the straight line knowing the two points of its intersection with the two coordinate axes (a , 0) and (0 , b) is:

1

by

aX

3)The equation of the straight line parallel to the X – axis and passes through the point (0 , L) is Y = L.

4)The equation of the S.t line // to the Y – axis and passes through the point (K, 0) is X = K.

5)The equation of S.t line passing through the origin point O (0,0) is Y = mx.

𝑟 = 𝐴 + 𝑘𝑈

X = 𝑎 + 𝑘𝑙 , Y = 𝑏 + 𝑘𝑚

𝑦−𝑏

𝑥−𝑎 =

𝑚

𝑙 = slope

The equation of the straight line:

he slope of the straight line:

1

3

4

2

1) Passing through (1 , 3) and its slope = 32

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

2) Passing through the point (3 , -2) and its slope is -2

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

3) Passing through the two points (3 , 1) and (5 , 4)

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

4) Passing through the point (0 , -5) and makes with the

positive direction of X – axis an angle of measure 1350.

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

Find the equation of the S.t line:

1

2

3

4

5) Passing through the point (-2 , 1) and parallel to the straight

line

𝑟 = (2, −3) + 𝑘(1,0)

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

6) Passing through the point (3 , 5) and perpendicular to the

vector 𝑈 = (-1,2)

……………………………………………………………………….

…………………………………………………………………………

…………………………………………………………………………

7) If A = (2 ,3) , B = (6 , 6) and C = (10 , 1) are the vertices of

∆ABC. Find:

a) the equation of s.t line 𝐴𝐷 where D is midpoint of 𝐵𝐶 .

b) the equation of s.t line passing through the point D

perpendicular to 𝐵𝐶 .

…………………………………………………………………………

………………………………………………………………………

………………………………………………………………………

5

6

7

Lesson (3)

tan θ = ± (2

m1

m12

m1

m

)

m1 = tan θ1 , m2 = tan θ2

1) Find the measure of the angle between the two straight lines

whose slopes are: 32

,43

tan θ = ± 617

32

43

1

32

43

θ = 700 34

\ or θ = 109

0 26

\

2)Find the measure of the angle between each of the following pairs of straight lines:

a) L1: X – 2y + 1 = 0 L2: X + 3y + 2 = 0

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

The angle between two

straight lines

y

y\

X\ X

L2 L1

θ1 θ2

θ

1

2

b) L1: 4X – 3y = 1 L2: 8y + 6X = 0

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

c) L1: 12Y

3X

L2: y – 4X = 2

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

3) Find the measure of the angle between the straight line

3X – 4y – 5=0 and the straight line which passes through the two

points (2 ,1) and (1 , 8).

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

4) If the measure of the angle between the two straight lines

2X–y+5 =0 and kx – y = 0 is θ where tan θ = 43 , find k.

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

4

3

5) If the measure of the angle between the two straight lines

X-2y+8=0 and X+ky+5=0 is 450, find the value of k.

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

6) Find the measures of the angles of ∆ABC whose vertices are

A= (4,7), B = (-2 , -1) and C = (2 , -4).

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

7)Find the equation of the straight line that passes through the

point of intersection of the two straight lines: X + Y = 4 and

2X-3y+7=0 and perpendicular to S the straight line whose

slope is 2

3

.

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

…………………………………………………………………………

5

6

7

Lesson (4)

Assuming that we have a given

Straight line, its equation is:

ax + by + c = 0

P (X\ , y

\) is a given point in the

Cartesian plane of this straight line.

The length of the perpendicular

L = \ \|ax by |

2 2a b

c

Ex1: Find the length of the perpendicular drawn from the point

P to the straight line L if:

a) P (-2 , -4) , L: 4X – 3y – 12 = 0

Answer

L = | 8 | 8|4 ( 2) 3 ( 4) 12|

1.616 9 5 5

unit length.

On the length of the perpendicular

from a point to a line

P(x\ , y\)

Ex 1

Ex2: Find the length of the perpendicular drawn from the point

of intersection of the two straight lines X = 1 and Y – 2 = 0

to the straight line whose equation is 3X + 4Y – 25 = 0

.....................................................................................................

.....................................................................................................

.....................................................................................................

.....................................................................................................

.....................................................................................................

.....................................................................................................

Ex3: Calculate the radius length of the circle whose centre is M

(3,-1) and touches the straight line whose equation

L: 4X+3Y+6=0.

.....................................................................................................

.....................................................................................................

Ex4: If the length of the perpendicular drawn from the point

(7 , -1) to the straight line ax + y = 0 equals 2 10 length unit,

find the possible value of a.

.....................................................................................................

.....................................................................................................

.....................................................................................................

.....................................................................................................

Ex 2

Ex 3

Ex 4

Ex5: If the points A = (-4 , 1) , B = (2 , 3) and C = (-2, 6) are the

vertices of a triangle find:

1)The length of 𝐵𝐶

2) The equation of s.t line 𝐵𝐶

3)The length of the perpendicular drawn from A to 𝐵𝐶

4) The area of ∆ABC.

.....................................................................................................

.....................................................................................................

.....................................................................................................

Ex6: Prove that the two points (1 ,1) and (-2, 3) lies on two

different sides of the straight line. 2X – y + 3 = 0 and at

equal distances from it.

.....................................................................................................

.....................................................................................................

.....................................................................................................

.....................................................................................................

Ex 5

Ex 6

Lesson (6)

If the two straight lines L1:a1x+b1y+c1=0 and L2:a2x+b2y+c2=0

intersect at a point, then the general equation of all straight

lines which pass through the point of intersection of them is:

Ex1: Find the equation of the straight line which passes through

the point of intersection of the two straight lines X+Y =3 and

2X–y=6 and passes through the point (2 , -1).

Answer

a1= 1 , b1 = 1 , c1 = -3 , a2 = 2 , b2 = -1 , c2 = -6 then the G.E is X + Y – 3 + k (2X – y – 6) = 0 at (2 , -1) 2 – 1 – 3 + k (2X2 + 1 – 6) = 0 -4 – k = 0

K = -4

G.E is X + Y – 3 – 4 (2X – Y – 6) = 0 -7 X + 4Y + 21 = 0

The general equation of the straight

line passing through the point of

intersection of two given lines

a1x + b1y + c1 + k (a2x + b2y + c2) = 0

Ex 1

Ex2: Find the equation of the straight line passing through the

point of intersection of the two straight lines: X–3y+5=0 and

2X-Y-5=0 and parallel to Y – axis.

…………………………………………………………………………

…………………………………………………………………………

Ex3: Find the equation of the straight line passing through the

point of intersection of the two straight lines: 4X–3Y–11=0 ,

2X-2Y-6=0 and perpendicular to the straight line:

3X+2Y–9=0.

…………………………………………………………………………

…………………………………………………………………………

Ex4: Find the equation of the straight line passing through the

point of intersection of the two straight lines X + Y = 4

and X – Y = 2 and length of the perpendicular drawn to it

from the origin point = 1 length unit.

…………………………………………………………………………

…………………………………………………………………………

Ex5: Find the equation of the S.t line passing through the point

of intersection of the two straight lines 2X+Y–1=0 and

X–Y+3=0 and cuts from the negative direction of Y – axis

a part of length 3 length unit.

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Ex 2

Ex 3

Ex 4

Ex 5