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3rd prep

1st term

Analytical geometry

Analytical geometry
➔ Distance between two points
Points : A ( x1 , y1 ) and B ( x2 , y2 )
AB = √(𝑥1 −

𝑥2)2

+ (𝑦1 −

B

𝑦2)2
A

Important to know
1 three points A , B , C are collinear (lie on one straight line)
if : AB ( the greatest distance) = AC + BC

2 three points A , B , C are vertices of triangle
if : AB ( the greatest length ) < AC + BC

3 four points A , B , C ,D vertices of parallelogram
if : AB = CD , BC = AD

4 four points A , B , C ,D vertices of rhombus
if

AB = BC = CD = AD

5 four points A , B , C ,D vertices of rectangle
if

AB = CD , BC = AD , AC = BD

6 four points A , B , C ,D vertices of square
if

AB = BC = CD = AD , AC = BD

7 M is a point which is center of a circle and A , B , C are points
lie on the same circle if : MA = MB = MC = r ( radius )

Notes
1 In triangle ABC :
▪ (𝐴𝐶 )2 > ( 𝐴𝐵 )2 + ( 𝐵𝐶 )2 , then triangle optuse at B
▪ ( 𝐴𝐶 )2 = ( 𝐴𝐵)2 + ( 𝐵𝐶 )2 , then triangle right at B
▪ ( 𝐴𝐶 )2 < (𝐴𝐵)2 + ( 𝐵𝐶 )2 , then triangle acute at B

2 Any figure with 4 points is called quadrilateral
3 In circle : area = π 𝑟 2 , circumference = 2 π r

Math πrates

Habiba Attar

3rd prep

1st term

Analytical geometry

4 In isosceles triangle :
▪ CD is axis of symmetry perpendicular on AB
▪ CA = CB

➔ Midpoint of a line segement
A ( x1 , y1 ) , B ( x2 , y2 ) has mid point M :
M=(

𝑥1+𝑥2
2

,

𝑦1+𝑦2
2

)

➔ Slope of a straight line
Straight line “L” has two points A ( x1 , y1 ) , B ( x2 , y2 ) and make angle θ positive
direction of x-axis :

Slope =

𝑦2−𝑦1
𝑥2−𝑥1

= tan (θ)

Important to know

θ

1 if θ is :
• Acute angle → slope is positive
• Obtuse angle → slope is negative
• = zero → L is parallel to x-axis → slope = zero
• = 90 ° → L is parallel to y-axis → slope is undefind → x2-x1=zero

2 Two straight lines L1 ( slope m1 , θ1 ) and L2 (slope m2 , θ2 ) :
• L1 parallel L2 → m1 = m2 →θ1 = θ2
• L1 perpendicular L2 → m1 X m2 = -1

To Prove using slopes ( as for parallel m1= m2 , for perpendicular m1 X m2=-1)
1 trapezium ABCD → AB parallel CD , AD not parallel BC
2 Parallelogram ABCD → AB parallel CD , AD parallel BC and AB = CD , AC = BC
3 rectangle ABCD → ABCD is parallelogram and AB prependicular BC ,
CD perpendicular AD , AC = BD

Math πrates

Habiba Attar

3rd prep

1st term

Analytical geometry

4 rhombus ABCD → ABCD is parallelogram and AB = BC = CD = AD , AC perpendicular BD
5 square ABCD → ABCD is rhombus and AB perpendicular BC,CD perpendicular AD

➔ Equation of a straight line
1st aX + bY + c = 0 is the first form which
has slope “m”
−𝑎
• m= 𝑏
• intercepted part from y-axis =

−𝑐
𝑏

2nd Y = m X + c is the second form which
has slope “m”
• intercepted part from y-axis = c

Important to know
• to find intercepted from y-axis put
x = 0 ( at y-axis x = 0 )
• to find intercepted from x-axis put y=0
(at x-axis y=0)
• Y=mX → the straight line passes through origin (0,0)
• Equation of x-axis → y=0
• Equation of y-asis →x=0

By Habiba Attar
Water mark by Sara Hatem
Mathpirates team are Habiba Hatem , Habiba Ghanem , Joliana Remon

Math πrates

Habiba Attar



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