AS level Math P1 Coordinate Geometry notes, Lecture notes of Mathematics

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Coordinate

Geometry

m

Coordinate geometry

• >

Equation of^ a straight line^ Y^ =^ ma^ +

• > (^) Distance

y-y.^ =^ m(x-^ x,)

• > (^) Gradients
• > (^) Points of (^) intersections
• > Equation of^ a^ circle

Y

1 · Midpoint of AB

· : I = (^) (- 1. (^5) ,^ 2)

↓ ·^ Gradient^ of^ line^ AB:

: Y2-Y-, - > Changein^ y

· Distance of AB : x2- x, Change ina #c - xi) +^ (Y2. Yil^ I^ -^1 -^3

• 4 - 1 : (^) - 1) + (^) (1-3)2 = (^) =
• 5 : 5 +^4

• > Y x = 9 = 5. (^3852) = 0. 4 · 5. 39(3 s. (^) f) ~ for every

1 unit increase in se,

y increases^ by 0.^4 units

0 : undefined^

8

: undefined

• >

Equation of^ a^ straight line:

& (^) Gradient-Intercept form (^) : y = (^) mx + c · (^) find the (^) equation of (^) a straight line^ that (^) passes

through the^ point^ A^ (2,^ 3)^ and

has gradient 4

y

= (^) mx + C

3 =^ -^8 +^ C

C =^1

so y

= -^ 4x +^11

· Every point^ on^ this^ line^ should^ satisfy this^ equation. · (^) find the (^) equation of (^) a straight line^ that (^) passes

through the^ point^ A^ (2,^ 3)^

and

B (^) (- (^5) , 1)

y

= (^) mx + (^2) 3 =^ T 2()

• (^) c = 1 - 3

• 5 - 2 : (^) = 2 21 =^4 +^ 7) &

7 =^17

c = 1 y = zx

• 1 7 & (^) Gradient-point form : y

• y,^ =^ m(x^ - xi) · (^) find the (^) equation of (^) a straight line^ that

passes through the^ point^ A^ (2,^ 3)^ and

has gradient 4 y

• 3 : (^) - 4(2- 2) - (can (^) be the final (^) answer if not (^) specified, y
• 3 = - 4x+ (^8) no need to simplify) y

= - 4x + 11

· (^) find the

equation of^ a straight line^ that^ passes through the^ point A^ (2,^ 3)^ and

B (^) (- (^5) , 1)

y - Y.^ =^ m(x^ -^ x)^ -yY

y -^3 =^ E(x^ -^ 2)^ -^ N^

• (^2) ,

y -^3 =^ =x^ -^7 = 1 -^3

y :^ zx^ +^7

• 5 - 2 = (^) - 2 &
• > Solving simultaneous (^) equations: Elimination : x +

2y

= (^) I^ -^ 2x^ + by =^25 - ② 2x+ 4y = 1 -^0 x +

2y

= · n^

• 1. (^) 9) =^ ) -^ >^ 1- 0. (^8) , 3. 9) 10 y = 39 X^ =^ -0^.^8 This^ is^ the^ point^ of^ intersection^ of^ the^2 lines y

= 3.^9

Substitution :

x + 2y = &

• 2x + by =^25

y

= = ,^ y^ :x =

25 +^2

6 42 - 6x = 50 +^42

• 10x = 8 x =^8 10

x =^ - 0. 8

• > finding angle^ between^ two (^) straight lines : · Angle blu^ line^ Ex-axis^.

tanc-E],

Q. (^) (22, (^) y2) - m 1 =^2 x2 - X,

2 Yz- Y,^ tan0^ =^ m

(2 1.%1) (^) * & &^ -v (^) x-axis angle between a line^ and x-axis

[ 22 - 2 .& tan 0 = 2 180-63. 43

0 =^63. 43° :^116.^57 :

Q.^ m^ =^0.^5 tan Q^ =^ m tan Q =^ -0. 5

& S 0 = - 26. 56

0 = (^26). (^568) · Angle blw^ any two^ lines:

tand =^1. 5 ,^ tan^ Q^ =^ -0.^2

Q. (^) m = 0. 2 m^ =^1.^5 0 = 56. 3) 0 =^11. 310

1. 31 +^11. (^31) or 180-67. (^6)

14.^310 / (^1563) , (^) = 67. Jo : (^112). 40 · General formula for

any 2 lines:

tan (^) (A-B) =^ tan A-tan B & (^) tand (^) = M

• I^ +^ tanA^ tan^ B 1 +^ m^ , mc

Ang

& le between any two lines.

Q. m = 1. 5 tanD^ = - - mz

m =-^0.^2 1 + m.mz

tand : E. (^5) -

1 +^ 1.0.^ 2)(1^. 5)

tand :

0 =^ - 67.^6 -^ >^67. 60

• > Finding areas^ using coordinates: · DABC A (^) ( , 5)^ B(2^ ,^ -3)^ , 2(-4,^ 2) Is
• (^5) A (^) (2, 5) Area^ =^ 2 x bxh : x8x

= 24 units^2

C (. 4 .2) -^2 > 2

• • 3 B(2, -3) · SPQR p (2. 5)

, Q13.^ 5), R16,^ 11)

1 Area^ =^ 2 x bxh

• (^) I^ · R(6, 11) : 2 + (^) 5x

n = G^ =^15 units

Q. b = (^5) p(2, 5) (- (^3) , 5)

• 3!^ >^2

• > Properties of^ diagonals of^ quadrilaterals: Equal Meet^ at^ midpoint^ Perpendicular · Square:

A B

• ~ (^) ~ ~

B C

· Rhombus: /

-^ -^ -^ X^ ~ ~ I ·

Rectangle:

~ (^) ~ X · Parallelogram: > (^) X ~ (^) X

3 · Kite: -1 X X ~

Y i · Trapezium :

X X X

Equation of a circle · y = - 2x +^5 · Circle:^ (centre on (^) origin) Y r^ =^5 1 Y 1

x2 + yz =^52

• (^5 7) ⑳ (x , (^) y) x^ + yz =^25 5 - Y > 1 0 >^ N · Centre not (^) on origin: Y General form for (^) equation of (^) a circle:

· (x^ , y) (x -a)^ + (y- b) =^ r

get ((a, b)o- - -^! centre : (a, b) , radius : r

G (^) an >^2 (^1) n-a · find equation of^ a^ circle^ with^ radius^ <^ units^ and^ centre^ at^17 ,^ 9) ↓ (^) * (x-^ 7) 2^ +^ (y-^ 9) 2^ =^42

(x - 7) + ly- 9) =^ 16 r

· find equation of^ a^ circle^ with^ radius [ units and centre^ at^ (3, 5)

(2+^ 3)2^ +^ (y-^ 6)^

72 (x + 3) + (y-6) =^ 99r^ lone form of (^) answer x2 +^ 6x + 9 + yz- 12y + 36 - 49 =^0

22 + yz + 6x - 12y - 4 = 0 - /expanded form)

• > Angle properties^ of (^) circles:

b)

c) a > b>^900 only from^ diameter = (^900) I ·^ B F

a PQ^ =^ PR 2x · Perpendicular bisector^ of^ every chord^ in^ a^ circle^ passes (^) through the^ centre^ of^ the circle.