**MATHEMATICS FORM ONE TOPIC 4- COORDINATE GEOMETRY(1)**

**UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU: 0787237719**

**ALSO READ**;

**MATHEMATICS FORM ONE TOPIC 4- COORDINATE GEOMETRY(1)**

**COORDINATE GEOMETRY**

**Coordinates of a point**

Solutions

The coordinates of the points are

A = (0,5)

B = (5,0)

C = (0,4)

D = (-5,5)

Exercise 10.1

**SLOP/GRADIENT OF A LINE**

Slope / gradient is the change in the vertical axis to the change in the horizontal axis.

(a) A (2, 4) and B (-2, 6)

=

m=

(b) (5,8), (4,1)

(c) (1,5), (4,7)

(d) (2,6), (5,3)

(e) (1,6), (3,-1)

(f) (3,6), (-2,-1)

(g) (0,2), (6,2)

(h) (2,3), (-1,-3)

(i) (2,10), (2,0)

(j) (

(k) (-2,1), (4,3)

(l) (-4,4),(-3,3)

(m) (0,0), (-3,4)

(n) (99,6), (119,1)

(o) (0.64,-1.62, (1.36,-0.62))

we can find the equation of the line having any point on a line say (x,y) and any point,

then from

let (x1,y1) = (x,y), and (x2,y2) = (5,0)

20-4x = -3y

∴y =

The point on the line (x,y) is called arbitrary point

let (x,y)=(12,-6),

(x2-y2)= (2,6)

5(y-6)=-6(x-2)

5y-30=-6x+12

5y=-6+12+30

5y=-6x+42

(2) Give that y = –

The gradient is

Example: find the equation of the line passing through the point (4, 6) and having a slope -1/2

i.

(x,y) = (1,4)

Y=

then -3 + 3x = 8 – 2y

(ii) Y – intercept

x =

The coordinate of X-intercept is (22/3,0)

(iii) If ax + by = 12 goes through points (1,-2) and (4, 1) find the value a and b-

solution

let the two collinear point be (x,y),(4,1) and gradient 1

then from

x-4 = y-1

x-y=–1+4

x-y=3

if the equation is multiple by 4 both side we have

4(x-y)=4×3

4x-4y=12

compare the equations.

x-y=3 and ax+by=12

ax+by=4(3)

x-y=3

then

a=1

b=-4

The value of a=1 and b=-4

EXERCISE 10.3

d) (-2, -4) with gradient

f) (-3 , -3) and y- intercept

h) (-1 , -1 ) and y – intercept –

(v) 8x – (

(vii)

**SIMULTANEOUS EQUATIONS**

(ii) Substitution method.

(iii) Graphical Method

here we can now omit x by subtracting.

+

13x=24

3x=6

6x+y=15

7x= 21

=

y= 4

-7y = -28

y= 4

x=3Therefore: x=3 and y = 4

=

5y=10

x = 3

Let find the value of x by taken one equation

8x + 5y = 9

8x = -16

x = -2

17x = 34

2(2) – 3y =7

-3y=7-4, -3y=3

=

y=-1

Therefore: x = 2, y = -1

2x + 3y = 8

6y=6

2x = 8-3

x=5

–

2x = 20

=

= –

5x= 15

Therefore :

-4x=-10

=

x = 2

**MATHEMATICS FORM ONE TOPIC 4- COORDINATE GEOMETRY(1)**

**method.**

3x + y = 9……….. (ii)

from equation 1

6x+y =15

y=15-6x……………..(iii)

-3x = 9 – 15

-3x=-6

x=2

From (i)

from (i)

2y=8-3x

2y=8

9x + 5y = 5

2. Solve the following simultaneous equations by using substitution method(i). 3x – 2y =

2x + y = 8

8x – 2y = 16

8x – 2y = 16

X + 2y = 2

2a – 4b = 2

4x + 3y =

5x- 2y = 19

x – y = 16. Solve the following by any method

**equations**.

x+ y = 60 …………(ii)

then

x + 4x = 60

5x=60

y = 4 (12) = 48

2x=14

**fraction**is decreased by 1 its value become2/3 but if it denominator is increased by 5 its value becomes ½ , what the fraction?

3a – 3 =2b

3a – 2b=3………………(iii)

from (ii)

**times**the original number

10y + x + 3 = 40x + 4y

=

= +

15x = 15

The original number is 16

Exercise 10.5

1. The sum of two number is 109 and the difference of the same numbers 29. find the numbers

2. Two number are such that the first number plus three times the second number is 1. And the first minus three times the second is 1/7. Find the two numbers

3. The sum of the number of boys and girls in a class is 36. If twice the number of girls exceeds the number of boys by 12, find the number of girls and that of boys in the class.

**length**and the width is one centimeter find the dimensions of the rectangle.

EXERCISE 10.1

(e) Octagon

EXERCISE 10.2

1. a) (0, 3), (2, 5)

let (x1,y1) be (0,3)

(x2,y2) be (2,5)

M = 1 it is positive gradient.

(c) (1, 5), (4, 7)