# THE STRAIGHT LINE AND MID-POINT OF A LINE SEGMENT

**FURTHER MATHEMATICS SS 1 SECOND TERM**

**WEEK 8**

** CLASS: SSS 1 **

**2 ^{ND} TERM**

**SUBJECT: Additional Maths **

**TOPIC: THE STRAIGHT LINE **

**CONTENT:** (I) mid- level of a section (ii) Gradient of a straight line (iii) Distances between two factors (iv) Situations for parallelism and perpendicularity (v) angles between two intersecting traces.

**SUB-TOPIC: MID-POINT OF A LINE SEGMENT **

x-x_{1}

x_{2}-x_{1}

Q(x_{2},y_{2})

x

P(x_{1},y_{1})

R(x,y)

O

y

S

y_{2}-y

y-y_{1}

Within the Cartesian aircraft above, let be the mid-point of the road section with the coordinate and .

As triangles are related;

Since R is the mid-point,

:.

Equally,

Therefore, the co-ordinates of the mid-point of the road becoming a member of and are:

Instance 1:- Discover the mid-point ‘R’ of the road section the place and .

Answer:- and

The

The mid-point of a line section is a particular into ratio; On this case, the ratio is //

The next are different circumstances of dividing a line section in given ratios :

- Inner division:

Let and be the 2 given factors on a line section which divides it within the given ratio . It’s required to search out the co-ordinates of P. Suppose they’re , as illustrated within the cartesian graph sketch beneath:

L

A(x_{1},y_{1})

P(x,y)

B(x_{2},y_{2})

L

M

N

x

y

0

ok

T

With the letterings on the graph and related triangles

Now, AP: PB=M_{1}:M_{2},

AK = MN = ON – OM = X_{2}-X

KP = MP- MK = MP- LA = y-y_{1}

TB = NB – NT = NB – MP = y_{2}-y

From (i) we now have

The primary two relation give

Or

Equally, from the relation we get

this finally offers

Therefore, the co-ordinate of the purpose diving a line becoming a member of () and within the are given as ,

Be aware that these outcomes applies to increase division, with both or taken as destructive

Instance 2:-

Discover the co-ordinates of the purpose which divides the road becoming a member of the factors (8,9) and (-7,4) internally within the ratio 2:3 .

SOLUTION:-

The co-ordinates of the purpose is obtained by substitution,

Instance 3: Discover the centroid of the triangle whose vertices have the coordinate (-4,6),(2,-2) and (2,5) respectively.

Answer:- Recall :

The centroid of a triangle is the purpose of intersection of its median.

Let AD be the median bisecting its base.

Then

(-4,6)A

C(2,5)

B(2,-2)

D

y

6

4

2

4

6

-2

2

-2

-4

-6

-4

x

The purpose on which divides it internally within the ratio 2:1 is the centroid. If are co-ordinates of the centroid, then

The centroid is (0,3)

One other case of inner division is as proven within the section beneath

P

3

A

22

Q2

We are saying A divides PQ internally within the ratio 3:2

- Everlasting division

P

Q

1

3

A

We are saying A divides PQ externally (i.e exterior PQ) within the ratio 3 : -1

i.e (1<3, subsequently A lies on the left of P)

A

1

Q

3

P

divides externally within the ratio -1:3

i.e

Instance 4

The purpose C divides the road the place the co-ordinates of respectivelly within the ratio 3:2 .Discover the co-ordinate

Answer

A(3,2)

B(4,1)

C(x,y)

Therefore, the co-ordinates of C are (6,1)

** EVALUATION: **

- Discover the co-ordinates of the mid-points of the traces becoming a member of the next pairs of factors;
- (3,6) and (5,8) (b) () and ()
- Discover the mid-points of the perimeters of triangle whose vertices are

**Sub-Subject 2**: Gradient of a straight line

The gradient of a line is outlined because the ratio enhance in in going from one level to a different on a line.

y_{1}

X^{1}

Y^{1}

Y

P_{2}(x_{2},y_{2})

P(x_{1},y_{1})

L

Is the change in x because the variable x will increase or decreases from x_{1} to x_{2} and is the change in y with respect to y_{1} and y_{2} .

The slope (gradient) m of a straight line L is outlined as

If is the angle of inclination to the slope of L,then ; is known as the angle of slope of the road.

Instance 5: Discover the slope and the angle of inclination of the L by means of factors and

y

x

P_{1}(1,2)

P_{3}(2,5)

P_{2}(3,8)

10

8

6

4

2

0

1

2

3

4

5

L

Answer:

The slope m of factors P_{1} and P_{2} on L is

The slope of the factors P1 and P3 on L is

Subsequently, implies that the slope of the road L is 3.

Since

Subsequently, the angle of inclination is 71.57

It may possibly subsequently be concluded from the instance above that any given line has one and just one slope.

Be aware: the factors and are mentioned to be **collinear** that means that the factors lie on a straight line, and the gradient of any two of the purpose is identical.

Instance 6: Discover the gradient of the road becoming a member of the pair of level

Answer:

Let symbolize the gradient

= -1

**EVALUATION: **

- Discover the angle between traces L
_{1}, with slope -7 and L2 which passes by means of (2,-1) and (5,3) - Discover the gradients of the traces becoming a member of the next pairs of factors :

**Sub-Subject 3:- DISTANCE BETWEEN TWO POINTS **

Let be two distinct factors. The gap between them can thus be calculated.

P_{2}(x_{2},y_{2})

y

x

P(x_{1},y_{1})

x_{2}-x_{1}

Y_{2}-y_{1}

0

Making use of Pythagoras theorem to right- angled triangle within the graph above,

Instance 7:

Calculated the space between the factors, (4,1) and (3,-2)

Answer:

The gap between the factors (4,1) and (3, -2) is ^{22}

**EVALUATION:**

Discover the distances between the given factors:

**SUB – TOPIC 4: CONDITIONS FOR PARALLELISM AND PERPENDICULARITY/ ANGLES BETWEEN TWO INTERSECTING LINEs**.

**A**

**C**

**D**

**B**

_{2}

_{1}

**Q**

Within the diagram above, the angles between the traces AB and DC are as indicated are associated as follows:

AQC = DQB = (vertically reverse angles)

_{21} in triangle ACQ (exterior angle of a triangle equals sum of reverse inside angles)

The traces AB (gradient m_{1}) and CD(gradient m_{2)} making angles _{1} and _{2} with the reveals that: _{1} = m_{1} and _{2} = m_{2}

Since _{2} = _{121}

⇒ _{21}) _{2} – _{1})_{21}) =

i.e. if ** is the angle between two intersecting traces** whose gradients are then the angle between them is

**If the 2 traces are parallel**, the angle between them is zero, therefore and therefore m_{2} – m_{1})/(1 + m_{2}m_{1}) ⇒m_{2} – m_{1}) = 0 ⇒ **m _{2} = m_{1}**

**If the 2 traces are perpendicular**, . Since any worth divided by zero⇒from m_{2} – m_{1})/(1 + m_{2}m_{1}), (1 + m_{2}m_{1}) = 0, ** m _{2}m_{1 }= – 1** or

**m**or

_{1}= -1/ m_{2}**m**

_{2}= -1/m_{1}Instance 8: Discover the acute angle between the pair of traces and

Answer: let the gradients of the traces be **m _{1} **and

**m**respectively.

_{2}Recall that ; subsequently, writing every of the traces on this kind we acquire

and

Therefore, **m _{1} ** and

**m**

_{2}Subsequently, the angle between the 2 traces is given by

** = **

** = **

** = **301.4^{0}

Instance 9: Discover the gradient of a line parallel to the road which passes by means of the pair of factors (-1,3) and (4,7)

Answer: let the gradient of the road passing by means of the factors (-1,3) and (4,7) be **m _{1 }**

Since, then

=

Therefore, the gradient, **m _{2}, **of a line parallel to the road which passes by means of the pair of factors (-1,3) and (4,7) is since

**m**

_{2}= m_{1}**EVALUATION:**

- Discover the gradient of a line parallel to the road whose gradient is 2.3
- Discover the gradient of a line perpendicular to the road which passes by means of every pair of the next factors
- (0,8) and (-5,2) (b). (-k,h) and (b, -f)
- Discover the inside angles of the triangle whose vertices are A(4,3), B(-2,2) and C(2,-8).

**PRACTICE QUESTIONS:**

- The road becoming a member of (-5,7) and (0,-2) is perpendicular to the road becoming a member of (1,-3) and (4,x). Discover x.
- The gradient of the road passing by means of the factors Discover the worth of
- The gap between is 13units. Discover the values of
- Utilizing gradients, decide which of the next units of three factors are collinear?
- (1,-1) , (-2,4), (0,1)
- (5,-2) , (7,6), (0,-2)
- (-2,3) , (8,-5), (5,4)
- (6,-1) , (5,0), (2,3)
- (-1,5) , (3,1), (5,7)
- Discover the gradient of a line parallel to the road whose gradient is – 3

**ASSIGNMENT:**

- Given the three factors
- The gradient of AB
- The equation of the road by means of C perpendicular to AB
- If M and N are the mid – factors of BC and AC respectively, calculate:
- The coordinates of the purpose of intersection G
- The worth of the ratio lAGl : lGMl.
- In parallelogram ABCD, slope of AB = -2, slope of BC = . State the slope of
- AD c. CD
- The altitude to AD d. the altitude to CD

**KEY WORDS**

**CORDINATE****GRADIENT****PARALLEL****PERPENDICULAR****COLLINEAR**