THE STRAIGHT LINE AND MID-POINT OF A LINE SEGMENT

 

 

 

 

 

FURTHER MATHEMATICS SS 1 SECOND TERM

 

 

WEEK 8

CLASS: SSS 1

2ND 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-x1

x2-x1

Q(x2,y2)

x

P(x1,y1)

R(x,y)

O

y

S

y2-y

y-y1

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 :

  1. 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(x1,y1)

P(x,y)

B(x2,y2)

L

M

N

x

y

0

ok

T

With the letterings on the graph and related triangles

Now, AP: PB=M1:M2,

AK = MN = ON – OM = X2-X

KP = MP- MK = MP- LA = y-y1

TB = NB – NT = NB – MP = y2-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

  1. 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:

  1. Discover the co-ordinates of the mid-points of the traces becoming a member of the next pairs of factors;
  2. (3,6) and (5,8) (b) () and ()
  3. 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.

y1

X1

Y1

Y

P2(x2,y2)

P(x1,y1)

L

 

Is the change in x because the variable x will increase or decreases from x1 to x2 and is the change in y with respect to y1 and y2 .

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

P1(1,2)

P3(2,5)

P2(3,8)

10

8

6

4

2

0

1

2

3

4

5

L

Answer:

The slope m of factors P1 and P2 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:

  1. Discover the angle between traces L1 , with slope -7 and L2 which passes by means of (2,-1) and (5,3)
  2. 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.

P2(x2,y2)

y

x

P(x1,y1)

x2-x1

Y2-y1

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 m1) and CD(gradient m2) making angles 1 and 2 with the reveals that: 1 = m1 and 2 = m2

Since 2 = 121

21) 21)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 m2 – m1)/(1 + m2m1) ⇒m2 – m1) = 0 ⇒ m2 = m1

If the 2 traces are perpendicular, . Since any worth divided by zero⇒from m2 – m1)/(1 + m2m1), (1 + m2m1) = 0, m2m1 = – 1 or m1 = -1/ m2 or m2 = -1/m1

Instance 8: Discover the acute angle between the pair of traces and

Answer: let the gradients of the traces be m1 and m2 respectively.

Recall that ; subsequently, writing every of the traces on this kind we acquire

and

Therefore, m1 and m2

Subsequently, the angle between the 2 traces is given by

 

=

=

= 301.40

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 m1

Since, then

=

Therefore, the gradient, m2, of a line parallel to the road which passes by means of the pair of factors (-1,3) and (4,7) is since m2 = m1

EVALUATION:

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

PRACTICE QUESTIONS:

  1. 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.
  2. The gradient of the road passing by means of the factors Discover the worth of
  3. The gap between is 13units. Discover the values of
  4. Utilizing gradients, decide which of the next units of three factors are collinear?
  5. (1,-1) , (-2,4), (0,1)
  6. (5,-2) , (7,6), (0,-2)
  7. (-2,3) , (8,-5), (5,4)
  8. (6,-1) , (5,0), (2,3)
  9. (-1,5) , (3,1), (5,7)
  10. Discover the gradient of a line parallel to the road whose gradient is – 3

ASSIGNMENT:

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

KEY WORDS

  • CORDINATE
  • GRADIENT
  • PARALLEL
  • PERPENDICULAR
  • COLLINEAR

 

 

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