# The straight line (cont’d)

WEEK 9

SUBJECT: FURTHER MATHEMATICS

CLASS: SSS 1

FURTHER MATHEMATICS SS 1 SECOND TERM

TOPIC: The straight line (cont’d)

CONTENT: (i) Equation of line

(iii)Transform relationship into linear form

(iv) Locus of points.

(v) Determination of law.

SUB-TOPIC: EQUATION OF LINE

Two point form: The equation of a straight line through points P1 () and P2 () is =

Point- slope form: The equation of a straight line through point P1 () with slope is

Slope – intercept form: Suppose a straight line cuts the at the point the distance of this point from the origin is called the intercept on the Therefore, the equation of a line, of the slope , making an intercept on the is . . The line is parallel to which is the equation of a straight line passing through the origin and slope, (as shown below)

(o, c)

C

0

x

y

Y = mx

Notice that the coefficient of , which in each case is is the slope of the line. Also the sign of determines whether the line cuts the above or below the origin, . For example, the equation is the equation of a straight line of slope 2 and as intercept

Intercept form:

y

L

(0,b)

(x,y)

P

B

A(a,0)

O

y

x

a-x

M

The line L can be considered as the Locus of a moving point P(x,y) . From similar triangles AMP and AOB, we got or or

Therefore, the equation of the straight line is parallel to (i.e a line with zero slope) the equation is . If the straight line is parallel to , the equation is

General form:

Generally, an equation are constants with at least one of A or B different from zero, represents a straight line.

. The line is therefore a vertical line

However if .

An equation like that contains only the first powers of is said to be linear in

Example 1: Find the equation of the straight line joining the points

Solution: The slope of the line is

Let be any point on the line other than, then the slope is

Since the slope of any point along any given line is equal, then

:.

. Thus, the required equation

Perpendicular form or Normal form:

To find the equation of a straight line in term of the length of the perpendicular ‘P’ from the origin to the line and the angle ‘ ’ which the perpendicular makes with the . Gradient of the line in the diagram below is . Therefore, the gradient of given line is . The coordinate of )

C

0

B

A

V

X

P(x, y)

Y

Applying the one-point form of the straight lines the equation required is

Since

Example2: Sketch and write down the equation of the line for which

Solution: (a) The equation of the line is

EVALUATION:

1. Find the equation of the line joining the origin to the point of intersection of
2. Determine the x-intercept ‘a’ and the y- intercept ‘h’ of the following lines. Sketch each (a) (b)

SUB-TOPIC 2: LOCUS OF POINTS

When a point moves in space or in a plane under a given set of conditions, it traces a path. This path is called the locus of the moving point. We may wish to find the equation of the locus.

Example 1: Find the equation of the locus of a point which moves so that its distance from (4,5) is twice its distance from (2,-3).

Solution:

Let A(4,5) and B(2,-3) be the two given points. Let P(x,y) be any point on the locus. Then the locus of P is such that

PA = 2PB

=

4

=

Since, it implies that

The required equation is

Example 2: A point P moves in the x – y plane in such a way that its distance from the point (1,-3) is equal to its distance from line y – 4 = 0. Find the equation of the locus of the point P.

Solution: A point P moving in the x – y plane in such a way that its distance from a point A(1,-3) to its distance from line y – 4 = 0 is a parabola. The diagram below illustrates the locus p.

y – 4 = 0

A

-3

1

4

B

x

y

From the diagram above, AP = PB since P moves in the x – y plane in such a way that its distance from the point (1,-3) is equal to its distance from line y – 4 = 0

AP = and PB = y – 4

Equating the two and solving we get

CLASS ACTIVITY:

1. L is a point (-2,4) and M is the point (3,-1). A variable point Q(x,y) is such that . Show that and describe the locus of Q. prove that the locus of Q divides the line segment LM in the ratio 3:2
2. The line meets the axes at the points A and B respectively. If M is the mid – point of AB and P has coordinate, find the equation of the locus of all points in the plane whose distance from P is .

SUB-TOPIC 3: AREAS OF TRIANGLE AND QUADRILATERALS

The graph below shows a triangle PQR with the coordinates PK, QL, RM drawn. The lines form three trapeziums from which the area of triangle can be found.

x

y

Q

P

K

M

L

R

O

Area of triangle PQR area of trapezium KPRM (area of trap. KPQL + area of trap. LQRP)

Area of trap (sum of parallel sides) (perpendicular dist. between // sides)

Example (1): Let the vertices of the triangle be

Then

Area of trap. KPRM 2

Area of trap. KPLM unit2

Area of trap LQRM 2

:. Area of of PQR 2

:. Changing the sign, the area of the triangle in magnitude 2

The following steps are important in finding the area of quadrilateral and area of triangle:

1. Make a sketch of the figure to locate the vertices
2. Mark in the coordinates to form trapezium or triangles
3. Calculate the area of the figure from the numerical areas of the trapezium and triangles.

Example (2): Find the area of the quadrilateral whose vertices are

B(0,5)

A(3,4)

D(3,-2)

C(2,-1)

O

K

L

x

y

SOLUTION:

Let the coordinates of the vertices be

=

2

Alternatively, we can use the following formulae:

Area of triangle with vertices and is and

Area of quadrilateral with vertices and is

EVALUATION:

1. The three vertices of a triangle are If the area of the triangle is 4 square units ,find the expression relating
2. If the area of a quadrilateral whose angular points A,B,C,D taken in order are be zero, find the value of

TRANSFORMING RELATIONSHIP INTO LINEAR FORM/DETERMINATION OF LAW

The relationship between two quantities may not necessarily be linear but it is often possible to transform such relations into linear forms. The method of reducing non – linear forms to linear forms is called Determination of Law

Some of such common relationships are given below:

1. If the relationship is of the form (where n is unknown)

Rewrite as

Comparing this with , Plot versus

giving a straight line of slope n and as the intercept.

1. If the relationship is of the form (where n is known)

Compare with ; plot versus giving a straight line of slope a and b as intercept.

1. If the relationship is of the form

Rewrite as

Comparing this with , Plot versus

giving a straight line of slope and as the intercept.

PRACTICE QUESTIONS:

1. The parallelogram RQRS has vertices. Find the coordinates of the point of intersection of the diagonal
2. (b) (c) (d)
3. Find the equation of the line passing through and parallel to the
4. Y= -1 (b) y=0 (c) x=0 (d) x=-1
5. Given that the straight lines are parallel, find a relationship connecting the constants (a) (b) (c) (d)
6. The points A, B and C are respectively. AB produced cuts the produced cuts the .Find the coordinates of the points.
7. A straight line passes through (-1,2) and makes angle 450 with the negative direction of the x – axis, find its equation.

ASSIGNMENT:

1. The following table gives corresponding values of two related variables x and y as obtained in an experiment:
 1 2 3 4 5 6 0.68 1.41 1.81 2.12 2.31 2.46

Theoretical considerations suggest that the relation between and is of the form

Where and are constants. Draw a suitable linear graph and use it to:

1. Decide whether or not the experiment confirms the theory reasonably well;
2. Determine, each correct to one decimal place, the values and , according to the theory.
3. Find the equation of the straight line at a distance of 3 units from the origin such that the perpendicular from the origin to the line makes an angle α, given by the equation with the positive direction of the axis of x.

KEY WORDS

• EQUATION
• LOCUS
• AREA

FURTHER MATHEMATICS SS 1 SECOND TERM

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