Chapter 2 : The Straight Line

2.1 The Gradient of Straight Line

Definition : The gradient of a straight line is defined as

Gradient of Straight Line, m = = ----- (1)

If the straight line makes an angle θ with the positive direction of the x-axis and this

angle is a measure of the slope of the line with respect to the horizontal

x-axis .

In general , the gradient of a line passing through the points A(x1,y1) and B(x2 , y2)

is m = tanθ = , where x1≠ x2 and 0 0 ≤ θ < 1800

2.2 Sign of Gradient

From the below diagrams , state whether the gradients of the straight lines are positive , negative , zero or undefined .

Example 1:
Find the gradient of the line joining the points and also its angle of inclination.

Example 2 :

Find the gradient of the line joining the points on the curve y = 2x2 – 3x + 1 whose abscissae are 1 and 3 .

[ans : 5]

*Example 3: ( Collinear Points : three points lie on the same straight line )

(a) Given that three points A(0, t-2 ) , B(t+1 , t- 4) , C(t-5,2) . Find the values of t if the three points are
collinear .

[ans : t = 2 , 3 ]

(b) Show that the points A (0,-3 ) , B (4,-2 ) , C (16,1 ) are lie on the same line .

2.3 The Equation of the Straight Line

General Form of the Equations of Straight Lines
In general , an equation of the form ax + by + c = 0 ( a , b not both zero ) represents a straight line in the
rectangular coordinate plane , see figure 1.

There are two special types of straight lines :
(a) (b)

In the following section , we will discuss other forms of straight lines when some conditions are given .

( I ) Gradient-intercept Form 斜截式

y-intercept : The point where the straight line cuts the y-axis , the coordinate is ( 0 , c ) .
1 : Given a straight line , gradient m and y-intercept c , find its equation .

Let P(x,y) be any point lie on the straight line passing through the point A ( 0,c ) .
gradient of AP , m =
y – c = mx
i.e. y = mx + c is the equation of straight line .

Example 4 :
Write down the gradient , m and y-intercept , c of the following equations .

Example 5 :

Find the equation of each of the following lines .

(II) Point-slope form 点斜式

Given a fixed point Q (x1 , y1 )

Let P ( x , y ) be any point on the line , then gradient of PQ , m =

i.e. y – y1 = m ( x – x1 )

(III) Two – Point Form 两点式

As the straight line l passing through two points A ( x1 , y1 ) and B (x2 , y2 ) .

let P(x,y) be any point on line l , the gradient AB = gradient BP .

gradient BP = m or gradient AB =

∴

Example 6:

Find the equation of each of the following straight lines given its gradient , m and the coordinates of a point on the line .

Example 7 :

Find the equation of the straight line which the line makes an angle of 450 with the positive x-axis and passing through the point P ( - 4 , 5 ) .

Example 8 : Find the equation of the straight line passing through the points A and B .

Example 9

The vertices of a triangle ABC are A(3,5) , B(-2,-5) and C(-5,-1) . Find the equation of the side AB and also the median of BC through vertex A .

Example 10

Find the equation of the straight line passing through point ( -1 , 2) and through the point of intersection of the lines

3x-4y=18 and 5x + 6y = - 8 .

Form IV : Intercept Form 截距式

Let a : the x – intercept

b : the y – intercept Example 11 :

from two point form of the straight line : Find the equation of line which through A(0,5) and B(-1,0).

ay = -bx + ab

bx + ay = ab

divide ab both sides :

Example 11:

Interpret the general equation of line 3x + 2y – 7 = 0 into the following form :

(a) intercept form (b) slope-intercept form

Example 12 :

Find the equation of the following lines .

Example 13:

Find equation of the line which passing through the point (2 , 2 ) and cutting off intercepts on the axes whose sum is 9 .

[ans: 3x + 6y -18=0 or 6x + 3y-18= 0]

Common Practise 1:

1. Write down the equations of the lines :

(i) gradient 5/6 , and through the point ( -3 , -1 ) (ii) inclined at 600 to the x-axis , and through ( 4 , -8 )

(iii) through the points ( 3 , 8 ) , ( 6 , 12 ) (iv) through the points

2. Find the equation of the chord joining the points on the curve y = 2x3 + 1 whose absicissae are 1 and -2 .

3. Find in terms of m , the equation of the straight line passing through the point (0,m ) with gradient m . Find the value of
m if this line passes through the point (2 , 6 ) .

4. If the mid-points of the sides BC , CA and AB of the triangle ABC be (1, 3 ) , ( 5,7 ) and ( -5 , 7 ) , find the equation of
the side AB .

5. A , B are the points ( 3 , 7 ) , ( -11 , -1 ) . Find the equation of the line through the point ( 4 , -6 ) and bisecting AB .

6. P ( a , b ) lies on the line 6x – y = 1 and Q ( b , a ) lies on the line 2x – 5y = 5 . Find the equation of PQ .

7. Find the equation of the line PQ which joining the points of intersection of the line y = x – 6 and the curve y2 = 8x .

Answer : 1. (i) 5x – 6y + 9 = 0 ; (ii) x – y = 4 (2 +) ; (iii) 4x – 3y + 12 = 0 ; (iv) x + 2t2 y = 3ct 2. y = 6x – 3 3. y = mx + m , m = 2 4. x - y + 12= 0
5. 9x + 8y + 12 = 0 6. x + y – 6 = 0 7. y = x – 6

2.4 Equations of Parallel Lines

Consider 2 lines L AB : y = m1 x + c1

L PQ : y = m2 x + c2

as shown in the diagram ,

since AB is parallel to PQ ,

∴

∵

∴ m1 = m2

Example 14: Find the equation of the line which parallel to the given line and passing through the given point .

Example 15

For what value of k is the line x – y + 2 + k(2x + 3y ) = 0 parallel to the line 3x + y = 0 ?

[ans : k = 4/7]

2.5 Perpendicular Lines

In the diagram , two straight lines l1 and l 2 with gradients m1 and m2

intersect at right angles at A and meet the x-axis at B and C respectively .

Since AD ⊥ BC , in ⊿ADB

as l1 has positive gradient , m1 = = tanθ --- (1)

but , the line l 2 has negative gradient ( i.e. m2 < 0 )
in ⊿ADC , we have m2 = − -------(2)
but ⊿ADB ~ ⊿CDA ⇒
from (1) and (2) ⇒ ( )
Therefore , m1 × m2 = -1 .

* Two non-vertical lines with gradient m1 and m2 are perpendicular ⇔ m1 m2 = -1

Example 16: Find the equation of the straight line which perpendicular to the given line and passing through the given
point .

Example 17

Show that A( 1, 2 ) , B ( 2, -5 ) , C ( 7, 0) and D(6,7) are vertices of a rhombus . to show parallelogram

show that diagonal bisect each
other

a rhombus is equal sides
parallelogram

Example 18

Find the equations of the straight line through the intersection of 2x + 3y + 4 = 0 and 3x + 4y – 5 = 0 and perpendicular to 6x – 7y + 8 = 0

[Ans : 7x + 6y = 85]

Example 19

(a) Given the points A (7,0 ) , B(6,2) , C(-1,-2 ) are (b) PQRS is a rhombus with P (0,5) and the equation of
the vertices of a triangle ABC , find the equation QS is y = 2x + 1 . Find the equation of the diagonal
of the altitude through A. PR .

[Ans : 7x + 4y – 49 = 0 ] [ans: x + 2y – 10 = 0 ]

2.5.1 Perpendicular Bisector

Key steps to find the equation of perpendicular bisector :

Step 1 : Find gradient of AB , m =

Step2 : Find gradient perpendicular to AB ,

Step 3 : Find midpoint of AB ,
Step 4 : Equation of perpendicular bisector of AB ,

Example 20

Find the equation of perpendicular bisector of AB given that A ( 1 , 2 ) and B ( 5 , -1 ) .

[ans : 6y = 8x – 21 ]

Example 21

A triangle has vertices A(1,2) , B(3,4 ) and C(0,3) , calculate the coordinates of circumcentre of the ⊿ABC .

Point of concurrency of triangle

Example 22:

Find the coordinates of the orthocentre of the triangle whose vertices are (0,0) , (2,-1) and (-1,3 ) .

[ans: (- 4 , -3)]

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