5.7 SPM Practice (Long Questions 3)

Question 5:

 
In the diagram above, a straight line 5x + 7y + 35 = 0 intersects with the x-axis and y-axis at R and S respectively. Determine
(a) the gradient of the straight line RS.
(b) the x-intercept of the straight line RS.
(c) the distance of RS.
 
Solution:
(a)
5 x + 7 y + 35 = 0 7 y = 5 x 35 y = 5 7 x 5 The gradient of the straight line R S = 5 7 .

(b)
At x-intercept, y = 0 0 = 5 7 x 5 5 7 x = 5 x = 7 x-intercept of the straight line R S = 7.

(c)
Point R = ( 7 , 0 ) and point S = ( 0 , 5 ) Distance of R S = ( 7 0 ) 2 + ( 0 ( 5 ) ) 2 Distance of R S = 49 + 25 Distance of R S = 74 units



Question 6:

 
In the diagram above, O is the origin of the Cartesian plane, AOB is a straight line and OA = AC. Find
(a) the coordinates of C.
(b) the value of h.
(c) the equation of BC.
 
Solution:
(a)
x-coordinate of C = –3 × 2 = –6
Therefore coordinates of C = (–6, 0).

(b)
Gradient of AO= Gradient of OB 0( 4 ) 0( 3 ) = h0 60 4 3 = h 6 h=8

(c)
Gradient of BC= 80 6( 6 ) = 8 12 = 2 3 At point C( 6,0 ), 0= 2 3 ( 6 )+c c=4 The equation of BC is, y= 2 3 x+4

5.2 Gradient of a Straight Line in Cartesian Coordinates

5.2 Finding the Gradient of a Straight Line

The gradient, m, of a straight line which passes through (x1, y1) and (x2, y2) is given by, 
 
mPQ  =  y 2 y 1 x 2 x 1




Example 1:


Find the gradient of the straight line joining two points and Q in the above diagram.

Solution:
P = (x1, y1) = (4, 3), Q = (x2, y2) = (10, 5)

Gradient of the straight line PQ
= y 2 y 1 x 2 x 1 = 5 3 10 4 = 2 6 = 1 3


Example 2:
Calculate the gradient of a straight line which passes through point A (7, -3) and point B (-3, 6).

Solution:
A = (x1, y1) = (7, -3), B = (x2, y2) = (-3, 6)

Gradient of the straight line AB
= y 2 y 1 x 2 x 1 = 6 ( 3 ) 3 7 = 9 10

5.7 SPM Practice (Long Questions 2)

Question 3:
Diagram below shows a straight line JK and a straight line ST drawn on a Cartesian plane. JK is parallel to ST.


Find
(a) the equation of the straight ST,
(b) the x-intercept of the straight line ST.

Solution:
(a) 
JK is parallel to ST, therefore gradient of JK = gradient of ST.
= 8 0 0 4 = 2
Substitute m = –2 and S (5, 6) into y = mx + c
6 = –2 (5) + c
c = 16
Therefore equation of ST: y = –2x + 16

(b)

For x-intercept, y = 0
0 = –2x + 16
2x = 16
x = 8
Therefore x-intercept of ST = 8



Question 4:


In the diagram above, PQRS is a parallelogram. Find
(a)  the gradient of SR,
(b) the equation of QR,
(c)  the x-intercept of QR.

Solution:
(a)
PQ is parallel to SR, gradient of PQ = gradient of SR.
Gradient of SR= 6 3 =2

(b)
Gradient of QR= 86 50 = 2 5 Substitute m= 2 5  and R (5,8) into y=mx+c 8= 2 5 ( 5 )+c c=6 Therefore equation of QR: y= 2 5 x+6

(c)
For x-intercept, y = 0 0 = 2 5 x + 6 x = 15
Therefore x-intercept of QR = –15.

5.6 SPM Practice (Short Questions)


Question 1:
Diagram below shows a straight line RS on a Cartesian plane.
 
Find the gradient of RS.

Solution:
Using gradient formula  y 2 y 1 x 2 x 1 Gradient of RS= 31 5( 1 ) = 2 6 = 1 3



Question 2:
In diagram below, PQ is a straight line with gradient  1 2 .


Find the x-intercept of the straight line PQ.

Solution:
Using the gradient formula, m= y-intercept x-intercept 1 2 =( 3 x-intercept ) x-intercept=3×( 2 )=6



Question 3:
Diagram below shows a straight line RS drawn on a Cartesian plane.

 
It is given that the distance of RS is 10 units.
Find the gradient of RS.

Solution:
RS=10 units, OS=6 units OR= 10 2 ( 6 ) 2 =8 units y-intercept of RS=6 x-intercept of RS=8 Using the gradient formula, m= y-intercept x-intercept  Gradient of RS=( 6 8 )= 3 4



Question 4:
The gradient of the straight line 3x – 4= 24 is

Solution:
Rearrange the equation in the form y = mx+ c
3x – 4y = 24
4y = 3x – 24
y = 3 4 x 6

Therefore, gradient of the straight line = 3 4 .



Question 5:
Determine the y-intercept of the straight line 3x + 2y = 5

Solution:
For y-intercept, x = 0
3(0) + 2y = 5
   y= 5 2 y-intercept= 5 2 .



5.7 SPM Practice (Long Questions 1)

Question 1:
In diagram below, ABCD is a trapezium drawn on a Cartesian plane. BC is parallel to AD and O is the origin. The equation of the straight line BC is 3y = kx+ 7 and the equation of the straight line AD is y = 1 2 x + 3


Find
(a) the value of k,
(b) the x-intercept of the straight line BC.

Solution:

(a)
Equation of BC :
3y = kx + 7
y= k 3 x+ 7 3 Gradient of BC= k 3 Equation of AD: y= 1 2 x+3 Gradient of AD= 1 2 Gradient of BC= gradient of AD k 3 = 1 2 k= 3 2

Gradient of BC = gradient of AD
k 3 = 1 2 k = 3 2

(b)

Equation of BC 3 y = 3 2 x + 7
For x-intercept, y = 0
3 ( 0 ) = 3 2 x + 7 3 2 x = 7 x = 14 3
Therefore x-intercept of BC 14 3



Question 2:
In diagram below, is the origin. Straight line MN is parallel to a straight line OK.


Find
(a) the equation of the straight line MN,
(b) the x-intercept of the straight line MN.

Solution:

(a)
Gradient of MN = gradient of OK
Gradient of MN
= 5 0 3 0 = 5 3

Substitute m = 5/3 and (–2, 5) into y = mx + c
5= 5 3 ( 2 )+c
15 = – 10 + 3c
3c = 25
c = 25/3

Therefore equation of MN y = 5 3 x + 25 3

(b) 
For x-intercept, y = 0
0 = 5 3 x + 25 3 5 3 x = 25 3
5x = –25
x = –5
Therefore x-intercept of MN = –5

5.3 Intercepts (Sample Questions)


Example 1:


The x-intercept of the line ST is

Solution:

The x-coordinate for the point of intersection of the straight line with x-axis is -0.4.
Therefore the x-intercept of the line ST is 0.4.



Example 2:
Find the x-intercept of the straight line 2x + 3y + 6 = 0.

Solution:
2x + 3y + 6 = 0
At x-intercept, y = 0
2x + 3(0) + 6 = 0
2x = –6
x = –3


Example 3
:
What is the y-intercept of the straight line 12x – 15y = 60?

Solution:
12x – 15y = 60
At y-intercept, x = 0
12(0) – 15y = 60
– 15y = 60
= –4

5.3 Intercepts

5.3 Intercepts
 
1. The x-intercept is the point of intersection of a straight line with the x-axis.
2. The y-intercept is the point of intersection of a straight line with the y-axis.

3. In the above diagram, the x-intercept of the straight line PQ is 6 and the y-intercept of PQ is 5.
 
4. If the x-intercept and y-intercept of a straight line are given,
  Gradient, m = y intercept x intercept

5.1 Gradient of a Straight Line


5.1 Gradient of a Straight Line
The gradient of a straight line is the ratio of the vertical distance to the horizontal distance between any two given points on the straight line.

Gradient, m = Vertical distance Horizontal distance
Example:












Find the gradient of the straight line above.

Solution
:
Gradient, m = Vertical distance Horizontal distance = 4 units 6 units = 2 3

5.2 Gradient of a Straight Line in Cartesian Coordinates (Sample Questions)


Example 1:
Given that a straight line passes through points (-3, -7) and (4, 14). What is the gradient of the straight line?

Solution:

Let (x1, y1) = (-3, -7) and (x2, y2) = (4, 14).

Gradient of the straight line 
= y 2 y 1 x 2 x 1 = 14 ( 7 ) 4 ( 3 ) = 21 7 = 3


Example 2:


The gradient of the straight line PQ in the diagram above is

Solution:
Let (x1, y1) = (12, 0) and (x2, y2) = (0, 7).

Gradient of the straight line PQ
= y 2 y 1 x 2 x 1 = 7 0 0 12 = 7 12


Example 3:
A straight line with gradient -3 passes through points (-4, 6) and (-1, p). Find the value of p.

Solution:
y 2 y 1 x 2 x 1 = 3 p 6 1 ( 4 ) = 3 p 6 3 = 3 p 6 = 9 p = 3

5.4 Equation of a Straight Line (Sample Questions)


Example 1:
Given that the equation of a straight line is 4x + 6y – 3 = 0. What is the gradient of the line?

Solution:
4x + 6y – 3 = 0
6y  = – 4x + 3
y = 4 x 6 + 3 6 y = 2 3 x + 1 2 y = m x + c g r a d i e n t , m = 2 3  


Example 2:
Given that the equation of a straight line is y = – 7x + 3. Find the y-intercept of the line?

Solution:
y = mx + c, c is y-intercept of the straight line.
Therefore for the straight line y = – 7x + 3,
y-intercept is 3


Example 3
:

 
 
 
 






Find the equation of the straight line MN if its gradient is equal to 3.
 
Solution:
Given m = 3
Substitute m = 3 and (-2, 5) into y = mx + c.
5 = 3 (-2) + c
5 = -6 + c
c = 11
Therefore the equation of the straight line MN is y = 3+ 11