5.3 Intercepts

Posted on April 23, 2020 by user

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 = - \frac{y - intercept}{x - intercept}

5.1 Gradient of a Straight Line

Posted on April 23, 2020 by user

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 = \frac{Vertical distance}{Horizontal distance}

Example:

Find the gradient of the straight line above.

Solution:

\begin{array}{l} Gradient, m = \frac{Vertical distance}{Horizontal distance} \\ = \frac{4 units}{6 units} \\ = \frac{2}{3} \end{array}

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

Posted on April 23, 2020 by user

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

\begin{array}{l} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{14 - (- 7)}{4 - (- 3)} \\ = \frac{21}{7} = 3 \end{array}

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

\begin{array}{l} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{7 - 0}{0 - 12} \\ = - \frac{7}{12} \end{array}

Example 3:

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

Solution:

\begin{array}{l} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = - 3 \\ \frac{p - 6}{- 1 - (- 4)} = - 3 \\ \frac{p - 6}{3} = - 3 \\ p - 6 = - 9 \\ p = - 3 \end{array}

5.4 Equation of a Straight Line (Sample Questions)

Posted on April 23, 2020 by user

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

\begin{array}{l} y = - \frac{4 x}{6} + \frac{3}{6} \\ y = - \frac{2}{3} x + \frac{1}{2} \\ y = m x + c \\ g r a d i e n t, m = - \frac{2}{3} \end{array}

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 = 3x + 11

5.5 Parallel Lines (Sample Questions)

Posted on April 23, 2020 by user

Example 1:

The straight lines MN and PQ in the diagram above are parallel. Find the value of q.

Solution:

If two lines are parallel, their gradients are equal.

m₁ = m₂

m_MN = m_PQ

\begin{array}{l} using gradient formula \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ \frac{9 - 4}{5 - (- 1)} = \frac{q - (- 5)}{5 - (- 7)} \\ \frac{5}{6} = \frac{q + 5}{12} \end{array}

60 = 6q + 30

6q = 30

q = 5

5.5 Parallel Lines (Part 1)

Posted on April 23, 2020 by user

5.5 Parallel Lines (Part 1)

(A) Gradient of parallel lines

1. Two straight lines are
parallel if they have
the same gradient.

If PQ // RS,

then m_PQ = m_RS

2. If two straight lines have

the same gradient, then

they are parallel.

If m_AB = m_CD

then AB // CD

Example 1:

Determine whether the two straight lines are parallel.

(a) 2y – 4x = 6

y = 2x – 5

(b) 2y = 3x –4

3y = 2x +12

Solution:

(a)

2y – 4x = 6

2y = 6 + 4x

y = 2x + 3, m₁= 2

y = 2x – 5, m₂ = 2

m₁= m₂

Therefore, the two straight lines are parallel.

(b)

\begin{array}{l} 2 y = 3 x - 4 \\ y = \frac{3}{2} x - 2, m_{1} = \frac{3}{2} \\ 3 y = 2 x + 12 \\ y = \frac{2}{3} x + 4, m_{2} = \frac{2}{3} \\ m_{1} \neq m_{2} \\ ∴ The two straight lines are not parallel . \end{array}

5.6 SPM Practice (Short Questions)

Posted on April 23, 2020 by user

5.6.2 The Straight Line, SPM Paper 1 (Short Questions)

Question 6:

Diagram below shows a straight line RS with equation 3y = –px – 12, where p is a constant.

It is given that OR: OS = 3 : 2.

Find the value of p.

Solution:

Method 1:

Substitute x= –6 and y = 0 into 3y = –px – 12:

3(0) = –p (–6) – 12

0 = 6p – 12

–6p = –12

p = 2

Method 2:

OR: OS = 3 : 2

\begin{array}{l} \frac{O R}{O S} = \frac{3}{2} \\ \frac{6}{O S} = \frac{3}{2} \\ O S = 6 \times \frac{2}{3} = 4 units \end{array}

Coordinates of S = (0, –4)

Gradient of the straight line RS =

- \frac{- 4}{- 6} = - \frac{2}{3}

Given 3y = –px – 12
Rearrange the equation in the form y = mx + c

\begin{array}{l} y = - \frac{p}{3} x - 4 \\ Gradient of the straight line R S = - \frac{P}{3} \\ ∴ - \frac{P}{3} = - \frac{2}{3} \\ P = 2 \end{array}

Question 7:

The above diagram shows two straight lines, KL and LM, on a Cartesian plane. The distance KL is 10 units and the gradient of LM is 2. Find the x-intercept of LM.

Solution:

Let point N be = (0, 2).

Using Pythagoras’ Theorem,

LN = √10² – 6² = 8

Point L = (0, 2 + 8) = (0, 10)

y-intercept of LM = 10

\begin{array}{l} Using the gradient formula, m = - \frac{y-intercept}{x-intercept} \\ 2 = - (\frac{10}{x-intercept}) \\ x-intercept of L M = - \frac{10}{2} = - 5 \end{array}

4.3 Operations on Statements (Sample Questions 1)

Posted on April 23, 2020 by user

Example 1:

Form a compound statement by combining two given statements using the word ‘and’.

(a) 3 × 12 = 36

7 × 5 = 35

(b) 5 is a prime number.

5 is an odd number.

(c) Rectangles have 4 sides.

Rectangles have 4 vertices.

Solution:

(a) 3 × 12 = 36 and 7 × 5 = 35

(b) 5 is a prime number and an odd number.

(c) Rectangles have 4 sides and 4 vertices.

Example 2:

Form a compound statement by combining two given statements using the word ‘or’.

(a) 16 is a perfect square. 16 is an even number.

(b) 4 > 3. -5 < -1

Solution:

(a) 16 is a perfect square or an even number.

(b) 4 > 3 or -5 < -1.

4.1 Statements (Sample Questions)

Posted on April 23, 2020 by user

Example 1:

Determine whether the following sentences are statements or not. Give a reason for your answer.

(a) 3 + 3 = 8

(b) 9 – 4 = 5

(d) 4 is a prime number.

(e) Is 40 divisible by 3?

(f) Find the perimeter of a square with each side of 4 cm.

(g) Help!

Answer:

(a) Statement; it is a false statement.

(b) Statement; it is a true statement.

(d) Statement; it is a false statement.

(e) Not a statement; it is a question.

(f) Not a statement; it is an instruction.

(g) Not a statement; it is an exclamation.

Deduction and Induction

Posted on April 23, 2020 by user

4.6 Deduction and Induction

(A) Reasoning by Deduction and Induction

1. Reasoning by deduction is a process of making a conclusion for a specific case based on a given general statement.

2. Reasoning by induction is a process of making a generalization based on specific cases.

Maths Tip

1. General statement → Special conclusion → Deduction

2. Specific cases → General conclusion → Induction

Example:

Determine whether the following conclusion is made based on a deductive reasoning or inductive reasoning.

(a) Area of triangle = ½ × Base × Height

(i)

Area of ∆ ABC

= ½ × 7cm × 5cm

= 17.5 cm²

(ii)

Area of ∆ DEF

= ½ × 7cm × 4cm

= 14 cm²

(b)

1 = 7 (1)² – 6

22 = 7 (2)² – 6

57 = 7 (3)² – 6

106 = 7 (4)² – 6

7n² – 6, n = 1, 2, 3, 4…

Solution:

(a)

The specific conclusion is made based on a general statement ~ Area of triangle = ½ × Base × Height. Therefore, the conclusion is made based on deductive reasoning.

(b)

The general conclusion 7n² – 6, n = 1, 2, 3, 4… is made based on specific cases. Therefore, the conclusion is based on inductive reasoning.