6.4 Equation of Straight Lines (Part 2)

Posted on April 21, 2020 by user

6.4.2 Equation of Straight Lines

Case 1
1. The gradient and coordinates of a point are given.

2. The equation of a straight line with gradient m passes through the point (x₁, y₁) is:

Example 1:

A straight line with gradient –3 passes through the point (–1, 5). Find the equation of this line.

Solution:

y – y₁= m (x – x₁)

y – 5= – 3 (x – (–1))

y – 5= – 3x – 3

y = – 3x + 2

Case 2
1. The coordinates of two points are given.

2. The equation of a straight line joining the points (x₁, y₁)
and (x₂, y₂) is:

Example 2:
Find the equation of the straight line joining the points (2, 4) and (5, 6).

Solution:

\begin{array}{l} \frac{y - y_{1}}{x - x_{1}} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ Let (x_{1}, y_{1}) = (2, 4) and (x_{2}, y_{2}) = (5, 6) \\ \frac{y - 4}{x - 2} = \frac{6 - 4}{5 - 2} \\ \frac{y - 4}{x - 2} = \frac{2}{3} \\ 3 y - 12 = 2 x - 4 \\ 3 y = 2 x + 8 \end{array}

Case 3

1. The equation of a straight line with x–intercept “a” and y–intercept“b” is:

Example 3:

Find the equation of the straight line joining the points (5, 0) and (0, –6).

Solution:

x–intercept, a = 5, y–intercept, b = –6

Equation of the straight line

\begin{array}{l} \frac{x}{a} + \frac{y}{b} = 1 \\ \frac{x}{5} + \frac{y}{(- 6)} = 1 \\ \frac{x}{5} - \frac{y}{6} = 1 \end{array}

The equation of a straight line can be expressed in three forms:

(a)

(b)

(c)

6.4 Equation of Straight Lines (Part 1)

Posted on April 21, 2020 by user

6.4.1 Axes Intercepts and Gradient

(A) Formula for gradient:

1. Gradient of the line joining (x₁, y_l) and (x₂, y₂) is:

2. Gradient of the line with knowing x–intercept and y–intercept is:

3. The gradient of the straight line joining P and Q is equal to the tangent of angle θ, where θ is the angle made by the straight line PQ and the positive direction of the x-axis.

(B) Collinear points

The gradient of a straight line is always constant i.e.the gradient of AB is equal to the gradient of BC.

Example 1:

The gradient of the line passing through point (k, 1 – k) and point (–3k, –3) is 5. Find the value of k.

Solution:

\begin{array}{l} Gradient, m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ \frac{- 3 - (1 - k)}{- 3 k - k} = 5 \\ \frac{- 3 - 1 + k}{- 4 k} = 5 \\ - 4 + k = - 20 k \\ 21 k = 4 \\ k = \frac{4}{21} \end{array}

Example 2:

Based on the diagram below, find the gradient of the line.

Solution:

Gradient, m = - (\frac{y intercept}{x intercept}) = - (\frac{- 5}{10}) = \frac{1}{2}

6.3 Areas of Polygons

Posted on April 21, 2020 by user

6.3 Areas of Polygons

(A) Area of Triangle

Area of Triangle

1. When the given points are taken in an anticlockwise direction the result is positive; taken in a clockwise direction the result is negative. The answer for the area must be given as a positive value.

Example 1:

Calculate the area of ∆ ABC with the vertices A (-5, 5), B (-2, -4),

C (4, -1).

Solution:

\begin{array}{l} Area of △ A B C \\ = \frac{1}{2} | \begin{array}{l} - 5 - 2 4 \\ 5 - 4 - 1 \end{array} \begin{array}{l} - 5 \\ 5 \end{array} | \\ = \frac{1}{2} | (- 5) (- 4) + (- 2) (- 1) + (4) (5) \\ - (5) (- 2) - (- 4) (4) - (- 1) (- 5) | \\ = \frac{1}{2} | 20 + 2 + 20 + 10 + 16 - 5 | \\ = \frac{1}{2} | 63 | \\ = 31 \frac{1}{2} {unit}^{2} \end{array}

(B) Area of Quadrilateral

Area of Quadrilateral

(C) If the points A, B and C are collinear, then the area of ∆ ABC is 0.

Example 2:

Find the values of k if the points P (2, 1), Q (6, k) and R (3k,

\frac{9}{2}

) are collinear.

Solution:

\begin{array}{l} Area of △ P Q R = 0 \\ \frac{1}{2} | \begin{array}{l} 2 63 k \\ 1 k \frac{9}{2} \end{array} \begin{array}{l} 2 \\ 1 \end{array} | = 0 \\ \frac{1}{2} | 2 k + 27 + 3 k - 6 - 3 k^{2} - 9 | = 0 \\ | - 3 k^{2} + 5 k + 12 | = 0 \\ 3 k^{2} - 5 k - 12 = 0 \\ (3 k + 4) (k - 3) = 0 \\ k = - \frac{4}{3} or 3 \end{array}

6.2 Division of a Line Segment

Posted on April 21, 2020 by user

6.2 Division of a Line Segment

(A) Midpoints of a Line Segment

Formula for the midpoint, M of A (x_l, y₁) and B (x₂, y₂) is

Example 1:

Given B (m – 4, 3) is the midpoint of the straight line joining A(–1, n) and C (5, 8). Find the values of m and n.

Solution:

\begin{array}{l} B is the midpoint of A C \\ (m - 4, 3) = (\frac{- 1 + 5}{2}, \frac{n + 8}{2}) \\ (m - 4, 3) = (2, \frac{n + 8}{2}) \\ m - 4 = 2 and \frac{n + 8}{2} = 3 \\ m = 6 and n + 8 = 6 \\ n = - 2 \end{array}

(B) Point that Internally Divides a Line Segment in the Ratio m : n

Formula for the point P that lies on AB such that AP : PB = m : n is

Example 2:

The coordinate of R(2, –1) divide internally the line of AB with the ratio 3 : 2. If coordinate of A is (–1, 2), find the coordinate of B.

Solution:

\begin{array}{l} Let point B = (p, q) \\ (\frac{2 (- 1) + 3 p}{3 + 2}, \frac{2 (2) + 3 q}{3 + 2}) = (2, - 1) \\ (\frac{- 2 + 3 p}{5}, \frac{4 + 3 q}{5}) = (2, - 1) \\ Equating the x -coordinates, \\ \frac{- 2 + 3 p}{5} = 2 \\ - 2 + 3 p = 10 \\ 3 p = 12 \\ p = 4 \\ Equating the y -coordinates, \\ \frac{4 + 3 q}{5} = - 1 \\ 4 + 3 q = - 5 \\ 3 q = - 9 \\ q = - 3 \\ ∴ The coordinates of point B = (4, - 3) . \end{array}

6.1 Distance between Two Points

Posted on April 21, 2020 by user

6.1 Distance between Two Points

A (x₁, y₁) and C (x₂, y₂)are two points on a coordinate plane as shown below. BC is parallel to the x-axis and AB is parallel to the y-axis. Hence ∆ ABC = 90°.

Distance between Point A and C =

Example:

Find the distance between the points P (2, –2) and Q (–4, –5).

Solution:

Let P (2, –2) = (x₁, y₁) and Q (–4, –5) = (x₂, y₂).

\begin{array}{l} Distance of P Q \\ = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \\ = \sqrt{{(- 4 - 2)}^{2} + {(- 5 - (- 2))}^{2}} \\ = \sqrt{36 + 9} \\ = \sqrt{45} \\ = \sqrt{9 \times 5} \\ = 3 \sqrt{5} or (6.71) \end{array}

Long Questions (Question 3 – 4)

Posted on April 21, 2020 by user

5.6.2 Indices and Logarithms, SPM Practice (Long Questions)

Question 3:

Given that p = 3^r and q = 3^t, express the following in terms of r and/ or t.

(a)

(a) \log_{3} \frac{p q^{2}}{27},

(b) log₉p – log₂₇ q.

Solution:

(a)

Given p = 3^r, log₃ p = r

q= 3^t, log₃ q =t

\log_{3} \frac{p q^{2}}{27}

= log₃ pq² – log₃27

= log₃ p + log₃ q² – log₃ 3³

= r + 2 log₃ q – 3 log₃ 3

= r + 2 log₃ q – 3(1)

= r + 2t – 3

(b)

log₉ p– log₂₇ q

\begin{array}{l} = \frac{\log_{3} p}{\log_{3} 9} - \frac{\log_{3} q}{\log_{3} 27} \\ = \frac{r}{\log_{3} 3^{2}} - \frac{t}{\log_{3} 3^{3}} \\ = \frac{r}{2 \log_{3} 3} - \frac{t}{3 \log_{3} 3} \\ = \frac{r}{2} - \frac{t}{3} \end{array}

Question 4:

(a) Simplify:

log₂(2x + 1) – 5 log₄ x²+ 4 log₂ x

(b) Hence, solve the equation:

log₂(2x + 1) – 5 log₄ x²+ 4 log₂ x= 4

Solution:

(a)

log₂ (2x + 1) – 5 log₄ x²+ 4 log₂ x

\begin{array}{l} = \log_{2} (2 x + 1) - \frac{5 \log_{2} x^{2}}{\log_{2} 4} + 4 \log_{2} x \\ = \log_{2} (2 x + 1) - \frac{5}{2} \log_{2} x^{2} + \log_{2} x^{4} \\ = \log_{2} (2 x + 1) - \log_{2} {(x^{2})}^{(\frac{5}{2})} + \log_{2} x^{4} \end{array}

\begin{array}{l} = \log_{2} (2 x + 1) - \log_{2} x^{5} + \log_{2} x^{4} \\ = \log_{2} \frac{(2 x + 1) (x^{4})}{x^{5}} \\ = \log_{2} \frac{2 x + 1}{x} \end{array}

(b)

log₂ (2x + 1) – 5 log₄ x²+ 4 log₂ x= 4

\begin{array}{l} \log_{2} \frac{2 x + 1}{x} = 4 \\ \frac{2 x + 1}{x} = 2^{4} \\ \frac{2 x + 1}{x} = 16 \\ 2 x + 1 = 16 x \\ 14 x = 1 \\ x = \frac{1}{14} \end{array}

Long Questions (Question 1 – 2)

Posted on April 21, 2020 by user

5.6.1 Indices and Logarithms, SPM Practice (Long Questions)

Question 1:

(a) Find the value of

i. 2 log₂ 12 + 3 log₂5 – log₂ 15 – log₂ 150.

ii. log₈32

(b) Shows that 5ⁿ+ 5ⁿ⁺¹+ 5ⁿ⁺²can be divided by 31 for all the values of n which are positive integer.

Solution:

(a)(i)

2 log₂ 12 + 3 log₂ 5 – log₂15 – log₂ 150

= log₂ 12² + log₂ 5³– log₂ 15 – log₂ 150

= \log_{2} \frac{12^{2} \times 5^{3}}{15 \times 150}

= log₂ 8

= log₂ 2³

= 3

(a)(ii)

\begin{array}{l} \log_{8} 32 = \frac{\log_{2} 32}{\log_{2} 8} \\ = \frac{\log_{2} 2^{5}}{\log_{2} 2^{3}} = \frac{5}{3} \end{array}

(b)

5ⁿ+ 5ⁿ⁺¹+ 5ⁿ⁺²

= 5ⁿ+ (5 × 5ⁿ) + (5² × 5ⁿ)

= 5ⁿ (1 + 5 + 5²)

= 31 × 5ⁿ

Therefore, 5ⁿ+ 5ⁿ⁺¹+ 5ⁿ⁺²can be divided by 31 for all the values of n which are positive integer.

Question 2:

(a) Given log₁₀ x = 3 and log₁₀y = –2. Shows that 2xy – 10000y² = 19.

(b) Solve the equation log₃ x = log₉(x + 6).

Solution:

(a)

log₁₀x = 3 → (x = 10³)

log₁₀y = –2 → (y = 10^-2)

2xy – 10000y² = 19

Left hand side:

2xy – 10000y²

= 2 × 10³× 10^-2– 10000 (10^-2)²

= 20 – 10000 (10^-4)

= 20 – 1

= 19

= right hand side

(b)

\begin{array}{l} \log_{3} x = \log_{9} (x + 6) \\ \log_{3} x = \frac{\log_{3} (x + 6)}{\log_{3} 9} \\ \log_{3} x = \frac{\log_{3} (x + 6)}{\log_{3} 3^{2}} \\ \log_{3} x = \frac{\log_{3} (x + 6)}{2} \end{array}

2log₃ x= log₃ (x + 6)

log₃ x²= log₃ (x + 6)

x²= x + 6

x²– x – 6 = 0

(x + 2) (x – 3) = 0

x = – 2 atau 3.

log₃(– 2) not accepted (logarithm of a negative number is undefined)

Jadi, x = 3.

Short Questions (Question 12 – 14)

Posted on April 21, 2020 by user

Question 12

Solve the equation,

\log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6

Solution:

\begin{array}{l} \log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{\log_{2} 4} = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{2} = 6 \\ 2 \log_{2} 5 \sqrt{x} + \log_{2} 16 x = 12 \\ \log_{2} {(5 \sqrt{x})}^{2} + \log_{2} 16 x = 12 \\ \log_{2} (25 x) + \log_{2} 16 x = 12 \\ \log_{2} (25 x) (16 x) = 12 \\ \log_{2} 400 x^{2} = 12 \\ 400 x^{2} = 2^{12} \\ x^{2} = 10.24 \\ x = 3.2 \end{array}

Question 13

Given that 2 log₂ (x – y) = 3 + log₂x + log₂ y.

Prove that x² + y²– 10xy = 0.

Solution:

2 log₂ (x – y) = 3 + log₂x + log₂ y

log₂ (x– y)² = log₂ 8 + log₂ x + log₂y

log₂ (x– y)² = log₂ 8xy

(x – y)² = 8xy

x²– 2xy + y² = 8xy

x² + y² – 10xy = 0 (proven)

Question 14 (2 marks):
Given 2^p + 2^p = 2^k. Express p in terms of k.

Solution:

\begin{array}{l} 2^{p} + 2^{p} = 2^{k} \\ 2 (2^{p}) = 2^{k} \\ 2^{p} = \frac{2^{k}}{2^{1}} \\ 2^{p} = 2^{k - 1} \\ p = k - 1 \end{array}

Short Questions (Question 9 – 11)

Posted on April 21, 2020 by user

Question 9

Solve the equation,

\log_{2} 4 x = 1 - \log_{4} x

Solution:

\begin{array}{l} \log_{2} 4 x = 1 - \log_{4} x \\ \log_{2} 4 x = 1 - \frac{\log_{2} x}{\log_{2} 4} \\ \log_{2} 4 x = 1 - \frac{\log_{2} x}{2} \\ 2 \log_{2} 4 x = 2 - \log_{2} x \\ \log_{2} 16 x^{2} = \log_{2} 4 - \log_{2} x \\ \log_{2} 16 x^{2} = \log_{2} \frac{4}{x} \\ 16 x^{2} = \frac{4}{x} \\ x^{3} = \frac{4}{16} = \frac{1}{4} \\ x = {(\frac{1}{4})}^{\frac{1}{3}} = 0.62996 \end{array}

Question 10

Solve the equation,

\log_{4} x = 25 \log_{x} 4

Solution:

\begin{array}{l} \log_{4} x = 25 \log_{x} 4 \\ \frac{1}{\log_{x} 4} = 25 \log_{x} 4 \\ \frac{1}{25} = {(\log_{x} 4)}^{2} \\ \log_{x} 4 = \pm \frac{1}{5} \\ \log_{x} 4 = \frac{1}{5} or \log_{x} 4 = - \frac{1}{5} \\ 4 = x^{\frac{1}{5}} 4 = x^{- \frac{1}{5}} \\ x = 4^{5} 4 = \frac{1}{x^{\frac{1}{5}}} \\ x = 1024 x^{\frac{1}{5}} = \frac{1}{4} \\ x = \frac{1}{1024} \end{array}

Question 11

Solve the equation,

2 \log_{x} 5 + \log_{5} x = \lg 1000

Solution:

\begin{array}{l} 2 \log_{x} 5 + \log_{5} x = \lg 1000 \\ 2. \frac{1}{\log_{5} x} + \log_{5} x = 3 \\ \times (\log_{5} x) \to 2 + {(\log_{5} x)}^{2} = 3 \log_{5} x \\ {(\log_{5} x)}^{2} - 3 \log_{5} x + 2 = 0 \\ (\log_{5} x - 2) (\log_{5} x - 1) = 0 \\ \log_{5} x = 2 or \log_{5} x = 1 \\ x = 5^{2} x = 5 \\ x = 25 \end{array}

Short Questions (Question 5 – 8)

Posted on April 21, 2020 by user

Question 5

Solve the equation,

\log_{9} (x - 2) = \log_{3} 2

Solution:

\begin{array}{l} \log_{9} (x - 2) = \log_{3} 2 \\ \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \Rightarrow \frac{\log_{3} (x - 2)}{\log_{3} 9} = \log_{3} 2 \\ \frac{\log_{3} (x - 2)}{2} = \log_{3} 2 \\ \log_{3} (x - 2) = 2 \log_{3} 2 \\ \log_{3} (x - 2) = \log_{3} 2^{2} \\ x - 2 = 4 \\ x = 6 \end{array}

Question 6

Solve the equation,

\log_{9} (2 x + 12) = \log_{3} (x + 2)

Solution:

\begin{array}{l} \log_{9} (2 x + 12) = \log_{3} (x + 2) \\ \frac{\log_{3} (2 x + 12)}{\log_{3} 9} = \log_{3} (x + 2) \\ \log_{3} (2 x + 12) = 2 \log_{3} (x + 2) \\ \log_{3} (2 x + 12) = \log_{3} {(x + 2)}^{2} \\ 2 x + 12 = x^{2} + 4 x + 4 \\ x^{2} + 2 x - 8 = 0 \\ (x + 4) (x - 2) = 0 \\ x = - 4 (not accepted) \\ x = 2 \end{array}

Question 7

Solve the equation,

\log_{4} x = \frac{3}{2} \log_{2} 3

Solution:

\begin{array}{l} \log_{4} x = \frac{3}{2} \log_{2} 3 \\ \frac{\log_{2} x}{\log_{2} 4} = \frac{3}{2} \log_{2} 3 \\ \frac{\log_{2} x}{2} = \frac{3}{2} \log_{2} 3 \\ \log_{2} x = 2 \times \frac{3}{2} \log_{2} 3 \\ \log_{2} x = 3 \log_{2} 3 \\ \log_{2} x = \log_{2} 3^{3} \\ x = 27 \end{array}

Question 8

Solve the equation,

\frac{2}{\log_{5} 2} = \log_{2} (2 - x)

Solution:

\begin{array}{l} \frac{2}{\log_{5} 2} = \log_{2} (2 - x) \\ 2 = \log_{5} 2. \log_{2} (2 - x) \\ 2 = \frac{1}{\log_{2} 5} . \log_{2} (2 - x) \\ 2 \log_{2} 5 = \log_{2} (2 - x) \\ \log_{2} 5^{2} = \log_{2} (2 - x) \\ 25 = 2 - x \\ x = - 23 \end{array}