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Short Questions (Question 3, 4 & 5)

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Question 3:
The equation of the straight lines CD and EF are 5x + y – 4 = 0 and x 7 − y h = 1 . If CD and EF are parallel, find the value of h.

Solution:
Two parallel lines have same gradient.
 5x + y – 4 = 0  
 y = –5x + 4, mCD = –5

 For the straight line EF,
x 7 − y h = 1 m E F = − ( y -intercept x -intercept ) = − ( − h 7 ) = h 7 m C D = m E F − 5 = h 7 h = − 35




Question 4:
The straight line x 5 + y p = 1 has a y-intercept of 3 and is parallel to the straight line y + qx = 0. Determine the value of p and of q.

Solution:
Given y-intercept of the straight line  x 5 + y p = 1  is  3 , ∴ p = 3 Gradient of the straight line =  − 3 5 For another straight line  y + q x = 0 ,   y = − q x As two parallel lines have same gradient, − q = − 3 5 q = 3 5



Question 5:
The equations of two straight lines   y 7 + x 4 = 1 y = 4x + 21. Determine whether the lines are perpendicular to each other.

Solution:
For the straight line  y 7 + x 4 = 1 ,  gradient  = − 7 4 For the straight line  7 y = 4 x + 21 , y = 4 7 x + 3 ,  gradient  = 4 7 − 7 4 × 4 7 = − 1

Hence, the two straight lines are perpendicular to each other.


Short Questions (Question 1 & 2)

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Question 1:
The vertices of a triangle are P (6, 1), Q (5, 6) and R (m, -1). Given that the area of the triangle is 31 unit2, find the values of m.

Solution:
Area of  △   P Q R  =  31 1 2   | 6  5    m   1  6 − 1    6 1 | = 31 | ( 6 ) ( 6 ) + ( 5 ) ( − 1 ) + ( m ) ( 1 ) − ( 1 ) ( 5 ) − ( 6 ) ( m ) − ( − 1 ) ( 6 ) | = 62 | 36 − 5 + m − 5 − 6 m + 6 | = 62 | 32 − 5 m | = 62 32 − 5 m = ± 62 − 5 m = 62 − 32  or  − 5 m = − 62 − 32 m = − 6  or  m = 94 5



Question 2:
The points P (3m, m), Q (t, u) and R (3t, 2u) lie on a straight line. Q divides PR in the ratio 3 : 2. Express t in terms of u.

Solution:
( ( 3 m ) ( 2 ) + ( 3 t ) ( 3 ) 3 + 2 , ( m ) ( 2 ) + ( 2 u ) ( 3 ) 3 + 2 ) = ( t , u ) 6 m + 9 t 5 = t 6 m + 9 t = 5 t 6 m = − 4 t m = − 2 3 t → ( 1 ) 2 m + 6 u 5 = u 2 m + 6 u = 5 u 2 m = − u From (1), 2 ( − 2 t 3 ) = − u t = 3 4 u

6.6 Equation of a Locus

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6.6 Equation of a Locus
1. The equation of the locus of a moving point P(x, y) which is always at a constant distance (r) from a fixed point (x1, y1) is:


2. The equation of the locus of a moving point P(x, y) which is always at a constant distance from two fixed points (x1, y1) and  (x1, y1) with a ratio is:



3. The equation of the locus of a moving point P (x, y) which is always equidistant from two fixed points A and B is the perpendicular bisector of the straight line AB.


Example 1
Find the equation of the locus of a moving point P(x, y) which is always at a distance of 5 units from a fixed point Q (2, 4).

Solution:
(x – x1)2+ (y – y1)2 = r2
(x – 2)2 + (y – 4)2 = 52
x2 – 4x + 4 + y2 – 8y + 16 = 25
x2 + y2– 4x – 8y – 5 = 0



Example 2
Find the equation of the locus of a moving point P(x, y) which is always equidistant from points A (-2, 3) and B (4, -1).

Solution:
Given  P A = P B ( x − ( − 2 ) ) 2 + ( y − 3 ) 2 = ( x − 4 ) 2 + ( y − ( − 1 ) ) 2 Square both sides to eliminate the square roots . ( x + 2 ) 2 + ( y − 3 ) 2 = ( x − 4 ) 2 + ( y + 1 ) 2 x 2 + 2 x + 4 + y 2 − 6 y + 9 = x 2 − 8 x + 16 + y 2 + 2 y + 1 10 x − 8 y − 4 = 0 Hence, the equation of the locus of point  P  is 10 x − 8 y − 4 = 0


Example 3
A (2, 0) and B (0, -2) are two fixed points. Point P moves with a ratio so that AP: PB = 1: 2.  Find the equation of the locus of point P.

Solution:
A P : P B = 1 : 2 A P P B = 1 2 2 A P = P B 2 ( x − 2 ) 2 + ( y − 0 ) 2 = ( x − 0 ) 2 + ( y − ( − 2 ) ) 2 Square both sides to eliminate the square roots . 4 [ ( x − 2 ) 2 + y 2 ] = x 2 + ( y + 2 ) 2 4 ( x 2 − 4 x + 4 + y 2 ) = x 2 + y 2 + 4 y + 4 4 x 2 − 16 x + 16 + 4 y 2 = x 2 + y 2 + 4 y + 4 3 x 2 + 3 y 2 − 16 x − 4 y + 12 = 0 Hence, the equation of the locus of point  P  is 3 x 2 + 3 y 2 − 16 x − 4 y + 12 = 0

6.5 Parallel Lines and Perpendicular Lines

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6.5 Parallel Lines and Perpendicular Lines

(A) Parallel Lines
1. If two straight lines are parallel, they have same gradient.


In the above diagram, if straight line L1 is parallel to straight line L2, gradient of L1 = gradient of L2
m 1 = m 2

Example 1:
Given that the equation of a straight line parallel to x + 8y= 40 and passes through the point A(2, 3k) and B (-6, 4k2), find the values of k.

Solution:
x + 8 y = 40 8 y = − x + 40 y = − 1 8 x + 5 gradient  m 1 = − 1 8 Given a straight line passes through  point  A  and point  B  is parallel to  x + 8 y = 40 , ∴ m 1 = m 2 − 1 8 = 4 k 2 − 3 k − 6 − 2 8 = 32 k 2 − 24 k 1 = 4 k 2 − 3 k 4 k 2 − 3 k − 1 = 0 ( 4 k + 1 ) ( k − 1 ) = 0 4 k + 1 = 0    or    k − 1 = 0 k = − 1 4   or k = 1




(B) Perpendicular Lines
1. If two lines are perpendicular to each other, the product of their gradients is –1.


In the above diagram, if straight line L1 is perpendicular to straight line L2,
gradient of L1  ×  gradient of L2 = –1 
m 1 × m 2 = − 1

Example 2:
Given that points P (–2, 4), Q (4, 2), R (–1, –3) and S (2, 6), show that PQ is perpendicular to RS.

Solution:
m P Q = 2 − 4 4 − ( − 2 ) = − 1 3 m R S = 6 − ( − 3 ) 2 − ( − 1 ) = 3 ( m P Q ) ( m R S ) = ( − 1 3 ) ( 3 ) = − 1

Hence, PQ is perpendicular to RS.

6.4 Equation of Straight Lines (Part 2)

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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:
y − y 1 x − x 1 = y 2 − y 1 x 2 − x 1 Let  ( x 1 , y 1 ) = ( 2 ,   4 )  and  ( x 2 , y 2 )  =  ( 5 ,   6 ) y − 4 x − 2 = 6 − 4 5 − 2 y − 4 x − 2 = 2 3 3 y − 12 = 2 x − 4 3 y = 2 x + 8


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
x a + y b = 1 x 5 + y ( − 6 ) = 1 x 5 − y 6 = 1


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

(a)



(b)



(c)




6.4 Equation of Straight Lines (Part 1)

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6.4.1 Axes Intercepts and Gradient
(A) Formula for gradient:
1. Gradient of the line joining (x1, yl) 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:
Gradient,  m = y 2 − y 1 x 2 − x 1 − 3 − ( 1 − k ) − 3 k − k = 5 − 3 − 1 + k − 4 k = 5 − 4 + k = − 20 k 21 k = 4 k = 4 21


Example 2:
Based on the diagram below, find the gradient of the line.
Solution:
Gradient,  m = − ( y  intercept x  intercept ) = − ( − 5 10 ) = 1 2

6.3 Areas of Polygons

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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:

Area of  △   A B C = 1 2 | − 5     − 2    4     5   − 4     − 1    − 5    5 | = 1 2 | ( − 5 ) ( − 4 ) + ( − 2 ) ( − 1 ) + ( 4 ) ( 5 )    − ( 5 ) ( − 2 ) − ( − 4 ) ( 4 ) − ( − 1 ) ( − 5 ) | = 1 2 | 20 + 2 + 20 + 10 + 16 − 5 | = 1 2 | 63 | = 31 1 2  unit 2


(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, 9 2 ) are collinear.

Solution:
Area of  △   P Q R = 0 1 2   | 2   6   3 k   1    k    9 2   2 1 | = 0 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 = − 4 3   or   3

6.2 Division of a Line Segment

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6.2 Division of a Line Segment

(A) Midpoints of a Line Segment


Formula for the midpoint, M of A (xl, 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:

B  is the midpoint of  A C ( m − 4 ,   3 ) = ( − 1 + 5 2 ,   n + 8 2 ) ( m − 4 ,   3 ) = ( 2 ,   n + 8 2 ) m − 4 = 2  and n + 8 2 = 3 m = 6 and    n + 8 = 6    n = − 2


(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:


Let point  B = ( p ,   q ) ( 2 ( − 1 ) + 3 p 3 + 2 ,   2 ( 2 ) + 3 q 3 + 2 ) = ( 2 , − 1 ) ( − 2 + 3 p 5 ,   4 + 3 q 5 ) = ( 2 , − 1 ) Equating the  x -coordinates, − 2 + 3 p 5 = 2 − 2 + 3 p = 10 3 p = 12 p = 4 Equating the  y -coordinates, 4 + 3 q 5 = − 1 4 + 3 q = − 5 3 q = − 9 q = − 3 ∴  The coordinates of point  B = ( 4 , − 3 ) .

6.1 Distance between Two Points

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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 ).

Distance of  P Q = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( − 4 − 2 ) 2 + ( − 5 − ( − 2 ) ) 2 = 36 + 9 = 45 = 9 × 5 = 3 5  or  ( 6.71 )



Long Questions (Question 3 – 4)

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5.6.2 Indices and Logarithms, SPM Practice (Long Questions)

Question 3:
Given that p = 3r and q = 3t, express the following in terms of r and/ or t.
(a)  (a)  log 3 p q 2 27 ,
(b)  log9p – log27 q.

Solution:
(a)
Given p = 3r, log3 p = r
q= 3t, log3 q =t

log 3 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
= log 3 p log 3 9 − log 3 q log 3 27 = r log 3 3 2 − t log 3 3 3 = r 2 log 3 3 − t 3 log 3 3 = r 2 − t 3   



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
= log 2 ( 2 x + 1 ) − 5 log 2 x 2 log 2 4 + 4 log 2 x = log 2 ( 2 x + 1 ) − 5 2 log 2 x 2 + log 2 x 4 = log 2 ( 2 x + 1 ) − log 2 ( x 2 ) ( 5 2 ) + log 2 x 4
= log 2 ( 2 x + 1 ) − log 2 x 5 + log 2 x 4 = log 2 ( 2 x + 1 ) ( x 4 ) x 5 = log 2 2 x + 1 x

(b)
log2 (2x + 1) – 5 log4 x2 + 4 log2 x = 4
log 2 2 x + 1 x = 4    2 x + 1 x = 2 4    2 x + 1 x = 16    2 x + 1 = 16 x  14 x = 1 x = 1 14