# SPM Additional Mathematics (Model Test Paper)

SPM Additional Mathematics (Model Test Paper)

Section A
[40 marks]
Answer all questions.

Question 5
Diagram 2 shows a triangle EFG.

$\text{It is given}ER:RF=1:2,\text{}FG:TG=3:1,\text{}\stackrel{\to }{ER}=4\underset{˜}{x}\text{and}\stackrel{\to }{EG}=6\underset{˜}{y}.$
$\begin{array}{l}\text{(a) Express in terms of}\underset{˜}{x}\text{and}\underset{˜}{y}:\\ \text{(i)}\stackrel{\to }{GR}\\ \text{(ii)}\stackrel{\to }{GT}\end{array}$
[3 marks]

(b) If line GR is extended to point K such that $\stackrel{\to }{GK}=h\stackrel{\to }{GR}\text{and}\stackrel{\to }{EK}=6\underset{˜}{x}-3\underset{˜}{y},\text{find the value of}h.$
[3 marks]
Answer and Solution:
(a)(i)
$\begin{array}{l}\stackrel{\to }{GR}=\stackrel{\to }{GE}+\stackrel{\to }{ER}\\ =-6\underset{˜}{y}+4\underset{˜}{x}\end{array}$

(a)(ii)
$\begin{array}{l}\stackrel{\to }{GT}=\frac{1}{3}\stackrel{\to }{GF}\\ \text{}=\frac{1}{3}\left(\stackrel{\to }{GR}+\stackrel{\to }{RF}\right)\\ \text{}=\frac{1}{3}\left(-6\underset{˜}{y}+4\underset{˜}{x}+8\underset{˜}{x}\right)←\overline{)\stackrel{\to }{RF}=2\stackrel{\to }{ER}}\\ \text{}=-2\underset{˜}{y}+4\underset{˜}{x}\end{array}$

(b)
$\begin{array}{l}\stackrel{\to }{GK}=h\stackrel{\to }{GR}\\ \stackrel{\to }{GE}+\stackrel{\to }{EK}=h\stackrel{\to }{GR}\\ -6\underset{˜}{y}+\left(6\underset{˜}{x}-3\underset{˜}{y}\right)=h\left(-6\underset{˜}{y}+4\underset{˜}{x}\right)\\ -9\underset{˜}{y}+6\underset{˜}{x}=-6h\underset{˜}{y}+4h\underset{˜}{x}\\ \text{Compare,}\\ -6h=-9\\ h=\frac{3}{2}\end{array}$

Question 6
Solution by scale drawing is not accepted.

Diagram 3 shows three points, A, B and D in a Cartesian plane. The straight line AB is perpendicular to straight line BD which intersect the y-axis at D. The equation of the line BDis y = 2x – 5.
(a)    Find the equation of the straight line AB.                                                             [2 Marks]
(b)   The straight line AB is extended to a point C such that AB : BC= 2 : 3. Find the coordinates of C.                                                                                                                           [3 Marks]

(c)    Point P moves such that its distance from A is equal to its distance from B. Find the equation of the locus P.                                                                                                       [2 Marks]

Answer and Solution:
(a)
y= 2x – 5
mBD = 2, mAB = –½
y– 6 = –½ (x – 3)
2y – 12 = –x + 3
2y = –x + 15

(b)
y= 2x – 5 ----- (1)
2y = –x + 15 ----- (2)
Substitute (1) into (2),
2(2x – 5) = –x + 15
4x – 10 = –x + 15
5x = 25
x = 5
When x = 5, From (1)
y= 2(5) – 5 = 5
B = (5, 5)

AB: BC = 2 : 3
Let coordinates of point C = (h, k)
$\begin{array}{l}\left(\frac{3\left(3\right)+2h}{3+2},\frac{3\left(6\right)+2k}{3+2}\right)=\left(5,5\right)\\ \frac{9+2h}{5}=5\text{or}\frac{18+2k}{5}=5\\ 9+2h=25\text{}18+2k=25\\ h=8\text{}k=3\frac{1}{2}\\ \text{Coordinates of point}C=\left(8,\text{}3\frac{1}{2}\right)\end{array}$

(c)
Let point P= (x, y)
√(x – 3)2+ (y – 6)2 = √(x – 5)2 + (y – 5)2
(x – 3)2+ (y – 6)2 = (x – 5)2 + (y – 5)2  ← (Square for the both sides)
x2– 6x + 9 + y2 – 12y + 36 = x2 – 10x + 25 + y2 – 10y + 25
4x – 2y– 5 = 0