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$Paradigm Specialising in O Level Mathernatics 5 In the diagram, ABC is a triangle where M is the midpoint of AC. The point P lies on the line BM such that BP=(1)/(3)BM. The point Q lies on the line BC such that BQ:QC=1:4. Given that vec(AB)=b and vec(AC)=c. (a) Express, as simply as possible, in terms of b and/or c, (i) vec(BC), (ii) vec(BM), (iii) vec(AP), (iv) vec(AQ). (b) Write down two facts about A,P and Q. (c) Find the numeral value of (" Area of "/_\ABP)/(" Area of quadrilateral "CMPQ), Ans: (a)(i) c-b (ii) (1)/(2)c-b (iii) (1)/(6)(4b+c) (iv) (1)/(5)(4b+c) (b) A,P and Q are collinear AP:AQ=5:6 (c) (5)/(14)$

Paradigm $\newline$ Specialising in O Level Mathernatics $\newline$ $5$ In the diagram, $A B C$ is a triangle where $M$ is the midpoint of $A C$ . $\newline$ The point $P$ lies on the line $B M$ such that $B P=\frac{1}{3} B M$ . $\newline$ The point $Q$ lies on the line $B C$ such that $B Q: Q C=1: 4$ . $\newline$ Given that $\overrightarrow{A B}=b$ and $M$ $0$ . $\newline$ (a) Express, as simply as possible, in terms of $M$ $1$ and/or $M$ $2$ , $\newline$ (i) $M$ $3$ , $\newline$ (ii) $M$ $4$ , $\newline$ (iii) $M$ $5$ , $\newline$ (iv) $M$ $6$ . $\newline$ (b) Write down two facts about $M$ $7$ and $Q$ . $\newline$ (c) Find the numeral value of $M$ $9$ , $\newline$ Ans: $\newline$ (a)(i) $A C$ $0$ $\newline$ (ii) $A C$ $1$ $\newline$ (iii) $A C$ $2$ $\newline$ (iv) $A C$ $3$ $\newline$ (b) $M$ $7$ and $Q$ are collinear $A C$ $6$ $\newline$ (c) $A C$ $7$

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Math Problems

Algebra 2

Write equations of cosine functions using properties

Full solution

Q. Paradigm

\newline

Specialising in O Level Mathernatics

\newline

5

In the diagram,

A B C

is a triangle where

M

is the midpoint of

A C

\newline

The point

P

lies on the line

B M

such that

B P=\frac{1}{3} B M

\newline

The point

Q

lies on the line

B C

such that

B Q: Q C=1: 4

\newline

Given that

\overrightarrow{A B}=b

and

M

0

\newline

(a) Express, as simply as possible, in terms of

M

1

and/or

M

2

\newline

(i)

M

3

\newline

(ii)

M

4

\newline

(iii)

M

5

\newline

(iv)

M

6

\newline

(b) Write down two facts about

M

7

and

Q

\newline

M

9

\newline

Ans:

\newline

(a)(i)

A C

0

\newline

(ii)

A C

1

\newline

(iii)

A C

2

\newline

(iv)

A C

3

\newline

(b)

M

7

and

Q

are collinear

A C

6

\newline

(c)

A C

7

Vector BC Calculation: For $\vec{BC}$ , since $M$ is the midpoint of $AC$ , $\vec{BC} = \vec{BM} + \vec{MC} = \vec{BM} + \vec{MA} = \vec{BM} - \vec{AM}$ . Since $\vec{AM} = \frac{1}{2}\vec{AC}$ , $\vec{BC} = \vec{BM} - \frac{1}{2}\vec{AC}$ . But $\vec{BM} = \vec{BA} + \vec{AM} = -\vec{AB} + \frac{1}{2}\vec{AC} = -b + \frac{1}{2}c$ . Therefore, $\vec{BC} = (-b + \frac{1}{2}c) - \frac{1}{2}c = -b$ .
Vector BM Calculation: For $\vec{BM}$ , since $M$ is the midpoint of $AC$ , $\vec{BM} = \vec{BA} + \vec{AM} = -\vec{AB} + \frac{1}{2}\vec{AC} = -b + \frac{1}{2}c$ .
Vector AP Calculation: For $\vec{AP}$ , since $P$ divides $BM$ in the ratio $1:2$ ( $BP:PM = 1:2$ ), \vec{AP} = \vec{AB} + \vec{BP} = \vec{AB} + \frac{$1$}{$3$}\vec{BM} = \mathbf{b} + \frac{$1$}{$3$}(-\mathbf{b} + \frac{$1$}{$2$}\mathbf{c}) = \mathbf{b} - \frac{$1$}{$3$}\mathbf{b} + \frac{$1$}{$6$}\mathbf{c} = \frac{$2$}{$3$}\mathbf{b} + \frac{$1$}{$6$}\mathbf{c}.
Vector AQ Calculation: For $\vec{AQ}$, since $Q$ divides $BC$ in the ratio $1:4$ ($BQ:QC = 1:4$), \vec{AQ} = \vec{AB} + \vec{BQ} = \vec{AB} + \left(\frac{ $1$ }{ $5$ }\right)\vec{BC} = b + \left(\frac{ $1$ }{ $5$ }\right)(-b) = b - \left(\frac{ $1$ }{ $5$ }\right)b = \left(\frac{ $4$ }{ $5$ }\right)b.
Collinearity of Points: For part (b), since $P$ lies on $BM$ and $Q$ lies on $BC$ , and $M$ is the midpoint of $AC$ , it follows that $A$ , $P$ , and $Q$ are collinear. Also, since $AP:AQ = BP:BQ = 1:4$ , $BM$ $0$ .
Area Ratio Calculation: For part (c), to find the area ratio, we use the fact that the areas of triangles with the same height are proportional to their bases. The area of triangle $ABP$ is proportional to $BP$ , and the area of quadrilateral $CMPQ$ is the sum of the areas of triangles $CMB$ and $CMQ$ , which are proportional to $BM$ and $MQ$ , respectively. Since $BP = \frac{1}{3}BM$ and $MQ = \frac{4}{5}BC$ , and $BC = 2BM$ , $BP$ $0$ . The total base for the quadrilateral is $BP$ $1$ . Therefore, the area ratio is $BP$ $2$ . But we made a mistake, we forgot to consider the triangle $CMB$ in the area of the quadrilateral, which should be added to $MQ$ to get the total base. So, the correct total base for the quadrilateral is $BP$ $5$ . Therefore, the correct area ratio is $BP$ $6$ . But we made another mistake, the correct ratio of $BP$ to the total base of the quadrilateral is $BP$ $8$ . This is the correct area ratio.