dan segi tiga $$ D E F $$. D ialah titik tengah bagi $$ A F $$ dan $$ E $$ ialah titik tengah bagi $$ C D $$. Diberi $$ \overrightarrow{A B}=\underset{\sim}{x} $$ dan $$ \overrightarrow{A D}=\underset{\sim}{y} $$, ungkapkan dalam sebutan $$ \underset{\sim}{x} $$ dan/atau $$ \underset{\sim}{y} $$ : The diagram shows a parallelogram $$ A B C D $$ and a triangle $$ D E F . D $$ is the midpoint of $$ A F $$ and $$ E $$ is the midpoint of $$ C D $$. Given $$ \overrightarrow{A B}=\underset{\sim}{x} $$ and $$ \overrightarrow{A D}=\underset{\sim}{y} $$, express in terms of $$ \underset{\sim}{x} $$ andlor $$ \underset{\sim}{y} $$ : $$ \begin{array}{ll} ext { (i) } \overrightarrow{A E} & ext { (ii) } \overrightarrow{A C}+\overrightarrow{C F}\end{array} $$ $$ [4 $$ markah/4 marks]

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dan segi tiga $$ D E F $$. D ialah titik tengah bagi $$ A F $$ dan $$ E $$ ialah titik tengah bagi $$ C D $$. Diberi $$ \overrightarrow{A B}=\underset{\sim}{x} $$ dan $$ \overrightarrow{A D}=\underset{\sim}{y} $$, ungkapkan dalam sebutan $$ \underset{\sim}{x} $$ dan/atau $$ \underset{\sim}{y} $$ : The diagram shows a parallelogram $$ A B C D $$ and a triangle $$ D E F . D $$ is the midpoint of $$ A F $$ and $$ E $$ is the midpoint of $$ C D $$. Given $$ \overrightarrow{A B}=\underset{\sim}{x} $$ and $$ \overrightarrow{A D}=\underset{\sim}{y} $$, express in terms of $$ \underset{\sim}{x} $$ andlor $$ \underset{\sim}{y} $$ : $$ \begin{array}{ll}	ext { (i) } \overrightarrow{A E} & 	ext { (ii) } \overrightarrow{A C}+\overrightarrow{C F}\end{array} $$ $$ [4 $$ markah/4 marks]

UpStudy AI · Accepted Answer

We are given a parallelogram $$ABCD$$ and a triangle $$DEF$$ with the following information:
- $$ \overrightarrow{AB} = \underset{\sim}{x} $$ and $$ \overrightarrow{AD} = \underset{\sim}{y} $$.
- In the parallelogram, using $$A$$ as the origin, we have:
  - $$ B $$ at position $$\underset{\sim}{x}$$.
  - $$ D $$ at position $$\underset{\sim}{y}$$.
  - $$ C $$ must satisfy $$ \overrightarrow{AC} = \underset{\sim}{x}+\underset{\sim}{y} $$ since $$ C $$ is reached by going from $$A$$ to $$B$$ then from $$B$$ to $$C$$ (and the parallelogram property gives $$ \overrightarrow{AD} = \overrightarrow{BC}$$).

- In the triangle $$DEF$$:
  - $$D$$ is the midpoint of $$AF$$. Since $$ \overrightarrow{AD} = \underset{\sim}{y} $$, we have:
    $$
    \underset{\sim}{y} = \frac{1}{2}\overrightarrow{AF} \quad\Longrightarrow\quad \overrightarrow{AF}=2\underset{\sim}{y}.
    $$
    Hence, $$ F $$ is at position $$ 2\underset{\sim}{y} $$ relative to $$ A $$.
  - $$E$$ is the midpoint of $$CD$$. With $$ C $$ at $$\underset{\sim}{x}+\underset{\sim}{y}$$ and $$ D $$ at $$\underset{\sim}{y}$$, the position vector of $$E$$ is:
    $$
    \overrightarrow{E}=\frac{\overrightarrow{C}+\overrightarrow{D}}{2}=\frac{(\underset{\sim}{x}+\underset{\sim}{y})+\underset{\sim}{y}}{2}=\frac{\underset{\sim}{x}+2\underset{\sim}{y}}{2}.
    $$
  - Therefore, 
    $$
    \overrightarrow{AE}=\frac{\underset{\sim}{x}+2\underset{\sim}{y}}{2}.
    $$

Now, we express each required vector in terms of $$\underset{\sim}{x}$$ and/or $$\underset{\sim}{y}$$:

1. (i) For $$\overrightarrow{AE}$$:

As determined,
   $$
   \overrightarrow{AE}=\frac{\underset{\sim}{x}+2\underset{\sim}{y}}{2}.
   $$

2. (ii) For $$\overrightarrow{AC}+\overrightarrow{CF}$$:

We first note that 
   $$
   \overrightarrow{AC} = \underset{\sim}{x}+\underset{\sim}{y}.
   $$
   Next, to find $$\overrightarrow{CF}$$, we calculate:
   $$
   \overrightarrow{CF} = \overrightarrow{F}-\overrightarrow{C}.
   $$
   Since $$F$$ is at $$2\underset{\sim}{y}$$ and $$C$$ is at $$\underset{\sim}{x}+\underset{\sim}{y}$$, we have:
   $$
   \overrightarrow{CF}=2\underset{\sim}{y} - (\underset{\sim}{x}+\underset{\sim}{y}) = -\underset{\sim}{x}+\underset{\sim}{y}.
   $$
   Therefore,
   $$
   \overrightarrow{AC}+\overrightarrow{CF} = (\underset{\sim}{x}+\underset{\sim}{y})+(-\underset{\sim}{x}+\underset{\sim}{y}) = 2\underset{\sim}{y}.
   $$

Thus, the final answers are:

- (i) $$\displaystyle \overrightarrow{AE}=\frac{\underset{\sim}{x}+2\underset{\sim}{y}}{2}$$
- (ii) $$\displaystyle \overrightarrow{AC}+\overrightarrow{CF}=2\underset{\sim}{y}$$
$$ \overrightarrow{AE} = \frac{\underset{\sim}{x} + 2\underset{\sim}{y}}{2} $$ $$ \overrightarrow{AC} + \overrightarrow{CF} = 2\underset{\sim}{y} $$

dan segi tiga . D ialah titik tengah bagi dan
ialah titik tengah bagi . Diberi dan ,
ungkapkan dalam sebutan dan/atau :
The diagram shows a parallelogram and a triangle
is the midpoint of and is the midpoint of .
Given and , express in terms of andlor :

markah/4 marks]

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dan segi tiga . D ialah titik tengah bagi dan ialah titik tengah bagi . Diberi dan , ungkapkan dalam sebutan dan/atau : The diagram shows a parallelogram and a triangle is the midpoint of and is the midpoint of . Given and , express in terms of andlor : markah/4 marks]