Arun wants to go to Anu's home. Anu's home is $5\text{km}$ North-West of Priya's home which is $\sqrt{12.5}\text{km}$ East of Tharun's home. Tharun's home is $5\text{km}$ North-West of Arun's home. In which direction is Arun's home from Anu's home?

Let's establish a 2D coordinate system $(x,y)$ where East is along the positive $x$ -axis and North is along the positive $y$ -axis.
1. Let Tharun's home be at the origin: $T=(0,0)$.
2. Priya's home is $\sqrt{12.5}\text{km}$ East of Tharun's home:

$$\sqrt{12.5}=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}}\text{km}$$

So, $P=\left(\frac{5}{\sqrt{2}},0\right)$.
3. Anu's home ($A$) is $5\text{km}$ North-West of Priya's home. Moving North-West means a displacement of $-\frac{5}{\sqrt{2}}$ in the $x$ -direction and $+\frac{5}{\sqrt{2}}$ in the $y$ -direction:

$$A=\left(\frac{5}{\sqrt{2}}-\frac{5}{\sqrt{2}},0+\frac{5}{\sqrt{2}}\right)=\left(0,\frac{5}{\sqrt{2}}\right)$$

4. Tharun's home $T(0,0)$ is $5\text{km}$ North-West of Arun's home ($Ar$). This means Arun's home is $5\text{km}$ South-East of Tharun's home. A South-East displacement means $+\frac{5}{\sqrt{2}}$ in the $x$ -direction and $-\frac{5}{\sqrt{2}}$ in the $y$ -direction:

$$Ar=\left(0+\frac{5}{\sqrt{2}},0-\frac{5}{\sqrt{2}}\right)=\left(\frac{5}{\sqrt{2}},-\frac{5}{\sqrt{2}}\right)$$

Now, let's find the position of Arun's home relative to Anu's home by computing vector $\overrightarrow{AAr}$:

$$\overrightarrow{AAr}=Ar-A=\left(\frac{5}{\sqrt{2}}-0,-\frac{5}{\sqrt{2}}-\frac{5}{\sqrt{2}}\right)=\left(\frac{5}{\sqrt{2}},-\frac{10}{\sqrt{2}}\right)$$

Since the $x$ -coordinate is positive (East) and the $y$ -coordinate is negative (South), Arun's home is in the South-East direction from Anu's home.