An electric dipole is formed by two charges + q and - q located in x y -plane at ( 0 ,   2 )   mm…

Electric potential at any general point for vertical dipole is written as, $V_1\propto \frac{p_1\cos {\theta }_1}{{\left(r_1\right)}^2}$.

Electric potential at any general point for $V_2\propto \frac{p_2\cos {\theta }_2}{{\left(r_2\right)}^2}$.

For first case: 

$$\begin{gathered} \tan {\theta }_1=\frac{100}{100}=1 \\ \Rightarrow {\theta }_1=45° \end{gathered}$$

For second case:

slope in this is $=\frac{-2-2}{1-\left(-1\right)}=-2$

which is equal to 

$$\begin{gathered} \tan \left(45°+{\theta }_2\right)=-2 \\ \Rightarrow \frac{1+\tan {\theta }_2}{1-\tan {\theta }_2}=-2 \\ \Rightarrow 1+\tan {\theta }_2=-2+2\tan {\theta }_2 \\ \Rightarrow \tan {\theta }_2=3 \\ \cos {\theta }_2=\frac{1}{\sqrt{3^2+1^2}}=\frac{1}{\sqrt{10}} \end{gathered}$$

also $r_1=r_2$.

Therefore, required ratio

$\frac{V_2}{V_1}=\frac{p_2\cos {\theta }_2}{p_1\cos {\theta }_1}$

$$\begin{gathered} \frac{V_2}{V_1}=\frac{q{l}_2\times {\displaystyle \frac{1}{\sqrt{10}}}}{q{l}_1\times {\displaystyle \frac{1}{\sqrt{2}}}} \\ =\frac{\sqrt{{\left(2-\left(-2\right)\right)}^2+{\left(-1-1\right)}^2}}{4\times \sqrt{5}} \\ =\frac{2\sqrt{5}}{4\times \sqrt{5}}=\frac{1}{2} \end{gathered}$$

$\Rightarrow V_2=\frac{V_0}{2}$