Show that the straight lines x+y=0, 3x+y–4=0, x + 3y – 4 = 0 form on Isosceles triangle.

Given
$\begin{array}{l}\mathrm{x}+\mathrm{y}=0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\dots \left(1\right)\\ 3\mathrm{x}+\mathrm{y}-4=0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\dots \left(2\right)\\ \mathrm{x}+3\mathrm{y}-4=0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\dots \left(3\right)\end{array}$
$\mathrm{Let} {\mathrm{\theta }}_1 \mathrm{be} \mathrm{the} \mathrm{angle} \mathrm{between} \mathrm{equation} \left(1\right) \& \left(2\right)$
$\begin{array}{l}\cos {\mathrm{\theta }}_1=\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2}{\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2}}\\ =\frac{3+1}{\sqrt{1+1}\sqrt{9+1}}=\frac{4}{\sqrt{2}\sqrt{10}}=\frac{4}{\sqrt{20}}=\frac{2}{\sqrt{5}}\\ \therefore {\mathrm{\theta }}_1={\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{array}$
$\mathrm{Let} {\mathrm{\theta }}_2 \mathrm{be} \mathrm{the} \mathrm{angle} \mathrm{between} \mathrm{equation} \left(2\right) \& \left(3\right)$
$\therefore \cos {\mathrm{\theta }}_2=\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2}}$
$\begin{array}{l}=\frac{3\left(1\right)+1\left(3\right)}{\sqrt{9+1}\sqrt{9+1}}=\frac{6}{\sqrt{10}\sqrt{10}}=\frac{6}{10}=\frac{3}{5}\\ {\mathrm{\theta }}_2={\cos }^{-1}\left(\frac{3}{5}\right)\end{array}$
$\text{ Let }{\mathrm{\theta }}_3\text{ be the angle between equation (1) \& (3) }$
$\begin{array}{l}\therefore \cos {\mathrm{\theta }}_3=\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2}}\\ =\frac{1\left(1\right)+3\left(1\right)}{\sqrt{1+1}\sqrt{1+9}}=\frac{4}{\sqrt{20}}=\frac{2}{\sqrt{5}}\\ {\mathrm{\theta }}_3={\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right);\text{ since }{\mathrm{\theta }}_1={\mathrm{\theta }}_3\end{array}$
$\therefore$The triangle formed by the given lines is an isosceles triangle