If OA and OB are two equal chords of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}=0$ perpendicular to each other and passing through the origin, the slopes of OA and OB are the roots of the equation

Given that the chords $\overleftrightarrow{OA},\overleftrightarrow{OB}$ are perpendicular

Equal chords are at equal distances from centre 

$$\begin{gathered} \begin{array}{l}\overleftrightarrow{\mathrm{OA}}\overleftrightarrow{\mathrm{OB}}\text{ be }\mathrm{y}=\mathrm{mx},\mathrm{x}+\mathrm{my}=0\\ \Rightarrow {\mathrm{d}}_1={\mathrm{d}}_2\Rightarrow \frac{|\mathrm{m}+2|}{\sqrt{{\mathrm{m}}^2}+1}=\frac{|1-2\mathrm{m}|}{\sqrt{{\mathrm{m}}^2}+1}\end{array} \\ \therefore {\left(m+2\right)}^2={\left(1-2m\right)}^2\Rightarrow m^2+4m+4=1-4m+4m^2 \\ \Rightarrow 3m^2-8m-3=0 \end{gathered}$$