Variable straight lines ${\mathrm{L}}_1:\mathrm{y}=2\mathrm{x}+{\mathrm{c}}_1$ and ${\mathrm{L}}_2:\mathrm{y}=2\mathrm{x}+{\mathrm{c}}_2$ meet the x-axis in A₁ and A₂ respectively and y-axis in B₁ and B₂ respectively locus of intersection point of A₁B₂ and A₂B₁ is

${\mathrm{L}}_1:\mathrm{y}=2\mathrm{x}+{\mathrm{c}}_1\Rightarrow {\mathrm{L}}_2:\mathrm{Y}=2\mathrm{x}+{\mathrm{c}}_2$

$\begin{array}{l}\Rightarrow \text{ meet }\mathrm{y}\text{-axis }{\mathrm{B}}_1\left(0,{\mathrm{c}}_1\right){\mathrm{B}}_2\left(0,{\mathrm{c}}_2\right)\\ \Rightarrow \text{ meet X-axis }{\mathrm{A}}_1\mathrm{\&}{\mathrm{A}}_2=\left(\frac{-{\mathrm{c}}_1}{2},0\right)\left(\frac{-{\mathrm{c}}_2}{2},0\right)\end{array}$

equation ${\mathrm{A}}_1{\mathrm{B}}_2\text{ in }\frac{2\mathrm{x}}{-{\mathrm{c}}_1}+\frac{\mathrm{y}}{{\mathrm{c}}_2}=1$…..(1)

equation ${\mathrm{A}}_1{\mathrm{B}}_2\text{ in }\frac{2\mathrm{x}}{-{\mathrm{c}}_2}+\frac{\mathrm{y}}{{\mathrm{c}}_1}=1$…..(2)

$\Rightarrow$ eliminate ${\mathrm{c}}_1 \mathrm{\&} {\mathrm{c}}_2$ then $\mathrm{y}+2\mathrm{x}=0$