if the lines joining origin and point of intersection of curves ax2+2hxy+by2+2gx=0 and a1x2+2h1xy+b1y2+g1x=0 are mutually perpendicular then proove that g(a1+b1)=g1(a+b)

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$$\begin{gathered} \mathrm{To} \mathrm{get} \mathrm{equation} \mathrm{of} \mathrm{pair} \mathrm{of} \mathrm{straight} \mathrm{lines}, \mathrm{make} \mathrm{any} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equations} \\ \mathrm{homogenous}. \\ \mathrm{We} \mathrm{have} \\ {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}=0 \\ \Rightarrow {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}\left(1\right)=0 ;\mathrm{equation}\left(\mathrm{i}\right) \\ {\mathrm{a}}_1{\mathrm{x}}^2+2{\mathrm{h}}_1\mathrm{xy}+{\mathrm{b}}_1{\mathrm{y}}^2+2{\mathrm{g}}_1\mathrm{x}=0 \\ \Rightarrow {\mathrm{a}}_1{\mathrm{x}}^2+2{\mathrm{h}}_1\mathrm{xy}+{\mathrm{b}}_1{\mathrm{y}}^2=-2{\mathrm{g}}_1\mathrm{x} \\ \Rightarrow \frac{{\mathrm{a}}_1{\mathrm{x}}^2+2{\mathrm{h}}_1\mathrm{xy}+{\mathrm{b}}_1{\mathrm{y}}^2}{-2{\mathrm{g}}_1\mathrm{x}}=1 \\ \mathrm{Putting} \mathrm{value} \mathrm{of} \mathrm{x} \mathrm{in} \mathrm{equation}\left(\mathrm{i}\right), \mathrm{we} \mathrm{get}, \\ {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}\left(\frac{{\mathrm{a}}_1{\mathrm{x}}^2+2{\mathrm{h}}_1\mathrm{xy}+{\mathrm{b}}_1{\mathrm{y}}^2}{-2{\mathrm{g}}_1\mathrm{x}}\right)=0 \\ \Rightarrow {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2-\mathrm{g}\left(\frac{{\mathrm{a}}_1{\mathrm{x}}^2+2{\mathrm{h}}_1\mathrm{xy}+{\mathrm{b}}_1{\mathrm{y}}^2}{{\mathrm{g}}_1}\right)=0 \\ \Rightarrow {\mathrm{x}}^2\left(\mathrm{a}-\frac{{\mathrm{a}}_1\mathrm{g}}{{\mathrm{g}}_1}\right)+2\left(\mathrm{h}-\frac{{\mathrm{h}}_1\mathrm{g}}{{\mathrm{g}}_1}\right)\mathrm{xy}+{\mathrm{y}}^2\left(\mathrm{b}-\frac{{\mathrm{b}}_1\mathrm{g}}{{\mathrm{g}}_1}\right)=0 \\ \mathrm{This} \mathrm{is} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{pair} \mathrm{of} \mathrm{straight} \mathrm{lines} \mathrm{and} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{lines} \mathrm{are} \mathrm{perpendicular}, \mathrm{hence} \\ \mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^2+\mathrm{Coefficient} \mathrm{of} {\mathrm{y}}^2=0 \\ \mathrm{a}-\frac{{\mathrm{a}}_1\mathrm{g}}{{\mathrm{g}}_1}+\mathrm{b}-\frac{{\mathrm{b}}_1\mathrm{g}}{{\mathrm{g}}_1}=0 \\ \Rightarrow \frac{{\mathrm{ag}}_1-{\mathrm{a}}_1\mathrm{g}+{\mathrm{bg}}_1-{\mathrm{b}}_1\mathrm{g}}{{\mathrm{g}}_1}=0 \\ \Rightarrow {\mathrm{g}}_1\left(\mathrm{a}+\mathrm{b}\right)-\mathrm{g}\left({\mathrm{a}}_1+{\mathrm{b}}_1\right)=0 \\ \mathbf{\Rightarrow }\mathbf{g}\left({\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\right)\mathbf{=}{\mathbf{g}}_{\mathbf{1}}\left(\mathbf{a}\mathbf{+}\mathbf{b}\right)\mathbf{ }\left(\mathbf{Hence}\mathbf{ }\mathbf{proved}\right) \end{gathered}$$

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