Show that the distance of the point (6,-2) from the line 4x+3y=12 is half the distance of the point (3,4) from the line 4x-3y=12.

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$$\begin{gathered} \mathrm{We} \mathrm{have}: 4\mathrm{x}+3\mathrm{y}=12 \\ \Rightarrow 4\mathrm{x}+3\mathrm{y}-12=0 \\ \mathrm{Distance} \mathrm{of} \mathrm{point} \left(6,-2\right) \mathrm{from} 4\mathrm{x}+3\mathrm{y}-12=0 \mathrm{is} \mathrm{given} \mathrm{by}: \\ {\mathrm{d}}_1=\frac{\left|4\times 6+3\times \left(-2\right)-12\right|}{\sqrt{4^2+3^2}} \\ =\frac{\left|24-6-12\right|}{\sqrt{16+9}} \\ =\frac{\left|6\right|}{\sqrt{25}} \\ \Rightarrow {\mathrm{d}}_1=\frac{6}{5} .......\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{have} 4\mathrm{x}-3\mathrm{y}=12 \\ \Rightarrow 4\mathrm{x}-3\mathrm{y}-12=0 \\ \mathrm{Distance} \mathrm{of} \mathrm{point} \left(3,4\right) \mathrm{from} 4\mathrm{x}-3\mathrm{y}-12=0 \mathrm{is} \mathrm{given} \mathrm{by}: \\ {\mathrm{d}}_2=\frac{\left|4\times 3+\left(-3\right)\times 4-12\right|}{\sqrt{4^2+3^2}} \\ =\frac{\left|12-12-12\right|}{\sqrt{16+9}} \\ =\frac{\left|-12\right|}{\sqrt{25}} \\ \Rightarrow {\mathrm{d}}_2=\frac{12}{5} .......\left(\mathrm{ii}\right) \\ \mathrm{By} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{have}: \\ {\mathbf{d}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{2}}}{\mathbf{2}}\mathbf{ }\left(\mathbf{Hence}\mathbf{ }\mathbf{proved}\right) \end{gathered}$$

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