Find the distance of the  point (-5,4) from the line 4x-5y-7=0 . Measured parallel to the line 2x+6y+5=0.

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$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{measure} \mathrm{distance} \mathrm{parallel} \mathrm{to} \mathrm{line} 2\mathrm{x}+6\mathrm{y}+5=0 \\ \mathrm{Now} \mathrm{equation} \mathrm{of} \mathrm{line} \mathrm{paralle} \mathrm{to} 2\mathrm{x}+6\mathrm{y}+5=0, \mathrm{is} \mathrm{given} \mathrm{by}: \\ 2\mathrm{x}+6\mathrm{y}=\mathrm{\lambda } \\ \mathrm{As} \mathrm{line} \mathrm{passes} \mathrm{through} \left(-5,4\right) \\ \Rightarrow -10+24=\mathrm{\lambda } \\ \Rightarrow \mathrm{\lambda }=14 \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{line} \mathrm{on} \mathrm{which} \mathrm{point} \mathrm{lies} \mathrm{is} \\ 2\mathrm{x}+6\mathrm{y}=14 \\ \Rightarrow 2\mathrm{x}+6\mathrm{y}-14=0 ;\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{measure} \mathrm{distance} \mathrm{from} \mathrm{the} \mathrm{line}: \\ 4\mathrm{x}-5\mathrm{y}-7=0 ;\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)\times 2-\left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ 4\mathrm{x}+12\mathrm{y}-28-4\mathrm{x}+5\mathrm{y}+7=0 \\ \Rightarrow 17\mathrm{y}=21 \\ \mathrm{y}=\frac{21}{17} \\ \mathrm{Put} \mathrm{this} \mathrm{value} \mathrm{in} \left(\mathrm{i}\right) \\ 2\mathrm{x}+\frac{126}{17}=14 \\ \Rightarrow 2\mathrm{x}=14-\frac{126}{17}=\frac{112}{17} \\ \Rightarrow \mathrm{x}=\frac{56}{17} \\ \Rightarrow \left(\mathrm{x},\mathrm{y}\right)=\left(\frac{56}{17},\frac{21}{17}\right) \\ \mathrm{Required} \mathrm{distance}\left(\mathrm{d}\right)=\sqrt{{\left(-5-\frac{56}{17}\right)}^2+\left(4-\frac{21}{17}\right)} \\ =\sqrt{\frac{{\left(-85-56\right)}^2+{\left(68-21\right)}^2}{{17}^2}} \\ =\frac{\sqrt{{141}^2+{47}^2}}{17} \\ =\frac{\sqrt{19881+2209}}{17} \\ =\frac{\sqrt{\mathbf{22090}}}{\mathbf{17}}\mathbf{ }\mathbf{units}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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