The locus of the image of the point (2,3) in the line (2x-3y+4) + k(x-2y+3) = 0, k belongs to R is a :
(A) straight line parallel to x-axis    (B) straight line parallel to y-axis  
(C) circle of radius sq.rt.2                (D) circle of radius sq.rt.3   

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$$\begin{gathered} 2\mathrm{x} - 3\mathrm{y} + 4 + \mathrm{k}\left(\mathrm{x} - 2\mathrm{y} + 3\right) = 0 \mathrm{is} \mathrm{family} \mathrm{of} \mathrm{lines} \mathrm{passing} \mathrm{through} \left(1, 2\right) \\ \mathrm{Now} \mathrm{look} \mathrm{at} \mathrm{the} \mathrm{given} \mathrm{figure}: \\ \mathrm{Taking} \mathrm{triangle} \mathrm{ABM} \mathrm{and} \mathrm{AB}\text{'}\mathrm{M}, \\ \mathrm{AM}=\mathrm{AM} \left(\mathrm{common}\right) \\ \angle \mathrm{AMB}=\angle \mathrm{AMB}\text{'}=90° \\ \mathrm{BM}=\mathrm{MB}\text{'} \left(\mathrm{Given} \mathrm{AM} \mathrm{is} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{BB}\text{'}\right) \\ ∆\mathrm{ABM}\cong ∆\mathrm{AMB}\text{'} \left(\mathrm{SAS}\right) \\ \mathrm{hence}, \\ \mathrm{AB}=\mathrm{AB}\text{'} \\ \mathrm{so}, \\ \mathrm{Let} \mathrm{M} \mathrm{is} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{BB}\text{'} \mathrm{and} \mathrm{AM} \mathrm{is} \perp \mathrm{bisector} \mathrm{of} \mathrm{BB}\text{'} \left(\mathrm{where} \mathrm{A} \mathrm{is} \mathrm{the} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{given} \mathrm{lines}\right) \\ \left(\mathrm{x} – 2\right)\left(\mathrm{x} – 1\right) + \left(\mathrm{y} – 2\right)\left(\mathrm{y} – 3\right) = 0 \\ \mathrm{M} \mathrm{will} \mathrm{lie} \mathrm{on} \mathrm{the} \mathrm{above} \mathrm{line}: \\ \mathrm{so}, \\ \left(\frac{\mathrm{h}+2}{2} – 2\right)\left(\frac{\mathrm{h}+2}{2} – 1\right) + \left(\frac{\mathrm{k}+3}{2} – 2\right)\left(\frac{\mathrm{k}+3}{2}– 3\right) = 0 \\ \Rightarrow \left(\mathrm{h}-2\right)\left(\mathrm{h}\right)+\left(\mathrm{k}-1\right)\left(\mathrm{k}-3\right)=0 \\ \Rightarrow {\mathrm{h}}^2-2\mathrm{h}+{\mathrm{k}}^2-4\mathrm{k}+3=0 \\ \Rightarrow {\left(\mathrm{h}-1\right)}^2+{\left(\mathrm{k}-2\right)}^2=2 \\ \mathrm{replacing} \mathrm{h} \mathrm{by} \mathrm{x} \mathrm{and} \mathrm{k} \mathrm{by} \mathrm{y}. \\ {\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-2\right)}^2=2 \\ \mathrm{correct} \mathrm{option} \mathrm{is} \left(\mathrm{c}\right) \end{gathered}$$

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