if one diagonal of a square lies on x - 2y + 2 =0 and one vertex of the square is (1,4). find the equation of all the sides and the other diagonal.

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$$\begin{gathered} \mathrm{Given} \mathrm{one} \mathrm{diagonal}=\mathrm{x}-2\mathrm{y}+2=0 \\ 2\mathrm{y}=\mathrm{x}+2 \\ \mathrm{y}=\frac{\mathrm{x}}{2}+1 \\ {\mathrm{m}}_1=\frac{1}{2} \\ \mathrm{one} \mathrm{vertex}=\left(1,4\right) \\ \mathrm{other} \mathrm{diagonal} \mathrm{will} \mathrm{be} \perp \mathrm{to} \mathrm{this} \mathrm{diagonal} \\ {\mathrm{m}}_2=\frac{-1}{{\displaystyle \frac{1}{2}}}=-2 \\ \mathrm{and} \mathrm{will} \mathrm{pass} \mathrm{through} \left(1,4\right) \\ \mathrm{y}-4=-2\left(\mathrm{x}-1\right) \\ \mathrm{y}-4=-2\mathrm{x}+2 \\ 2\mathrm{x}+\mathrm{y}=6 \\ \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{two} \mathrm{diagonals} \mathbf{M}: \\ 2\mathrm{y}=\mathrm{x}+2 \\ 2\mathrm{x}+\mathrm{y}=6 \\ \mathrm{y}=6-2\mathrm{x} \\ 2\left(6-2\mathrm{x}\right)=\mathrm{x}+2 \\ 12-4\mathrm{x}=\mathrm{x}+2 \\ 5\mathrm{x}=10,\mathrm{x}=2 \\ \mathrm{y}=6-2\mathrm{x}=6-4=2 \\ \mathrm{M}\left(2,2\right) \\ \mathrm{Vertex} \mathrm{C} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}: \\ \frac{1+\mathrm{x}}{2}=2 \\ 1+\mathrm{x}=4 \\ \mathrm{x}=3 \\ \frac{4+\mathrm{y}}{2}=2 \\ \mathrm{y}=0 \\ \mathrm{C}\left(3,0\right) \\ \mathrm{angle} \mathrm{between} \mathrm{BD} \mathrm{and} \mathrm{DC}=45° \\ \mathrm{let} \mathrm{slope} \mathrm{of} \mathrm{DC}=\mathrm{m} \\ \tan 45°=\frac{{\displaystyle \frac{1}{2}}-\mathrm{m}}{1+{\displaystyle \frac{\mathrm{m}}{2}}} \\ 1+\frac{\mathrm{m}}{2}=\frac{1}{2}-\mathrm{m} \\ \frac{3\mathrm{m}}{2}=-\frac{1}{2} \\ \mathrm{m}=\frac{-1}{3} \\ \mathrm{eqn} \mathrm{DC}: \\ \mathrm{given} \mathrm{C}\left(3,0\right) \\ \mathrm{y}-0=\frac{-1}{3}\left(\mathrm{x}-3\right) \\ 3\mathrm{y}=-\mathrm{x}+3 \\ \mathrm{eqn} \mathrm{AB}: \\ \mathrm{Given} \mathrm{A}\left(1,4\right) \\ \mathrm{since} \mathrm{AB}\parallel \mathrm{DC} \\ \mathrm{y}-4=\frac{-1}{3}\left(\mathrm{x}-1\right) \\ 3\mathrm{y}-12=-\mathrm{x}+1 \\ 3\mathrm{y}+\mathrm{x}=13 \\ \mathrm{Angle} \mathrm{between} \mathrm{AC} \mathrm{and} \mathrm{AD}=45° \\ \mathrm{let} \mathrm{slope} \mathrm{AD}=\mathrm{m} \\ \tan 45°=\frac{-2-\mathrm{m}}{1-2\mathrm{m}} \\ 1-2\mathrm{m}=-2-\mathrm{m} \\ 3=\mathrm{m} \\ \mathrm{eqn} \mathrm{AD} \mathrm{given} \mathrm{A}\left(1,4\right) \\ \mathrm{y}-4=3\left(\mathrm{x}-1\right) \\ \mathrm{y}-4=3\mathrm{x}-3 \\ \mathrm{y}=3\mathrm{x}+1 \\ \mathrm{eqn} \mathrm{BC}\parallel \mathrm{AD} \\ \mathrm{given} \mathrm{C}\left(3,0\right) \\ \mathrm{y}-0=3\left(\mathrm{x}-3\right) \\ \mathrm{y}=3\mathrm{x}-9 \\ \mathrm{now} \mathrm{you} \mathrm{can} \mathrm{find} \mathrm{all} \mathrm{vertices} \mathrm{since} \mathrm{you} \mathrm{know} \mathrm{all} \mathrm{sides}. \end{gathered}$$

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