If the equations of the pairs of opposite sides of a parallelogram are \(x^2-5 x+6=0\) and \(y^2-6 y+5=0\),…
If the equations of the pairs of opposite sides of a parallelogram are \(x^2-5 x+6=0\) and \(y^2-6 y+5=0\), then equations of its diagonals are
\(x+4 y=13, y=4 x-7\)
\(4 x+y=13,4 y=x-7\)
\(4 x+y=13, y=4 x-7\)
\(y-4 x=13, y+4 x=7\)
Solution
Equations of the sides of the parallelogram are
\((x-3)(x-2)=0\) and \((y-5)(y-1)=0\)
i.e. \(x=3, x=2 ; y=5, y=1\)
Hence its vertices are : \(\mathrm{A}(2,1) ; \mathrm{B}(3,1)\); \(\mathrm{C}(3,5) ; \mathrm{D}(2,5)\)
Equation of the diagonal \(\mathrm{AC}\) is
\(y-1=\frac{4}{1}(x-2) \Rightarrow y=4 x-7\)
Equation of the diagonal \(\mathrm{BD}\) is
\(y-1=\frac{4}{-1}(x-3) \Rightarrow 4 x+y=13\)