Q51 / 125 · BITSAT 2010 · Mathematics

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

  • A
    \(x+4 y=13, y=4 x-7\)
  • B
    \(4 x+y=13,4 y=x-7\)
  • Correct answer
    C
    \(4 x+y=13, y=4 x-7\)
  • D
    \(y-4 x=13, y+4 x=7\)

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 \(y-1=\frac{4}{-1}(x-3) \Rightarrow 4 x+y=13\)

Hence, the correct answer is option: 'C'.

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