question_answer

The equations to a pair of opposite sides of a parallelogram are \[{{x}^{2}}-5x+6=0\]and \[{{y}^{2}}-6y+5=0\]. The equations to its diagonals are                 [IIT 1994; Pb. CET 2003]

A)            \[x+4y=13\]and \[y=4x-7\]

B)            \[4x+y=13\]and \[4y=x-7\]

C)            \[4x+y=13\]and \[y=4x-7\]    

D)            \[y-4x=13\]and \[y+4x=7\]

Correct Answer: C

Solution :

           Equation of diagonal \[{{d}_{1}}\] is \[y-1=\frac{5-1}{3-2}(x-2)\]            \[\Rightarrow y-1=\frac{4}{1}(x-2)\]\[\Rightarrow y=4x-7\] Equation of diagonal\[{{d}_{2}}\]is \[y-1=\frac{5-1}{2-3}(x-3)\]            \[\Rightarrow y-1=-4(x-3)\Rightarrow 4x+y=13\]            So equations are, \[4x+y=13\]and \[y=4x-7\].