The equation to a pair of opposite sides of a parallelogram are `x^2-5x+6=0` and `y^2+5=0` . The equations to its diagonals are `x+4y=13 ,y=4x-7` (b) `4x+y=13 ,4y=x-7` `4x+y=13 ,y=4x-7` (d) `y-4x=13 ,y+4x-7`

Options

A. `x+4y=13 and y=4x-7`
B. `4x+y =13 and 4y=x-7`
C. `4x+y=13 and y=4x-y`
D. `y-4x =13 and y+4x=7`

Answer

C

Explanation

The sides of the parallelogram are
x = 2, x = 3, y = 1
y = 5
`therefore ` Angular points are
A(2, 1), B(2, 5), C(3, 5), D(3, 1)
Equation of AC is
`(x-2)/(3-2)=(y-1)/(5-1)`
`or 4x+y=13`