1. Let us assume that the common line is y=mx. So, it satisfies the two homogeneous equations.

So, am² + 2m + 1 = 0

and , m² + 2m + a = 0

Solving them,

    m²      =      m                =          1              

2(1 - a)        a² - 1                   2(1 - a)

or, m² = 1 and m = -( a + 1)/2

or, (a + 1)² = 4m² = 4

So, a = -3 ( bcoz when a = 1 the two pairs have both the lines common)

and therefore, m = 1

Now given pairs of equations become,

x² + 2xy - 3y² = 0

or, (x - y) (x + 3y) = 0

and -3x² + 2xy + y² = 0

or, (x - y) (-3x - y) = 0

So, required equation is - (x + 3y) (3x + y) = 0

or, 3x² + 10xy + 3y² = 0