if the pair of lines x² + 2xy + ay² = 0 and ax² + 2xy + y² = 0 have exactly one line in common then the joint equation of the other two lines is given by........

please tell me the procedure how to solve it....!!!

2nd question,

the equation x - y = 4 and x² + 4xy + y² = 0 represent the sides of

the answer of second is equilateral triangle...

please please help mee.....i know its a silly question...

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

2. Acute angle between lines x² + 4xy + y² = 0 is tan^-1(3)^1/2 = pi / 3

Angle bisectors of x² + 4xy + y² = 0 are given by

x² - y² = xy

 1 - 1       2

or, x = + y

As, x + y = 0 is perpendicular to x - y = 4, given triangle is isosceles with vertical angle equal to pi / 3 and hence it is equilateral.

Both The Above Answers Are Absolutely Correct. Thanks For Making Me Revise Pair Of Lines.