prove the equation 3x^2+7xy+2y^2+5x+5y+2=0 represents a pair of straight lines.find the point of intersection.also find the angle between them
Answers
Given : 3x²+7xy+2y²+5x+5y+2=0
To Find : prove the equation represents a pair of straight lines.
find the point of intersection.
also find the angle between them
Solution:
3x²+7xy+2y²+5x+5y+2=0
=> 3x² + xy + 6xy+ 2y² + 3x + 2x + 4y + y + 2 = 0
=> 3x² + xy + + 2x + 6xy+ 2y² + 4y + 3x + y + 2 = 0
=> x(3x + y + 2) + 2y(3x + y + 2) + 1( 3x + y + 2 ) = 0
=> (3x + y+ 2)(x + 2y + 1) = 0
3x + y+ 2 = 0 , x + 2y + 1 = 0
3x + y+ 2 = 0 is Equation of straight line
x + 2y + 1 = 0 is also Equation of straight line
Hence 3x²+7xy+2y²+5x+5y+2=0 represents a pair of straight lines.
3x + y+ 2 = 0
x + 2y + 1 = 0
=> x = -3/5 , y = -1/5
the point of intersection ( -3/5 , - 1/5 )
3x + y+ 2 = 0 => slope = - 3
x + 2y + 1 = 0 => slope = - 1/2
Angle between two lines = α
Tan α = ± (-3 - (-1/2) )/ ( 1 + (-3)(-1/2))
=> Tan α = ± (-5) )/ ( 5)
=> Tan α = ± (-1)
=> Tan α = ± 1
=> α = 45° , 135°
the angle between them = 45° , 135°
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