Question Statement
Solve the quadratic equation 3 x 2 + 7 x y + 2 y 2 = 0 3x^2 + 7xy + 2y^2 = 0 3 x 2 + 7 x y + 2 y 2 = 0 to determine the lines it represents and the angle between them.
Background and Explanation
This question involves solving a homogeneous quadratic equation in x x x and y y y . To determine the lines it represents, we rewrite the equation in terms of y x = m \frac{y}{x} = m x y β = m , which corresponds to the slopes of the lines. We then use the formula for the angle between two lines:
tan β‘ ΞΈ = β£ m 1 β m 2 β£ 1 + m 1 m 2 , \tan\theta = \frac{|m_1 - m_2|}{1 + m_1 m_2}, tan ΞΈ = 1 + m 1 β m 2 β β£ m 1 β β m 2 β β£ β ,
or equivalently for a homogeneous equation:
tan β‘ ΞΈ = 2 h 2 β a b a + b , \tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b}, tan ΞΈ = a + b 2 h 2 β ab β β ,
where a a a , h h h , and b b b are coefficients of x 2 x^2 x 2 , x y xy x y , and y 2 y^2 y 2 , respectively.
Solution
Step 1: Rewrite the equation
Rewrite the quadratic equation in terms of y x = m \frac{y}{x} = m x y β = m :
2 ( y x ) 2 + 7 ( y x ) + 3 = 0 2\left(\frac{y}{x}\right)^2 + 7\left(\frac{y}{x}\right) + 3 = 0 2 ( x y β ) 2 + 7 ( x y β ) + 3 = 0
This becomes:
2 m 2 + 7 m + 3 = 0 2m^2 + 7m + 3 = 0 2 m 2 + 7 m + 3 = 0
Step 2: Solve for m m m
Using the quadratic formula, m = β B Β± B 2 β 4 A C 2 A m = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} m = 2 A β B Β± B 2 β 4 A C β β :
m = β 7 Β± 7 2 β 4 ( 2 ) ( 3 ) 2 ( 2 ) m = \frac{-7 \pm \sqrt{7^2 - 4(2)(3)}}{2(2)} m = 2 ( 2 ) β 7 Β± 7 2 β 4 ( 2 ) ( 3 ) β β
Simplify step by step:
Calculate the discriminant:
49 β 24 = 25 = 5 \sqrt{49 - 24} = \sqrt{25} = 5 49 β 24 β = 25 β = 5
Substitute into the formula:
m = β 7 Β± 5 4 m = \frac{-7 \pm 5}{4} m = 4 β 7 Β± 5 β
Find the two roots:
m = β 7 + 5 4 = β 2 4 = β 1 2 m = \frac{-7 + 5}{4} = -\frac{2}{4} = -\frac{1}{2} m = 4 β 7 + 5 β = β 4 2 β = β 2 1 β
m = β 7 β 5 4 = β 12 4 = β 3 m = \frac{-7 - 5}{4} = -\frac{12}{4} = -3 m = 4 β 7 β 5 β = β 4 12 β = β 3
Thus, m 1 = β 1 2 m_1 = -\frac{1}{2} m 1 β = β 2 1 β and m 2 = β 3 m_2 = -3 m 2 β = β 3 .
Step 3: Find the equations of the lines
The lines are given by:
y = m 1 x and y = m 2 x y = m_1x \quad \text{and} \quad y = m_2x y = m 1 β x and y = m 2 β x
Substitute m 1 m_1 m 1 β and m 2 m_2 m 2 β :
Line 1: y = β 1 2 x β x + 2 y = 0 y = -\frac{1}{2}x \quad \Rightarrow \quad x + 2y = 0 y = β 2 1 β x β x + 2 y = 0
Line 2: y = β 3 x β 3 x + y = 0 y = -3x \quad \Rightarrow \quad 3x + y = 0 y = β 3 x β 3 x + y = 0
Step 4: Find the angle between the lines
The formula for the angle between two lines in a homogeneous quadratic equation is:
tan β‘ ΞΈ = 2 h 2 β a b a + b \tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b} tan ΞΈ = a + b 2 h 2 β ab β β
Substitute a = 3 a = 3 a = 3 , h = 7 2 h = \frac{7}{2} h = 2 7 β , and b = 2 b = 2 b = 2 :
tan β‘ ΞΈ = 2 ( 7 2 ) 2 β ( 3 ) ( 2 ) 3 + 2 \tan\theta = \frac{2\sqrt{\left(\frac{7}{2}\right)^2 - (3)(2)}}{3 + 2} tan ΞΈ = 3 + 2 2 ( 2 7 β ) 2 β ( 3 ) ( 2 ) β β
Simplify step by step:
Square h h h :
( 7 2 ) 2 = 49 4 \left(\frac{7}{2}\right)^2 = \frac{49}{4} ( 2 7 β ) 2 = 4 49 β
Subtract a b ab ab :
49 4 β 6 = 25 4 \frac{49}{4} - 6 = \frac{25}{4} 4 49 β β 6 = 4 25 β
Take the square root:
25 4 = 5 2 \sqrt{\frac{25}{4}} = \frac{5}{2} 4 25 β β = 2 5 β
Multiply by 2 and divide by a + b a + b a + b :
2 ( 5 2 ) 5 = 5 5 = 1 \frac{2\left(\frac{5}{2}\right)}{5} = \frac{5}{5} = 1 5 2 ( 2 5 β ) β = 5 5 β = 1
Thus:
tan β‘ ΞΈ = 1 β ΞΈ = tan β‘ β 1 ( 1 ) = 4 5 β \tan\theta = 1 \quad \Rightarrow \quad \theta = \tan^{-1}(1) = 45^\circ tan ΞΈ = 1 β ΞΈ = tan β 1 ( 1 ) = 4 5 β
Quadratic Formula :
m = β B Β± B 2 β 4 A C 2 A m = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} m = 2 A β B Β± B 2 β 4 A C β β
Angle Between Two Lines in Homogeneous Form :
tan β‘ ΞΈ = 2 h 2 β a b a + b \tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b} tan ΞΈ = a + b 2 h 2 β ab β β
Summary of Steps
Rewrite the quadratic equation in terms of m = y x m = \frac{y}{x} m = x y β .
Solve the quadratic equation for m m m using the quadratic formula.
Write the equations of the lines using the roots of m m m .
Apply the formula for the angle between lines to calculate ΞΈ \theta ΞΈ .
Conclude that the angle between the lines is 4 5 β 45^\circ 4 5 β .