Introduction
A general equation of the second degree in $x$ and $y$ is represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. When this equation represents a pair of straight lines, we can break it down into two linear equations. In this post, we will solve the problem for $2x^2 + 3xy + y^2 + 5x + 2y - 3 = 0$.
1. Finding the Equations of the Lines
We look for factors of the form $(ax + by + c_1)(dx + ey + c_2) = 0$.
First, we factorize the homogeneous part $2x^2 + 3xy + y^2$: $2x^2 + 2xy + xy + y^2 = 2x(x+y) + y(x+y) = (2x+y)(x+y)$.
Let the lines be $(2x+y+c_1)(x+y+c_2) = 0$. Expanding this: $2x^2 + 2xy + 2xc_2 + xy + y^2 + yc_2 + c_1x + c_1y + c_1c_2 = 0$ $2x^2 + 3xy + y^2 + x(2c_2 + c_1) + y(c_2 + c_1) + c_1c_2 = 0$
Comparing with $2x^2 + 3xy + y^2 + 5x + 2y - 3 = 0$:
- $2c_2 + c_1 = 5$
- $c_2 + c_1 = 2$
- $c_1c_2 = -3$
Subtracting (2) from (1): $c_2 = 3$. Substituting back: $3 + c_1 = 2 \Rightarrow c_1 = -1$. Check condition 3: $(-1)(3) = -3$. It matches!
The two lines are:
- $L_1: 2x + y - 1 = 0$
- $L_2: x + y + 3 = 0$
2. Finding the Point of Intersection
To find the intersection, solve the system of linear equations:
- $2x + y = 1$
- $x + y = -3$
Subtract (2) from (1): $(2x - x) + (y - y) = 1 - (-3)$ $x = 4$
Substitute $x=4$ into (2): $4 + y = -3 \Rightarrow y = -7$. The point of intersection is $(4, -7)$.
3. Calculating the Angle Between the Lines
The angle $\theta$ between lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by: $\tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2}|$
- Slope of $L_1$ ($m_1$): $-A/B = -2/1 = -2$
- Slope of $L_2$ ($m_2$): $-A/B = -1/1 = -1$
$\tan \theta = |\frac{-2 - (-1)}{1 + (-2)(-1)}| = |\frac{-1}{1+2}| = |-\frac{1}{3}| = 1/3$
$\theta = \tan^{-1}(1/3)$.