General Equation of Second Degree
The most general form of a quadratic equation x and y is ax² + 2hxy + by² + 2gx + 2fy + c = 0. Since it is an equation in x and y, it must represent the equation of a locus in a plane. It may represent a pair of straight lines, circles, or other curves in different cases. Now, we will consider the case when the above equation represents two straight lines.
Condition for General Second – Degree Equation in x and y Represent a Pair of Straight Lines:
The given equation is
ax² + 2hxy + by² + 2gx + 2fy + c = 0 … (1)
if a ≠ 0, then writing (1) as a quadratic equation in x, we get
ax² + 2x (hy + g) + (by² + 2fy + c) = 0
Solving, we have
\(x=\frac{-2(hy+g)\pm \sqrt{4{{(hy+g)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}\).
\(=\frac{-(hy+g)\pm \sqrt{({{h}^{2}}-ab){{y}^{2}}+2(gh-afy)y+({{g}^{2}}-ac)}}{a}\).
Equation (1) will represent two straight lines if the LHS of (i) can be resolved into two linear factors. therefore, the expression under the square roots should be a perfect square.
Hence,
4(gh – af)² – 4(h² – ab)(g² – ac) = 0
(gh – af)² – (h² – ab)(g² – ac) = 0
g²h² + a²f² – 2afgh – h²g² + abg² + ach² – a²bc = 0
a (af² + bg² + ch² – 2fgh – abc) = 0
abc + 2fgh – af² – bg² – ch² = 0
Example: Find the lines whose combined equation is 6x² + 5xy – 4y² + 7x + 13y – 3 = 0
Solution: Given that,
6x² + 5xy – 4y² + 7x + 13y – 3 = 0
6x² + (5y + 7)x – (4y² – 13y + 3) = 0
\(x=\frac{-(5y+7)\pm \sqrt{{{(5y+7)}^{2}}+24(4{{y}^{2}}-13y+3)}}{12}\).
\(x=\frac{-(5y+7)\pm \sqrt{121{{y}^{2}}-242y+121}}{12}\).
\(x=\frac{-(5y+7)\pm 11(y-1)}{12}\).
\(x=\frac{6y-18}{12},\frac{-16y+4}{12}\).
\(x=\frac{y-3}{2},\frac{-4y+1}{3}\).
\(x=\frac{y-3}{2}\).
2x – y + 3 = 0,
\(x=\frac{-4y+1}{3}\).
3x + 4y – 1 = 0.