**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.