**General Equation of Second Degree**

The most general from of a quadratic equation in 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 to represent a Pair of Straight Lines:**

The given condition is ax² + 2hxy + by² + 2gx + 2fy + c = 0 … (i)

If a ≠ 0, then writing (i) as a quadratic equation in x we get

ax² +2x (hy + g) + by² + 2fy + c = 0

Solving we get

\(x=\frac{-2(hy+g)\pm \sqrt{4{{(hy+g)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}\).

\(x=\frac{-(hy+g)\pm \sqrt{{{(hy+g)}^{2}}-a(b{{y}^{2}}+2fy+c)}}{a}\).

\(x=\frac{-(hy+g)\pm \sqrt{({{h}^{2}}-ab){{y}^{2}}+2(gh-af)y+({{g}^{2}}-ac)}}{a}\).

Equation (i) 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

g²h² +a²f² – 2afgh – h²g² + abg² + ach² – a²bc = 0

a (af² + bg² + ch² – 2fgh – abc) = 0

abc + 2fgh – af² – bg² – ch² = 0

**Example: **Does equation x² + 2y² – 2√3 x – 4y + 5 = 0 satisfies the condition abc + 2gh – af² – bg² – ch² = 0. Does it represent a pair of straight lines?

**Solution: **Given equation x² + 2y² – 2√3 x – 4y + 5 = 0

Here a = 1, b = 2, h = 0, g = -√3, f = -2, c = 5

Clearly, abc + 2gh – af² – bg² – ch² = 0

10 + 0 – 4 – 6 = 0

The given equation can be written as (x – √3)² + 2 (y – 1)² = 0

Hence, it denotes only a point P (√3, 1) bot not a pair of straight lines.