# General Equation of Second Degree

## 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.