Point of Intersection of Pair of Straight Lines

Let (x₁, y₁) be the point of intersection of the lines represented by the given equation.

Shifting the origin at (x₁, y₁), we have x = X + x₁ and y = Y + y₁ … (i)

Putting the values of x and y in the given equation, we obtain

a (X + x₁)² + 2h (X + x₁) (Y + y₁) + b (Y + y₁)² + 2g (X + x₁) + 2f (Y + y₁) + c = 0

⇒ aX² + 2hXY + bY² + 2 (ax₁ + hy₁ + g) X + 2(hx₁ + by₁ + f) + ax²₁ + 2hx₁y₁ + by²₁ + 2gx₁ + 2fy₁ + c = 0

This equation represents a pair of straight lines passing through the new origin (X = 0, Y=0). So, it must be a homogeneous equation in X and Y i.e., it should not contain terms involving X, Y and constant terms.

∴ ax₁ + hy₁ + g = 0 … (ii)

hx1 + by1 + f = 0 … (iii)

And, ax²₁ + 2hx₁y₁ + by²₁ + 2gx₁ + 2fy₁ + c = 0 … (iv)

Solving (ii) and (iii), we get

x₁/ (hf - bg) = y₁/ (gh - af) = 1/ (ab - h²)

⇒ x₁ = (hf - bg)/ (ab - h²) and y₁ = (gh - af) /(ab - h²)

Thus, the two lines intersect at [(hf - bg)/ (ab - h²), (gh - af) / (ab - h²)]

REMARKS:

If the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of straight lines, then abc + 2fgh - af² – bg² – ch² = 0

⇒ 2fgh = af² + bg² + ch² – abc

∴ (hf - bg)/ (ab - h²) = √ [[(hf - bg)/ (ab - h²)]²

= √ [(h²f² + b²g² - 2bfgh)/ (ab - h²)²]

= √ [(h²f² + b²g² - abf² - b²g² - bch² + ab²c)/ (ab - h²)²]

= √ (h² - ab) f² - bc (h² - ab)/ (h² - ab)²

= √ (f² - bc)/ (h² - ab)

Similarly,

(hg – af)/ (ab - h²) = √ (g² - ac)/ (h² - ab)

Thus, the coordinates of the point of intersection of the lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0 are {√ (f² - bc)/ (h² - ab), √ (g² - ac)/ (h² - ab)}