If the diag

Given:

l₁x + m₁y + n₁ = 0 … (1)

l₂x + m₂y + n₂ = 0 … (2)

l₁x + m₁y + n’₁ = 0 … (3)

l₂x + m₂y + n₂‘ = 0 … (4)

To find:

The value of l¹² – l₂² + m¹₂ – m²₂.

Explanation:

Assuming:

(1), (2), (3) and (4) represent the sides AB, BC, CD and DA, respectively.

The equation of diagonal AC passing through the intersection of (2) and (3) is given by
l₁x + m₁y + n’₁+ λ(l₂x + m₂y + n₂) = 0

⇒ (l₁ + λl₂)x + (m₁ + λm₂)y + (n₁’ + λn₂) = 0

⇒ Slope of diagonal AC

Also, the equation of diagonal BD, passing through the intersection of (1) and (2), is given by
l₁x + m₁y + n₁ + μ(l₂x + m₂y + n₂) = 0

⇒ l₁ + μl₂x + m₁ + μm₂y + n₁ + μn₂ = 0

⇒ Slope of diagonal BD

The diagonals are perpendicular to each other.

∴

⇒ (l₁ + λl₂)(l₁ + λl₂) = (-m₁ + λm₂)(m₁ + μm₂)

Let λ = -1, μ = 1

⇒ (l₁ – l₂)(l₁ + l₂) = (-m₁-m₂)(m₁ + m₂)

⇒ (l₁²-l₂²⁾ = (-m₁²-m₂²⁾

⇒ (l₁²-l₂²) + (m₁²-m₂²) = 0

Hence, (l₁²-l₂²) + (m₁²-m₂²) = 0