Distance of a point from a line

The length of the perpendicular from a point (x₁, y₁) to a line ax + by + c = 0 is |ax₁ + by₁ + c/ √ (a² + b²)|.

The length of the perpendicular from the origin to the line ax + by + c = 0 is |c|/ √ (a² + b²)

Example:

If P is the length of the perpendicular from the origin to the line x/ a + y/ b = 1 then prove that 1/ p² = 1/ a² + 1/ b²

Solution:

The given line is bx + ay – ab = 0 … (1) 

It is given that

Length of the perpendicular from the origin to (1)

= |b (0) + a (0) – ab|/ √ (b² + a²)

= ab/ √ (a² + b²)

⇒ p² = a²b²/ (a² + b²)

⇒ 1/ p² = (a² + b²)/ a²b²

⇒ 1/ p² = 1/ a² + 1/ b²

⇒ The distance between two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is |c₂ - c₁|/ √ (a² + b²).

⇒ The area of the parallelogram ABCD whose sides AB, BC, CD and DA are a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₁x + b₁y + d₁ = 0 and a₂x + b₂y + d₂ = 0 is 

Example:

Prove that the four straight lines x/ a = y/ b = 1, x/ b + y/ a = 1, x/ a + y/ b = 2 and x/ b + y/ a = 2 form a rhombus. Find its area.

Solution:

The equations of the four sides are

x/ a + y/ b = 1 … (1)

x/ b + y/ a = 1 … (2)

x/ a + y/ b = 2 … (3)

x/ b + y/ a = 2 … (4)

Clearly, (1), (3) and (2), (4) form two sets of parallel lines.

So, the four lines form a parallelogram.

Let p₁ be the distance between parallel lines (1) and (3) and p₂ be the distance between (2) and (4). Then,

Clearly,  so, the given lines form a rhombus.