Distance of a Point from a Line

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 \(\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\).
    Distance of a Point from a Line
  • The length of the perpendicular from the origin to the line ax + by + c = 0   is \(\frac{\left| c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\).

Example: If P is the length of the perpendicular from the origin to the line, \(\frac{x}{a}+\frac{y}{b}=1,\) then prove that \(\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\).

Solution: The given line is bx + ay – ab = 0 … 1

It is given that

P = length of the perpendicular from the origin to (1)

\(=\frac{\left| b\left( 0 \right)+a\left( 0 \right)-ab\right|}{\sqrt{{{b}^{2}}+{{a}^{2}}}}\),

\(=\frac{ab}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\),

⇒ \({{p}^{2}}=\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\),

⇒ \(\frac{1}{{{p}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}\),

⇒ \(\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\).

  • The distance between two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is \(\frac{\left| {{c}_{2}}-{{c}_{1}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\).
  • 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 a is \(\left| \frac{\left( {{c}_{1}}-{{d}_{1}} \right)\left( {{c}_{2}}-{{d}_{2}} \right)}{\left| \begin{matrix}{{a}_{1}} & {{b}_{1}}  \\{{a}_{2}} & {{b}_{2}}  \\\end{matrix} \right|} \right|\)

Example: Prove that the four straight lines \(\frac{x}{a}=\frac{y}{b}=1\), \(\frac{x}{b}+\frac{y}{a}=1\), \(\frac{x}{a}+\frac{y}{b}=2\) and \(\frac{x}{b}+\frac{y}{a}=2\) form a rhombus. Find its area.

Solution: The equations of the four sides are

\(\frac{x}{a}+\frac{y}{b}=1\) … (1)

\(\frac{x}{b}+\frac{y}{a}=1\) … (2)

\(\frac{x}{a}+\frac{y}{b}=2\) … (3)

\(\frac{x}{b}+\frac{y}{a}=2\) … (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,

\({{P}_{1}}=\left| \frac{2-1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}} \right|=\frac{ab}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\),

\({{P}_{2}}=\left|\frac{2-1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}}\right|=\frac{ab}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\),

Clearly, P₁ = P₂ So, the given lines form a rhombus.

Area of rhombus \(=\left| \frac{\left( 2-1 \right)\left( 2-1 \right)}{\left|\begin{matrix}   \frac{1}{a} & \frac{1}{b}  \\   \frac{1}{b} & \frac{1}{a}  \\\end{matrix} \right|} \right|\),

\(=\left| \frac{1}{\left( \frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}} \right)} \right|=\frac{{{a}^{2}}{{b}^{2}}}{\left| {{b}^{2}}-{{a}^{2}} \right|}\).