# 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 $$\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$$. • 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 $$=\left| \frac{\left( 2-1 \right)\left( 2-1 \right)}{\left|\begin{matrix} \frac{1}{a} & \frac{1}{b} \\ \frac{1}{b} & \frac{1}{a} \\\end{matrix} \right|} \right|$$,

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