Distance of a Point from a Straight Line

We will learn how to find the perpendicular distance of a point from a straight line.

Prove that the length of the perpendicular from a point (x$_{1}$, y$_{1}$) to a line ax + by + c = 0 is $\frac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2} + b^{2}}}$

Let AB be the given straight line whose equation is ax + by + c = 0 ………………… (i) and P (x$_{1}$, y$_{1}$) be the given point.

To find the length of the perpendicular drawn from P upon the line (i).

Firstly, we assume that the line ax + by + c = 0 meets x-axis at y = 0.

Therefore, putting y = 0 in ax + by + c = 0 we get ax + c = 0 ⇒ x = -$\frac{c}{a}$.

Therefore, the coordinate of the point A where the line ax + by + c = 0 intersect at x-axis are (-$\frac{c}{a}$, 0).

Similarly, putting x = 0 in ax + by + c = 0 we get by + c = 0 ⇒ y = -$\frac{c}{b}$.

Therefore, the coordinate of the point B where the line ax + by + c = 0 intersect at y-axis are (0, -$\frac{c}{b}$).

From P draw PM perpendicular to AB.

Now find the area of ∆ PAB.

Area of ∆ PAB = ½|$x_{1}(0 + \frac{c}{b}) - \frac{c}{a}(-\frac{c}{b} - y_{1}) + 0(y_{1} - 0)$|

= ½|$\frac{cx_{1}}{b} + \frac{cy_{1}}{b} + \frac{c^{2}}{ab}$|

= |$(ax_{1} + by_{1} + c)\frac{c}{2 ab}$| ……………………………….. (i)

Again, area of PAB = ½ × AB × PM = ½ × $\sqrt{\frac{c^{2}}{a^{2}} + \frac{c^{2}}{b^{2}}}$ × PM = $\frac{c}{2ab}\sqrt{a^{2} + b^{2}}$ × PM ……………………………….. (ii)

Now from (i) and (ii) we get,

|$(ax_{1} + by_{1} + c)\frac{c}{2 ab}$| = $\frac{c}{2ab}\sqrt{a^{2} + b^{2}}$ × PM

⇒ PM = $\frac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2} + b^{2}}}$

Note: Evidently, the perpendicular distance of P (x$_{1}$, y$_{1}$) from the line ax + by + c = 0 is $\frac{ax_{1} + by_{1} + c}{\sqrt{a^{2} + b^{2}}}$ when ax$_{1}$ + by$_{1}$ + c   is positive; the corresponding distance is $\frac{ax_{1} + by_{1} + c}{\sqrt{a^{2} + b^{2}}}$ when ax$_{1}$ + by$_{1}$ + c is negative.

(ii) The length of the perpendicular from the origin to the straight line ax + by + c = 0 is $\frac{|c|}{\sqrt{a^{2} + b^{2}}}$.

i.e.,

The perpendicular distance of the line ax + by + c = 0 from the origin $\frac{c}{\sqrt{a^{2} + b^{2}}}$  when c > 0 and - $\frac{c}{\sqrt{a^{2} + b^{2}}}$ when c < 0.

Algorithm to find the length of the perpendicular from a point (x$_{1}$, y$_{1}$) upon a given line ax + by + c = 0.

Step I: Write the equation of the line in the from ax + by + c = 0.

Step II: Substitute the coordinates x$_{1}$ and y$_{1}$ of the point in place of x and y respectively in the expression.

Step III: Divide the result obtained in step II by the square root of the sum of the squares of the coefficients of x and y.

Step IV: Take the modulus of the expression obtained in step III.

Solved examples to find the perpendicular distance of a given point from a given straight line:

1. Find the perpendicular distance between the line 4x - y = 5 and the point (2, - 1).

Solution:

The equation of the given straight line is 4x - y = 5     

or, 4x - y - 5 = 0

If Z be the perpendicular distance of the straight line from the point (2, - 1), then

Z = $\frac{|4\cdot 2 - (-1) - 5|}{\sqrt{4^{2} + (-1)^{2}}}$

= $\frac{|8 + 1 - 5|}{\sqrt{16 + 1}}$

= $\frac{|4|}{\sqrt{17}}$

= $\frac{4}{\sqrt{17}}$

Therefore, the required perpendicular distance between the line 4x - y = 5 and the point (2, - 1)= $\frac{4}{\sqrt{17}}$ units.

2. Find the perpendicular distance of the straight line 12x - 5y + 9 from the point (2, 1)

Solution:

The required perpendicular distance of the straight line 12x - 5y + 9 from the point (2, 1) is |$\frac{12\cdot 2 - 5\cdot 1 + 9}{\sqrt{12^{2} + (-5)^{2}}}$| units.

= $\frac{|24  - 5 + 9|}{\sqrt{144 + 25}}$ units.

= $\frac{|28|}{\sqrt{169}}$ units.

= $\frac{28}{13}$ units.

3. Find the perpendicular distance of the straight line 5x - 12y + 7 = 0 from the point (3, 4).

Solution:

The required perpendicular distance of the straight line 5x - 12y + 7= 0 from the point (3, 4) is

If Z be the perpendicular distance of the straight line from the point (3, 4), then

Z = $\frac{|5\cdot 3 - 12 \cdot 4 + 7|}{\sqrt{5^{2} + (-12)^{2}}}$

= $\frac{|15 - 48 + 7|}{\sqrt{25 + 144}}$

= $\frac{|-26|}{\sqrt{169}}$

= $\frac{26}{13}$

= 2

Therefore, the required perpendicular distance of the straight line 5x - 12y + 7 = 0 from the point (3, 4) is 2 units.

● The Straight Line

• Straight Line
• Slope of a Straight Line
• Slope of a Line through Two Given Points
• Collinearity of Three Points
• Equation of a Line Parallel to x-axis
• Equation of a Line Parallel to y-axis
• Slope-intercept Form
• Point-slope Form
• Straight line in Two-point Form
• Straight Line in Intercept Form
• Straight Line in Normal Form
• General Form into Slope-intercept Form
• General Form into Intercept Form
• General Form into Normal Form
  
• Point of Intersection of Two Lines
  
• Concurrency of Three Lines
• Angle between Two Straight Lines
• Condition of Parallelism of Lines
• Equation of a Line Parallel to a Line
• Condition of Perpendicularity of Two Lines
• Equation of a Line Perpendicular to a Line
• Identical Straight Lines
• Position of a Point Relative to a Line
• Distance of a Point from a Straight Line
• Equations of the Bisectors of the Angles between Two Straight Lines
• Bisector of the Angle which Contains the Origin
• Straight Line Formulae
• Problems on Straight Lines
• Word Problems on Straight Lines
• Problems on Slope and Intercept

11 and 12 Grade Math

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