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

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