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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 (x1, y1) to a line ax + by + c = 0 is |ax1+by1+c|a2+b2

Let AB be the given straight line whose equation is ax + by + c = 0 ………………… (i) and P (x1, y1) 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 = -ca.

Therefore, the coordinate of the point A where the line ax + by + c = 0 intersect at x-axis are (-ca, 0).

Similarly, putting x = 0 in ax + by + c = 0 we get by + c = 0 ⇒ y = -cb.

Therefore, the coordinate of the point B where the line ax + by + c = 0 intersect at y-axis are (0, -cb).

From P draw PM perpendicular to AB.

Now find the area of ∆ PAB.

Area of ∆ PAB = ½|x1(0+cb)ca(cby1)+0(y10)|

= ½|cx1b+cy1b+c2ab|

= |(ax1+by1+c)c2ab| ……………………………….. (i)

Again, area of PAB = ½ × AB × PM = ½ × c2a2+c2b2 × PM = c2aba2+b2 × PM ……………………………….. (ii)

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

|(ax1+by1+c)c2ab| = c2aba2+b2 × PM

⇒ PM = |ax1+by1+c|a2+b2

 

Note: Evidently, the perpendicular distance of P (x1, y1) from the line ax + by + c = 0 is ax1+by1+ca2+b2 when ax1 + by1 + c   is positive; the corresponding distance is ax1+by1+ca2+b2 when ax1 + by1 + c is negative.

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

i.e.,

The perpendicular distance of the line ax + by + c = 0 from the origin ca2+b2  when c > 0 and - ca2+b2 when c < 0.

Algorithm to find the length of the perpendicular from a point (x1, y1) 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 x1 and y1 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 = |42(1)5|42+(1)2

= |8+15|16+1

= |4|17

= 417

Therefore, the required perpendicular distance between the line 4x - y = 5 and the point (2, - 1)= 417 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 |12251+9122+(5)2| units.

= |245+9|144+25 units.

= |28|169 units.

= 2813 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 = |53124+7|52+(12)2

= |1548+7|25+144

= |26|169

= 2613

= 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




11 and 12 Grade Math

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