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
11 and 12 Grade Math
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