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 xaxis 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 xaxis 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 yaxis 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.
11 and 12 Grade Math
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