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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|βa2+b2
Let AB be the given straight line whose equation is ax + by + c = 0 β¦β¦β¦β¦β¦β¦β¦ (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 β 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 β
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 β PAB.
Area of β PAB = Β½|x1(0+cb)βca(βcbβy1)+0(y1β0)|
= Β½|cx1b+cy1b+c2ab|
= |(ax1+by1+c)c2ab| β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (i)
Again, area of PAB = Β½ Γ AB Γ PM = Β½ Γ βc2a2+c2b2 Γ PM = c2abβa2+b2 Γ PM β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (ii)
Now from (i) and (ii) we get,
|(ax1+by1+c)c2ab| = c2abβa2+b2 Γ PM
β PM = |ax1+by1+c|βa2+b2
Note: Evidently, the perpendicular distance of P (x1, y1) from the line ax + by + c = 0 is ax1+by1+cβa2+b2 when ax1 + by1 + c is positive; the corresponding distance is ax1+by1+cβa2+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|βa2+b2.
i.e.,
The perpendicular distance of the line ax + by + c = 0 from the origin cβa2+b2 when c > 0 and - cβa2+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 = |4β 2β(β1)β5|β42+(β1)2
= |8+1β5|β16+1
= |4|β17
= 4β17
Therefore, the required perpendicular distance between the line 4x - y = 5 and the point (2, - 1)= 4β17 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 |12β 2β5β 1+9β122+(β5)2| units.
= |24β5+9|β144+25 units.
= |28|β169 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 = |5β 3β12β 4+7|β52+(β12)2
= |15β48+7|β25+144
= |β26|β169
= 2613
= 2
Therefore, the required perpendicular distance of the straight line 5x - 12y + 7 = 0 from the point (3, 4) is 2 units.
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