Online Lengths of Perpendiculars Tutor & Homework Help

Solution Lengths of Perpendiculars Tutoring Online, Tutor Help
LENGTHS OF PERPENDICULARS:
THEOREM:
To find the length of the perpendicular let fall from a given point upon a given straight line.
1) Let the equation of the straight line be
x cos ∝ + y sin ∝ - p = 0
(1)
If p be the perpendicular on it, then
ON = p and ∠
XON = ∝.
Let the give point be P(x1, y1).
Draw a parallel PR to the given line to meet ON produced in R and draw PQ
(which is the required perpendicular).
If OR be p1, then the equation to PR will be,
x cos ∝ + y sin ∝ - p1 = 0
Since this passes through the point (x1, y1), we get
x1 cos ∝ + y1 sin ∝ - p1 = 0
or p1 = x1 cos ∝ + y1 sin ∝
But the required perpendicular = PQ = NR = OR - ON = p1 - p
= x1 cos ∝ + y1 sin ∝ - p (2)
Thus the length of the required perpendicular is obtained by substituting x1 and y1 for x and y
in the given equation.
2) If the equation of the straight line is
Ax + By + C = 0 (3)
Dividing the above equation by √(A2 )+ B2 we get,
Now comparing the above equation with eq. (1) we get,
Hence the perpendicular from the point (x1, y1) on equation (3) = x1 cos ∝+ y1 sin ∝ - p
Corollary:
The length of perpendicular from the origin i.e. from (0, 0) =
Example:
Find the length of the perpendicular drawn from the point (4, 5) upon the straight line 3x + 4y = 10.
Solution:
We know that for the equation of straight line Ax + By + C = 0 the perpendicular from the point (x1, y1) on it =
Here (x1, y1) = (4, 5) and
the equation of the straight line is 3x + 4y = 10 or 3x + 4y - 10 = 0 so A = 3, B = 4 and C = -10

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