HOW TO FIND LENGTH OF PERPENDICULAR FROM A POINT TO A LINE

About "How to Find Length of Perpendicular From a Point to a Line"

How to Find Length of Perpendicular From a Point to a Line :

Here we are going to see how to find the length of perpendicular from a point to a line.

Distance between the point and a line

=  |Ax + By + C|/√A2 + B2

How to Find Length of Perpendicular From a Point to a Line - Practice questions

Question 1 :

If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x sec θ + y cosec θ = 2a and x cos θ − y sin θ = a cos 2θ, then prove that p12+ p22= a2.

Solution :

Length of perpendicular from origin (0, 0)  to the straight lines x sec θ + y cosec θ = 2a   :

=  |((0)sec θ + (0)cosec θ - 2a)| /  √(secθ)2 + (cosecθ)2

p1  =  2a /  √(sec2θ) + (cosec2θ)

Taking squares on both sides, we get

p12  =  4a2 / (sec2θ + cosec2θ)  ----(1)

Length of perpendicular from origin (0, 0)  to the straight lines  x cos θ − y sin θ = a cos 2θ  :

=  |(0) cos θ - (0) sin θ - a cos 2θ)| /  √(sinθ)2 + (cosθ)2

p2  =  a cos 2θ /  √(sin2θ + cos2θ)

p2  =  a cos 2θ /  1

p22  =  a2 cos22θ   ----(2)

p12  + p22 =  4a2 / (sec2θ + cosec2θ) + (a2 cos22θ)

sec θ =  1/cos θ, coseθ =  1/sin θ &

cos 2θ = cos2θ - sin2θ

=  4a2 (sin2θ cos2θ) + a2 (cos2θ - sin2θ)2

=  4a2 (sin2θ cos2θ) + a2 [cos4θ + sin4θ - 2sin2θ cos2θ]

=  4a2 sin2θ cos2θ + a2 cos4θ + a2sin4θ - 2a2sin2θ cos2θ

=  a2 cos4θ + a2sin4θ + 2a2sin2θ cos2θ

=  a2 [sin2θ + cos2θ]2

=  a2 (1)

=  a2

Hence proved.

Question 2 :

Find the distance between the parallel lines

(i) 12x + 5y = 7 and 12x + 5y+7 = 0

(ii) 3x − 4y+5 = 0 and 6x − 8y − 15 = 0.

Solution :

Distance between two parallel lines  =  |C1 - C2|/√A2 + B2

Since the given lines are parallel, coefficients of x and y terms will be equal in both the equations.

C1  =  -7, C2  =  7, A  =  12, B  =  5

=  |-7 - 7|/√122 + 52

=  |-14|/√144 + 25

=  14/√169

=  14/13

Solution :

Distance between two parallel lines  =  |C1 - C2|/√A2 + B2

In order to convert the coefficients of x and y terms as same as first equation, we have to divide the second equation by 2.

C1  =  5, C2  =  -15/2, A  =  3, B  =  -4

=  |5 + (15/2)|/√32 + (-4)2

=  |25/2|/√9 + 16

=  (25/2)/√25

=  (25/2)/5  =  5/2

Question 3 :

Find the family of straight lines (i) Perpendicular (ii) Parallel to 3x + 4y − 12 = 0.

Solution :

Line which is perpendicular to the given line 3x + 4y − 12 = 0.

4x - 3y + k  =  0  and K ∈ R

Line which is parallel to the given line 3x + 4y − 12 = 0.

3x + 4y + k  =  0  and K ∈ R

After having gone through the stuff given above, we hope that the students would have understood "How to Find Length of Perpendicular From a Point to a Line".

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