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Condition of Perpendicularity of Two Lines

We will learn how to find the condition of perpendicularity of two lines.

If two lines AB and CD of slopes m1 and m2 are perpendicular, then the angle between the lines θ is of 90°.

Therefore, cot θ = 0

1+m1m2m2m1 = 0

⇒ 1 + m1m2 = 0

m1m2 = -1.

Thus when two lines are perpendicular, the product of their slope is -1. If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

Let us assume that the lines y = m1x + c1 and y = m2 x + c2 make angles α and β respectively with the positive direction of the x-axis and θ be the angle between them.

Therefore, α = θ + β = 90° + β [Since, θ = 90°]

Now taking tan on both sides we get,

tan α = tan (θ + β)

tan α = - cot  β

tan α = - 1tanβ

or,  m1 =  - 1m1    

or, m1m2 = -1

Therefore, the condition of perpendicularity of the lines y = m1x + c1, and y = m2 x + c2 is m1m2 = -1.

Conversely, if m1m2 = - 1 then

tan ∙ tan β = - 1      

sinαsinβcosαcosβ = -1

sin α sin β = - cos α cos β

cos α cos β + sin α sin β = 0

cos (α - β) = 0        

Therefore, α - β = 90°

Therefore, θ = α - β = 90°

Thus, the straight lines AB and CD are perpendicular to each other.

 

Solved examples to find the condition of perpendicularity of two given straight lines:

1. Let P (6, 4) and Q (2, 12) be the two points. Find the slope of a line perpendicular to PQ.

Solution:

Let m be the slope of PQ.

Then m = 12426 = 84 = -2

Therefore the slope of the line perpendicular to PQ = - 1m = ½


2. Without using the Pythagoras theorem, show that P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled triangle.

Solution:

In ∆ ABC, we have:

m1 = Slope of the side PQ = 4543 = -1

m2 = Slope of the side PR = 4(1)4(1) = 1

Now clearly we see that m1m2 = 1 × -1 = -1

Therefore, the side PQ perpendicular to PR that is ∠RPQ = 90°.

Therefore, the given points P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled triangle.


3. Find the ortho-centre of the triangle formed by joining the points P (- 2, -3), Q (6, 1) and R (1, 6).

Solution:       

The slope of the side QR of the ∆PQR is  611655 = -1∙

Let PS be the perpendicular from P on QR; hence, if the slope of the line PS be m then,

m × (- 1) = - 1        

or, m  = 1.

Therefore, the equation of the straight line PS is

y + 3 = 1 (x + 2)         

 or, x - y = 1     …………………(1)  

Again, the slope of the side RP of the ∆ PQR is 6+31+2 = 3∙

Let QT be the perpendicular from Q on RP; hence, if the slope of the line QT be m1 then,

m1 × 3  = -1  

or, m1 =  -13

Therefore, tile equation of the straight line QT is

y – 1 = -13(x - 6)                        

or,  3y – 3 = - x + 6 

Or,  x + 3y = 9 ………………(2)

Now, solving equations (1) and (2) we get, x = 3, y = 2.

Therefore, the co-ordinates of the point of intersection of the lines (1) and (2) are (3, 2).

Therefore, the co-ordinates of the ortho-centre of the ∆PQR = the co-ordinates of the point of intersection of the straight lines PS and QT = (3, 2).

 The Straight Line




11 and 12 Grade Math

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