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 m$_{1}$ and m$_{2}$ are perpendicular, then the angle between the lines θ is of 90°.

Therefore, cot θ = 0

⇒ $\frac{1 + m_{1}m_{2}}{m_{2} - m_{1}}$ = 0

⇒ 1 + m$_{1}$m$_{2}$ = 0

⇒ m$_{1}$m$_{2}$ = -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 = m$_{1}$x + c$_{1}$ and y = m$_{2}$ x + c$_{2}$ 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 = - $\frac{1}{tan β}$

or,  m$_{1}$ =  - $\frac{1}{m_{1}}$    

or, m$_{1}$m$_{2}$ = -1

Therefore, the condition of perpendicularity of the lines y = m$_{1}$x + c$_{1}$, and y = m$_{2}$ x + c$_{2}$ is m$_{1}$m$_{2}$ = -1.

Conversely, if m$_{1}$m$_{2}$ = - 1 then

tan ∙ tan β = - 1      

$\frac{sin α sin β}{cos α cos β}$ = -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 = $\frac{12 - 4}{2 - 6}$ = $\frac{8}{-4}$ = -2

Therefore the slope of the line perpendicular to PQ = - $\frac{1}{m}$ = 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:

m$_{1}$ = Slope of the side PQ = $\frac{4 - 5}{4 - 3}$ = -1

m$_{2}$ = Slope of the side PR = $\frac{4 - (-1)}{4 - (-1)}$ = 1

Now clearly we see that m$_{1}$m$_{2}$ = 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  $\frac{6 - 1}{1 - 6}$ =  $\frac{5}{-5}$ = -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 $\frac{6 + 3}{1 + 2}$ = 3∙

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

m$_{1}$ × 3  = -1  

or, m$_{1}$ =  -$\frac{1}{3}$

Therefore, tile equation of the straight line QT is

y – 1 = -$\frac{1}{3}$(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

• Straight Line
• Slope of a Straight Line
• Slope of a Line through Two Given Points
• Collinearity of Three Points
• Equation of a Line Parallel to x-axis
• Equation of a Line Parallel to y-axis
• Slope-intercept Form
• Point-slope Form
• Straight line in Two-point Form
• Straight Line in Intercept Form
• Straight Line in Normal Form
• General Form into Slope-intercept Form
• General Form into Intercept Form
• General Form into Normal Form
  
• Point of Intersection of Two Lines
  
• Concurrency of Three Lines
• Angle between Two Straight Lines
• Condition of Parallelism of Lines
• Equation of a Line Parallel to a Line
• Condition of Perpendicularity of Two Lines
• Equation of a Line Perpendicular to a Line
• Identical Straight Lines
• Position of a Point Relative to a Line
• Distance of a Point from a Straight Line
• Equations of the Bisectors of the Angles between Two Straight Lines
• Bisector of the Angle which Contains the Origin
• Straight Line Formulae
• Problems on Straight Lines
• Word Problems on Straight Lines
• Problems on Slope and Intercept

11 and 12 Grade Math

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