# 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

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