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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