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 xaxis 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 orthocentre 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 coordinates of the point of intersection of the lines (1) and (2) are (3, 2).
Therefore, the coordinates of the orthocentre of the ∆PQR = the coordinates of the point of intersection of the straight lines PS and QT = (3, 2).
11 and 12 Grade Math
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