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
11 and 12 Grade Math
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