Condition of Parallelism of Lines

We will learn how to find the condition of parallelism of lines.

If two lines of slopes m\(_{1}\) and m\(_{2}\) are parallel, then the angle θ between them is of 90°.

Therefore, tan θ = tan 0° = 0

⇒ \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\) = 0, [Using tan θ = ± \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)]

⇒ \(m_{2} - m_{1}\) = 0

⇒ m\(_{2}\) = m\(_{1}\)

⇒ m\(_{1}\) = m\(_{2}\)

Thus when two lines are parallel, their slopes are equal.

Let, the equations of the straight lines AB and CD are y = m\(_{1}\)x+ c1 and y = m\(_{2}\)x + c\(_{2}\) respectively.

If the straight lines AB and CD be parallel, then we shall have m\(_{1}\) = m\(_{2}\).

That is the slope of line y = m\(_{1}\) x+ c\(_{1}\)  = the slope of the line y = m\(_{2}\)x + c\(_{2}\)

Conversely, if m\(_{1}\) = m\(_{2}\) then the lines y = m\(_{1}\) x+ c\(_{1}\) and y = m\(_{2}\)x + c\(_{2}\) make the same angle with the positive direction of x-axis and hence, the lines are parallel.

Solved examples to find the condition of parallelism of two given straight lines:

1. What is the value of k so that the line through (3, k) and (2, 7) is parallel to the line through (-1, 4) and (0, 6)?

Solution:

Let A(3, k), B(2, 7), C(-1, 4)and D(0, 6) be the given points. Then,

m\(_{1}\) = slope of the line AB = \(\frac{7 - k}{2 - 3}\) = \(\frac{7 - k}{-1}\) = k -7

m\(_{2}\) = slope of the line CD = \(\frac{6 - 4}{0 - (-1)}\) = \(\frac{2}{1}\) = 2

Since, Ab and CD are parallel, therefore = slope of the line AB = slope of the line CD i.e., m\(_{1}\) = m\(_{2}\).

Thus,

k - 7 = 2

Adding 7 on both sides we get,

K - 7 + 7 = 2 + 7

K = 9

Therefore, the value of k = 9.

Solution:

Let A(-4, 2), B(2, 6), C(8, 5) and D(9, -7) be the vertices of the given quadrilateral. Let P,Q, R and S be the mid-points of AB, BC, CD and DA respectively. Then the coordinates of P, Q, R and S are P(-1, 4), Q (5, 11/2), R(17/2, -1) and S(5/2, -5/2).

In order to prove that PQRS is a parallelogram, it is sufficient to show that PQ is parallel to RS and PQ =RS.

We have, m\(_{1}\) = Slope of the side PQ = \(\frac{\frac{11}{2} - 4}{5 - (-1)}\)= ¼

m\(_{2}\) = Slope of the side RS = \(\frac{\frac{-5}{2} + 1}{\frac{5}{2} - \frac{17}{2}}\) = ¼

Clearly, m\(_{1}\) = m\(_{2}\). This shows that PQ is parallel to RS.

Now, PQ = \(\sqrt{(5 + 1)^{2} + (\frac{11}{2} - 4)^{2}}\) = \(\frac{√153}{2}\)

RS = \(\sqrt{(\frac{5}{2} - \frac{17}{2})^{2} + (-\frac{5}{2} + 1)^{2}}\) = \(\frac{√153}{2}\)

Therefore, PQ = RS

Thus PQ ∥ RS and PQ = RS.

Hence, PQRS is a parallelogram.

● 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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