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.
2. A quadrilateral has the vertices at the points (-4, 2), (2, 6), (8, 5) and (9, -7). Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram.
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
11 and 12 Grade Math
From Condition of Parallelism of Lines to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Feb 24, 24 04:33 PM
Feb 24, 24 04:11 PM
Feb 24, 24 04:10 PM
Feb 24, 24 04:09 PM
Feb 24, 24 10:59 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.