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
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