# 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.

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, m1 = Slope of the side PQ = $$\frac{\frac{11}{2} - 4}{5 - (-1)}$$= ¼

m2 = Slope of the side RS = $$\frac{\frac{-5}{2} + 1}{\frac{5}{2} - \frac{17}{2}}$$ = ¼

Clearly, m1 = m2. 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.