Question

Which of the following pairs of the lines are parallel and which of them are perpendicular to each other?
i) $y = 3x + 4,2y = 6x + 5$
ii) $y - 2x = 7,4x - 2y = 5$
iii) $x + 2y + 5 = 0,2x - y + 20 = 0$
iv) $y = 3x - 7,3x + 9y - 2 = 0$

Answer 1

Hint: Hint: Two lines are parallel if they have the same slope. If two lines are perpendicular, then the product of their slopes will be $ - 1$. Using these two results we can find which pairs are parallel and which are perpendicular.

Formula used:
Standard equation of a line is $y = mx + c$, where $m$ is the slope of the line.
Consider two lines.
$y = {m_1}x + {c_1}$ and $y = {m_2}x + {c_2}$
If two lines are parallel, then they have the same slope. $ \Rightarrow {m_1} = {m_2}$
If two lines are perpendicular, then the product of their slopes is $ - 1$. $ \Rightarrow {m_1}{m_2} = - 1$

Complete step-by-step answer:
We are given four pairs of lines.
Let us check each pair one by one.
i) $y = 3x + 4,2y = 6x + 5$
Standard equation of a line is $y = mx + c$, where $m$ is the slope of the line.
Writing these lines in standard form we get,
 $y = 3x + 4,y = 3x + \dfrac{5}{2}$
So slope of these line are ${m_1} = 3,{m_2} = 3$
If two lines are parallel, then they have the same slope. $ \Rightarrow {m_1} = {m_2}$
If two lines are perpendicular, then the product of their slopes is $ - 1$. $ \Rightarrow {m_1}{m_2} = - 1$
Here, ${m_1} = {m_2} = 3$
So these lines are parallel.
ii) $y - 2x = 7,4x - 2y = 5$
Writing in standard form we get,
$y = 2x + 7,4x - 5 = 2y$
Dividing second equation by $2$,
$ \Rightarrow y = 2x + 7,y = 2x - \dfrac{5}{2}$
This gives slopes ${m_1} = 2,{m_2} = 2$ $ \Rightarrow {m_1} = {m_2}$
So the lines are parallel.

iii) $x + 2y + 5 = 0,2x - y + 20 = 0$
Converting to standard form we get,
$2y = - x - 5,2x + 20 = y$
Dividing the first equation by $2$
$ \Rightarrow y = - \dfrac{x}{2} - \dfrac{5}{2},y = 2x + 20$
This gives slopes ${m_1} = - \dfrac{1}{2},{m_2} = 2$ $ \Rightarrow {m_1}{m_2} = - 1$
So the lines are perpendicular.

iv) $y = 3x - 7,3x + 9y - 2 = 0$
Writing in standard form we get,
$y = 3x - 7,9y = - 3x + 2$
Dividing by $9$ on the second equation gives,
$ \Rightarrow y = 3x - 7,y = - \dfrac{1}{3}x + \dfrac{2}{9}$
This gives slopes ${m_1} = 3,{m_2} = - \dfrac{1}{3}$ $ \Rightarrow {m_1}{m_2} = - 1$
So the lines are perpendicular.
$\therefore $ Pairs (i) and (ii) are parallel and pairs (iii) and (iv) are perpendicular.

Note: It is important to write the given equations in standard form. If they are given in standard form we can simply take the slope as coefficient of $x$. Otherwise doing so will lead to the wrong answer.