Parallel Lines and Perpendicular Lines

 7.2

Parallel Lines and Perpendicular Lines

 
 

 

If two straight lines are parallel, then their gradients are equal, that is \(m_1 = m_2\) , and vice versa.

 

Example:

Given that the following pair of straight lines are parallel, find the value of \(k\).

 

\(\begin{aligned} &3x+ky=2 \\\\ &\text{and} \\\\ &4y+x-8=0 \end{aligned}\)

 
Write in the gradient form.
 

\(\begin{aligned} 3x+ky&=2 \\\\ ky&=-3x+2 \\\\\ y&=-\dfrac{3}{k}x+\dfrac{2}{k} \\\\ \therefore m_1&=-\dfrac{3}{k}.\end{aligned}\)

 
and
 

\(\begin{aligned} 4y+x-8&=0 \\\\ 4y&=-x+8 \\\\ y&=-\dfrac{1}{4}x+2 \\\\ \therefore m_2&=-\dfrac{1}{4}. \end{aligned}\)

 
Since the following pair of straight lines are parallel, hence
 
\(m_1 = m_2.\)
 
Therefore,
 

\(\begin{aligned} -\dfrac{3}{k}&=-\dfrac{1}{4} \\\\ k&=12. \end{aligned}\)

 
 

 
If two straight lines are perpendicular, then the product of their gradients is \(-1\), that is \(m_1m_2=-1\), and vice versa.
 

Example:

Given that the following pair of straight lines are perpendicular, find the value of \(p\).

 

\(\begin{aligned} &px+6y=8 \\\\ &\text{and} \\\\ &y-6x=24 \end{aligned}\)

 
Write in gradient form.
 

\(\begin{aligned}px+6y&=8\\\\ 6y&=-px+8 \\\\ y&=-\dfrac{p}{6}x+\dfrac{4}{3}\\\\ \therefore m_1&=-\dfrac{p}{6}. \end{aligned}\)

 
and
 

\(\begin{aligned} y-6x&=24 \\\\ y&=6x+24\\\\ \therefore m_2&=6. \end{aligned}\)

 
Since the following pair of straight lines are perpendicular, hence
 

\(m_1m_2=-1.\)

 
Therefore,
 

\(\begin{aligned} \left( -\dfrac{p}{6} \right) \left(6 \right) &=-1\\\\-p&=-1 \\\\ p&=1. \end{aligned}\)

 

Parallel Lines and Perpendicular Lines

 7.2

Parallel Lines and Perpendicular Lines

 
 

 

If two straight lines are parallel, then their gradients are equal, that is \(m_1 = m_2\) , and vice versa.

 

Example:

Given that the following pair of straight lines are parallel, find the value of \(k\).

 

\(\begin{aligned} &3x+ky=2 \\\\ &\text{and} \\\\ &4y+x-8=0 \end{aligned}\)

 
Write in the gradient form.
 

\(\begin{aligned} 3x+ky&=2 \\\\ ky&=-3x+2 \\\\\ y&=-\dfrac{3}{k}x+\dfrac{2}{k} \\\\ \therefore m_1&=-\dfrac{3}{k}.\end{aligned}\)

 
and
 

\(\begin{aligned} 4y+x-8&=0 \\\\ 4y&=-x+8 \\\\ y&=-\dfrac{1}{4}x+2 \\\\ \therefore m_2&=-\dfrac{1}{4}. \end{aligned}\)

 
Since the following pair of straight lines are parallel, hence
 
\(m_1 = m_2.\)
 
Therefore,
 

\(\begin{aligned} -\dfrac{3}{k}&=-\dfrac{1}{4} \\\\ k&=12. \end{aligned}\)

 
 

 
If two straight lines are perpendicular, then the product of their gradients is \(-1\), that is \(m_1m_2=-1\), and vice versa.
 

Example:

Given that the following pair of straight lines are perpendicular, find the value of \(p\).

 

\(\begin{aligned} &px+6y=8 \\\\ &\text{and} \\\\ &y-6x=24 \end{aligned}\)

 
Write in gradient form.
 

\(\begin{aligned}px+6y&=8\\\\ 6y&=-px+8 \\\\ y&=-\dfrac{p}{6}x+\dfrac{4}{3}\\\\ \therefore m_1&=-\dfrac{p}{6}. \end{aligned}\)

 
and
 

\(\begin{aligned} y-6x&=24 \\\\ y&=6x+24\\\\ \therefore m_2&=6. \end{aligned}\)

 
Since the following pair of straight lines are perpendicular, hence
 

\(m_1m_2=-1.\)

 
Therefore,
 

\(\begin{aligned} \left( -\dfrac{p}{6} \right) \left(6 \right) &=-1\\\\-p&=-1 \\\\ p&=1. \end{aligned}\)