# Thread: Parallel and Perpendicular lines...?

1. ## Parallel and Perpendicular lines...?

Hi folks, yr 10 student dilemna here!
need to know this 4 test tomoro, just some pointers for finding the equations of straight lines which pass through a given point and perpendicular to the line whose equation is given by:...... and so on,, same with parallel.
how to show that lines are perpendicular, with 2 equations???
any help welcome,
thanks a lot!

2. the lines are perpendicular when the gradient of the first line multiplied by the gradient of the second line gives you -1

$
1^{st} Line, y = m_1 x+ c_1
$

$
2^{nd} Line, y = m_2 x+ c_2
$

$
m_1 m_2 = -1
$

3. Originally Posted by mooch
Hi folks, yr 10 student dilemna here!
need to know this 4 test tomoro, just some pointers for finding the equations of straight lines which pass through a given point and perpendicular to the line whose equation is given by:...... and so on,, same with parallel.
how to show that lines are perpendicular, with 2 equations???
any help welcome,
thanks a lot!
The general equation of a straight line is:

$y = \underbrace{m}_{direction} \cdot x + \underbrace{c}_{y-intercept}$

Two lines are parallel if they have the same direction (=slope, gradient):

$l_1: y = m_1 \cdot x + c_1$ ... and ... $l_2: y = m_2 \cdot x + c_2$

$l_1\parallel l_2~\implies~m_1=m_2$

Two lines are perpendicular if

$m_1 \cdot m_2 = -1$

$l_1\perp l_2~\implies~m_2=-\frac1{m_1}$

If you have the slope (=direction) of a straight line and a point $P(x_P, y_P)$ which is placed on this line (or the line passes through the point P) then the coordinates of P must satisfy the equation:

$y-y_P = m(x-x_P)$ ... This is the point-slope-formula of a straight line.