coordinate geometry

**mathaddict** · November 15th 2008, 10:42 PM

Find the perpendicular distance between the pair of parallel lines 3x-4y+12 = 0 and 3x-4y+2=0

How do i start ??

**earboth** · November 15th 2008, 11:19 PM

Originally Posted by mathaddict

Find the perpendicular distance between the pair of parallel lines 3x-4y+12 = 0 and 3x-4y+2=0

How do i start ??

With your question there are two different ways to find the answer:

1. Choose any arbitrary point on the 1rst line; determine the equation of a line perpendicular to the 1rst line passing through this point; calculate the intersection point of the perpendicular line and the 2nd line; calculate the distance between the choosen point and the intersection point.

2. The equations of the two lines are given in normal form:

$3x-4y+12=0~\implies~(3, -4)\cdot (x, y) +12 = 0$ with

$(3, -4)$ indicating the perpendicular direction to the line and

12 indicating the perpendicular distance of the origin to the line.

If the normal vector is a unit vector then the distance of origin to the line is measured in units too. The difference between these two distances must be the distance between the two parallels:

$|(3, -4)| = 5$ Therefore the equation of the two lines become:

$\frac35x-\frac45y+\frac{12}5=0\ and\ \frac35x-\frac45y+\frac{2}5=0$

Therefore the distance between the parallels is:

$d = \frac{12}5 - \frac25 = 2$

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