##### Parallel and Perpendicular lines

Two lines in a plane when are extended either meet or they don’t meet at all.

Parallel lines

Two lines on the same plane are parallel if they never intersect with each other even if we extend them. The distance between them is always the same and they have exactly the same steepness which means their slopes are identical. The only difference between two parallel lines is the y-intercept.

Perpendicular lines

Two lines in the same plane are perpendicular if they intersect and they form a right angle (90°) at the intersection point. Perpendicular lines do not have the same steepness. The slopes of perpendicular lines are different from one another. The slope of one line is the negative reciprocal of the slope of the other line.

We mention the fact about how the steepness and the slope of two lines changes depending on their position.

The fact if two lines are parallel to each other, perpendicular to each other or either is based on their slopes, so the value of their gradient.

If we have two lines in a plane $\displaystyle {{l}_{1}}$ and $ \displaystyle {{l}_{2}}$ with their gradient respectively $\displaystyle {{m}_{1}}$ and $ \displaystyle {{m}_{2}}$

**1.** The lines $\displaystyle {{l}_{1}}$ and $ \displaystyle {{l}_{2}}$ are parallel if they have the same slope $\displaystyle {{m}_{1}}={{m}_{2}}$

**2.** The lines $\displaystyle {{l}_{1}}$ and $ \displaystyle {{l}_{2}}$ are perpendicular if their slopes are negative reciprocals to each other $ \displaystyle {{m}_{1}}=-\frac{1}{{{{m}_{2}}}}$

Don’t forget!

When we multiply the gradients of two perpendicular lines the product is always minus one.$ \displaystyle {{m}_{1}}\cdot {{m}_{2}}=-1$

Example 1: Identify which of the lines are parallel and which perpendicular.

a) $ \displaystyle y=3x+1$

b) $ \displaystyle y=\frac{1}{4}x-5$

c) $ \displaystyle y=3x+12$

d)$ \displaystyle y=-4x+1$

Solution

Parallel lines have the same slope. Since the functions $\displaystyle y=3x+1$ and $\displaystyle y=3x+12$ each have the same slope 3, they represent parallel lines.

Perpendicular lines have negative reciprocal slopes. Since -4 and -1/4 are negative reciprocals the equations $ \displaystyle y=\frac{1}{4}x-5$ and $\displaystyle y=-4x+1$ , they represent perpendicular lines.

**Example 2**: Find the equation of a line that is perpendicular to the line $ \displaystyle y=2x-2$ and goes through the point (1,3).

**Solution**

We know that the general equation of a straight line is $ \displaystyle y=mx+c$

Firstly we need to find the gradient of the line $ \displaystyle y=2x-2$

If we label the gradient of our line as $ \displaystyle {{m}_{1}}$ and the gradient of the line that is perpendicular with our line $ \displaystyle {{m}_{2}}$ then we know that the product of those two gradient should be -1.

The gradient of our line is $ \displaystyle {{m}_{1}}=2$.

Meanwhile the gradient of the line perpendicular to our line is: $ \displaystyle {{m}_{2}}=-\frac{1}{{{{m}_{1}}}}$

$\displaystyle {{m}_{2}}=-\frac{1}{2}$

After finding the gradient the equation of the line we want to find takes the form $ \displaystyle y=-\frac{1}{2}x+c$ .

To find the value of c, we substitute the point (1,3) on our equation, since the graph of this line passes through this point.

$\displaystyle y=-\frac{1}{2}x+c$

$\displaystyle 3=-\frac{1}{2}\cdot 1+c$

$\displaystyle 3=-\frac{1}{2}+c$

$ \displaystyle c=3+\frac{1}{2}$

$ \displaystyle c=\frac{6}{2}+\frac{1}{2}=\frac{7}{2}$

The final form of our line that is perpendicular with the given line is: $\displaystyle y=-\frac{1}{2}x+\frac{7}{2}$

Example 3: Find the equation of a line that is parallel to the line $ \displaystyle x+y-1=0$ and goes through the point (-1,1).

Solution

Firstly we rewrite our line $ \displaystyle x+y-1=0$ on the correct form $ \displaystyle y=-x+1$

Then we find the gradient of our line that is $\displaystyle {{m}_{1}}=-1$

We know that parallel lines have the same gradient so the gradient of the line we are going to find is

$ \displaystyle {{m}_{1}}={{m}_{2}}=-1$

The line takes the form $ \displaystyle y=-x+c$

To find the value of c, we substitute the point (-1,1) on our equation, since the graph of this line passes through this point.

$ \displaystyle y=-x+c$

$ \displaystyle 1=-(-1)+c$

$ \displaystyle 1=1+c$

$\displaystyle c=1-1=0$

The final form of our line that is parallel with the given line is: $\displaystyle y=-x$