**Parallel and perpendicular lines :**

In this section, we are going to study about parallel and perpendicular lines.

**Parallel Lines : **

Parallel lines are two lines that never intersect. In the coordinate plane, that would look like this :

If we take a closer look at these two lines, the slope of both the lines is 1/2.

This can be generalized to any pair of parallel lines. Parallel lines always have the same slope and different *y*−intercepts.

**Postulate (Slopes of parallel lines) : **

In a coordinate plane, two lines are parallel if and only if they have the same slope.

**Perpendicular Lines : **

Recall that the definition of perpendicular is two lines that intersect at a 90°, or right angle. In the coordinate plane, that would look like this :

If we take a closer look at these two lines, we see that the slope of one is 2 and the other is -1/2.

This can be generalized to any pair of perpendicular lines in the coordinate plane. The slopes of perpendicular lines are opposite signs and reciprocals of each other.

or

The product of slopes of any two perpendicular lines is always equal to -1.

In the above example, we have

(-1/2) x 2 = -1

**Postulate (Slopes of perpendicular lines) : **

In a coordinate plane, two lines are perpendicular if and only if the product of their slopes is -1.

**Example 1 :**

Think of each segment in the diagram as part of a line.

Which of the lines appear to fit the descriptions given below ?

(i) Parallel to AB and contains D

(ii) Perpendicular to AB and contains D

(iii) Skew to AB and contains D

(iv) Name the plane(s) that contain D and appear to be parallel to plane ABE.

**Solution (i) :**

CD, GH and EF are all parallel to AB. But, only CD passes through D and is parallel to AB.

**Solution (ii) :**

BC, AD, AE and BF are all perpendicular to AB. But, only AD passes through D and is perpendicular to AB.

**Solution (iii) :**

DG, DH and DE all pass through D and are skew to AB.

**Solution (iv) :**

Only plane DCH contains D and is parallel to plane ABE.

**Example 2 : **

In the diagram given below, lines m, n and k represent three of the oars. If m||n and n||k, then prove m||k.

**Solution :**

m||n ∠1 ≅ ∠2 n||k ∠2 ≅ ∠3 ∠1 ≅ ∠3 m||k |
Given Corresponding angles postulate Given Corresponding angles postulate Transitive property of congruence Corresponding angle converse |

**Example 3 :**

In the diagram given below, find the slope of each line. Determine whether the lines j_{1 }and j_{2 }are parallel.

**Solution : **

Line j_{1} has a slope of

m_{2} = 4/2 = 2

Line j_{2} has a slope of

m_{2} = 2/1 = 2

Since the slope of the lines j_{1 }and j_{2 }are equal, the lines j_{1 }and j_{2 }are parallel.

**Example 4 :**

In the diagram given below,

Line n_{1} has the equation y = -x/3 -1.

Line n_{2} is parallel to the line n_{1} and passes through the point (3, 2).

Write the equation of the line n_{2}.

**Solution : **

The slope of the line n_{1} is -1/3. Because the lines n_{1} and n_{2} are parallel, they have the same slope. So, the slope of the line n_{2} is also -1/3.

Slope-intercept form equation of a line :

y = mx + b ------(1)

Because the line n_{2} is passing through (3, 2), substitute aa(x, y) = (3, 2) amd m = -1/3

2 = (-1/3)(3) + b

Simplify

2 = -1 + b

Add 1 to both sides.

3 = b

The equation of the required line is

(1) ------> y = (-1/3)x + 3

y = -x/3 + 3

**Example 5 :**

Decide whether the lines are perpendicular.

Line 1 : y = 3x/4 + 2

Line 2 : y = -4x/3 - 3

**Solution : **

When we compare the given equations to slope intercept equation of a line y = mx + b, we get

slope of line 1 = 3/4

slope of line 2 = -4/3

Multiply the slopes :

The product is

= (3/4) x (-4/3)

= - 1

Since the product of slopes of the lines is -1, the given lines are perpendicular.

**Example 6 :**

In the diagram given below, the equation y = 3x/2 + 3 represents a mirror. A ray of light hits the mirror at (-2, 0). What is the equation of the line p that is perpendicular to the mirror at this point ?

**Solution : **

The slope of the mirror is 3/2. So, the slope of the line p is -2/3.

Let y = mx + b be the equation of the line p.

Substitute (x, y) = (-2, 0) and m = -2/3 to find the value of b.

0 = (-2/3)(-2) + b

0 = 4/3 + b

Subtract 4/3 from both sides.

-4/3 = b

So, the equation of the line p is

y = -2x/3 - 4/3

After having gone through the stuff given above, we hope that the students would have understood "Parallel and perpendicular lines".

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