# Slope of a Straight Line

### Slope (or gradient) of a straight line

If ( 90°) is the inclination of a
straight line, then tan is called its **slope** (or
**gradient**). The slope of a line is usually denoted by m.

**Remark.** Since tan is not defined when
= 90°, therefore, the slope of a vertical line is not
defined.

### Slope of the line joining two points

The slope m of a non-vertical line passing through the points P(x_{1}
, y_{1}) and Q(x_{1}, y_{1}) is given by

**slope = m = (y _{2} -y_{1)/(}x_{2}
-x_{1)}**

### Remarks

- Two (non-vertical) lines are parallel iff their slopes are equal.
- Two (non-vertical) lines are perpendicular iff the product of their slopes = -1.
- Slope of a perpendicular line is the negative reciprocal of the slope of the given line.

## Illustrative Examples

### Example

Without using Pythagoras theorem, show that the points A (1, 2), B (4, 5) and C (6, 3) are the vertices of a right-angled triangle.

### Solution

In ABC, we have

m_{1} = slope of AB = (5 -2)/(4-1) = 3/3 = 1 and

m_{2} = slope of BC = (3 -5)/(6-4) = 2/2 = -1

m_{1} m_{2} = 1. (-1) = -1 => ABBC

Hence, the given points are the vertices of a right-angled triangle.

### Example

Using slopes, show that the points A (6, -1), B (5, 0) and C (2, 3) are collinear.

### Solution

Slope of AB = [0 -(-1)]/(5-6) = 1/(-1) = -1

and slope of AC = [3 -(-1)]/(2-6) = 4/(-4) = -1

=> slope of AB = slope of AC

=> AB and AC are parallel.

But AB and AC have point A is common, therefore, the given points A, B and C are collinear.

## Exercise

- Find the slope of a line whose inclination is:

(i) 30° (ii) 2 /3 (iii) /3 - Find the inclination of a line whose gradient is:

(i) 1/ 3 (ii) -1 (iii) - 3 - Find the gradient of the line containing the points

(i) (-2, 3) and (5, -7) (ii) (3, -7) and (0, 2) - A line passes through the points (4, -6) and (-2, -5). Does it make an acute angle with the positive direction of x-axis?
- Find the equation of the locus of all points P such that the slope of the line joining origin and P is - 2.
- Show that the line joining (2, -3) and (-5, 1) is

(i) parallel to the line joining (7, -1) and (0, 3)

(ii) perpendicular to the line joining (4, 5) and (0, - 2) - State, whether the two lines in each of the following problems are
parallel, perpendicular or neither:

(i) through (2, -5) and (-2, 5); through (6, 3) and (1, 1)

(ii) through (5, 6) and (2, 3); through (9, -2) and (6, - 5)

(iii) through (8, 2) and (-5, 3); through (16, 6) and (3, 15) - Find y if the slope of the line joining (-8, 11), (2, y) is - 4/3.
- Find the value of x so that the line through (x, 9) and (2, 7) is parallel to the line through (2, - 2) and (6, 4).
- Without using Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.

## Answers

**1.**(i) 1/ 3 (ii) - 3 (iii) not defined

**2.**(i) 30° (ii) 135° (iii) 120°

**3.**(i)- 10/7 (ii) -3

**4.**No

**5.**2x +y = 0

**7.**(i) Perpendicular (ii) parallel (iii) neither

**8.**-7/3

**9.**10./3