**Rise over run formula :**

The formula for slope is referred to as "**rise over run**",

Because the fraction consists of the "**rise**" (the change in y, going up or down) divided by the "**run**" (the change in x, going from left to the right).

The figure given below illustrates this.

From the above figure, the slope of the straight line joining the points A (x₁, y₁) and B (x₂, y₂) is

That is,

If the equation of a straight line given in general form

ax + by + c = 0,

then, the formula to find slope of the line is

**Let **θ be the angle between the straight line "l" and the positive side of x - axis.

The figure given below illustrates this.

Then, the formula to find slope of the line is

**m = tan θ**

In the general form of equation of a straight line

ax + by + c = 0,

(i) if "x" term is missing, then the line will be parallel to x - axis and its slope will be zero.

We know that slope = change in y / change in x

In the above figure, the value of "y" is fixed and that is "k"

So, there is no change in "y" and change in y = 0

Slope = 0 / change in x

**Slope = 0 **

(i) if "y" term is missing, then the line will be parallel to y - axis and its slope will be undefined.

We know that slope = change in y / change in x

In the above figure, the value of "x" is fixed and that is "c"

So, there is no change in "x" and change in x = 0

Slope = change in y / 0

**Slope = Undefined **

Slope of the coordinate axes "x" and "y".

(i) Slope of "x" axis zero.

(ii) Slope of "y" axis undefined.

When we look at a straight line visually, we can come to know its slope easily.

To know the sign of slope of a straight line, always we have to look at the straight line from left to right.

The figures given below illustrate this.

**Problem 1 :**

Find the angle of inclination of the straight line whose slope is 1/√3

**Solution :**

Let θ be the angle of inclination of the line.

Then, slope of the line, m = tan θ

Given : Slope = 1/√3

So, we have

tan θ = 1/√3

θ = 30°

**Hence, the angle of inclination is 30° **

**Problem 2 :**

Find the slope of the straight line passing through the points (3, -2) and (-1, 4).

**Solution :**

Let (x₁, y₁) = (3, -2) and (x₂, y₂) = (-1, 4)

Then, the formula to find the slope,

m = (y₂ - y₁) / (x₂ - x₁)

Plug (x₁, y₁) = (3, -2) and (x₂, y₂) = (-1, 4)

m = (4 + 2) / (-1 - 3)

m = - 6 / 4

m = - 3 / 2

**Hence, the slope is -3/2**

**Problem 3 :**

Using the concept of slope, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

**Solution :**

Slope of the line joining (x₁, y₁) and (x₂, y₂) is,

m = (y₂ - y₁) / (x₂ - x₁)

Using the above formula,

Slope of the line AB joining the points A (5, - 2) and B (4- 1) is

= (-1 + 2) / (4 - 5)

= - 1

Slope of the line BC joining the points B (4- 1) and C (1, 2) is

= (2 + 1) / (1 - 4)

= - 1

Thus, slope of AB = slope of BC.

Also, B is the common point.

**Hence, the points A , B and C are collinear.**

**Problem 4 :**

Find the slope of the line 3x - 2y + 7 = 0.

**Solution :**

When the general form of equation of a straight line is given, the formula to find slope is

m = - coefficient of x / coefficient of y

In the given line 3x - 2y + 7 = 0,

coefficient of x = 3 and coefficient of y = - 2

Slope, m = (-3) / (-2) = 3/2

**Hence, slope of the given line is 3/2. **

**Problem 5 :**

If the straight line 5x + ky - 1 = 0 has the slope 5, find the value of "k"

**Solution :**

When the general form of equation of a straight line is given, the formula to find slope is

m = - coefficient of x / coefficient of y

In the given line 3x - 2y + 7 = 0,

coefficient of x = 3 and coefficient of y = k

Slope, m = -5 / k

Given : Slope = 5

So, we have 5 = -5/k

5k = -5

k = -1

**Hence, the value of "k" is -1. **

After having gone through the stuff given above, we hope that the students would have understood "Rise over run formula".

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