[Maths Class Notes] on Slope Pdf for Exam

The slope or gradient of a line specifies both the direction and steepness of a line. The slope is often represented by the letter ‘m’. The slope is determined by finding the ratio of “vertical change” to the “horizontal change” between any two different points on a line. The steepness or grade of a line is measured by the absolute value of the shape. A slope with a greater absolute value represents a steeper line. The direction of a line either rises, falls, and is horizontal or vertical.

• A line that extends from left to right has a positive run and positive rise, and also yielding a positive slope i.e. m > 0

• A line that declines from left to right has a negative run and negative fall, and also yielding a negative slope i.e. m < o

• Horizontal lines have a zero positive slope, as they have zero rise and a positive run. 

• The slope is undefined if the line is vertical as the vertical line has zero rise and any amount of run.

Slope Equation

As we know, Tan θ = [frac{Height}{Base}]

And, we know that between any two given points 

[frac{Height}{Base}=frac{y_{2}-y_{1}}{x_{2}-x_{1}}]

Finally, we get slope equations as: 

[m=Tantheta =frac{Delta y}{Delta x}]

Therefore, this becomes our final slope equation at any given point.

Equation of a Straight Line

The general equation of a straight line is represented in the form of y = mx + c, where m is the gradient and coordinates of the y-intercepts are (0,c). 

We can determine the equation of a straight line when the gradient and point on the line are given by using the formula: y – b = m ( x – a)

Here, m represents the gradients and (a, b) is on the line.

Equation of a Line Example

Q. Find the Equation of a Line With Gradient 5, Passing Through the Point ( 4,1).

Solution: 

Using the formula y – b = m ( x – a), and substituting the values: m = 5, a = 4, and b = 1

We get, y  – 1 = 5 ( x – 4)

y – 1 = 5x – 20

y =  5x – 20 + 1

y = 5x – 19

y =  5x – 19

Therefore, the equation of a line with gradient 5, passing through ( 4,1) is y =  5x – 19

How to Find Equation of Line Passing Through Two Given Points?

If a line passes  through two points M (x₁, y₁) and N (x₂, y₂) such that x₁ is not equal to x₂ and y₁ is not equal to y₂, the equation of a line can be found using the formula mentioned below:

[frac{y-y_{1}}{y_{2}-y_{1}}=frac{x-x_{1}}{x_{2}-x_{1}}]

Here, the gradient (m) can be calculated as :

[m=frac{y_{2}-y_{1}}{x_{2}-x_{1}}]

Line Perpendicular

Perpendicular lines are two lines that meet at right angles (90⁰). The slopes of the two lines are negative reciprocals of each other. This means when one line has a slope of m, a perpendicular line has a slope of -1/m.

When we get multiply slope m by perpendicular slope -1/m., we get the answer -1. 

What Does the Slope of a Velocity Time Graph Give?

Velocity is a term that measures both speed and direction of a moving body. A change in velocity is known as acceleration. When the velocity and time are graphed on the y – axis and the x-axis respectively, then the slope of the velocity-time graph gives the acceleration of the object.

The slope is the ratio of change in the y-axis and change in the x-axis.

Therefore, we can determine the slope by using the following formula:

[m=frac{y_{2}-y_{1}}{x_{2}-x_{1}}]

Here,

Solved Example:

1. Find the Equation of a Line that Passes Through the Two Points (2,3) and ( 6,-5)

Solution:

The equation of a line through the point (2,3) and ( 6,-5) can be determined using the formula:

[frac{y-y_{1}}{y_{2}-y_{1}}=frac{x-x_{1}}{x_{2}-x_{1}}]

As the gradient (m) is not given, we will find the gradient by using the formula:

[m=frac{y_{2}-y_{1}}{x_{2}-x_{1}}]

Substituting the values x₁ = 2, x₂ = 6, y₁ = 3, and y₂ = -5 in the above formula, we get,

[m=frac{-5-3}{6-2}]

[m=frac{-8}{4}]

m = -2

Using the formula y – y₁ = m (x – x₁), and substituting the values: m = -2 , x₁ = 2 and y₁  = 3

We get,  y – 3 = -2 ( x – 2)

y – 3 = -2x + 4

Therefore, the equation of a line passing through the point (-1,2) and ( 2,4) is 2x + y + 1 = 0.

2. Find the Equation of a Line That Passes Through the Point (2,0) and Has a Gradient -2.

Solution: 

Using the formula y – b = m ( x – a), and substituting the values: m = -2, a = 2 and b = 0

We get, y – 0 = – 2 (x – 2)

y – 0 = -2x + 4

y  =  -2x + 4.