Introduction to gradient of straight lines:-

In mathematics, the line rises to straight lines at which the rate is vertically for every unit of the straight lines is called the gradient. In gradient of straight lines follows:

Gradient = `(Rise)/(Run)`

= `( x)/ (y)`

= `(Y_2 -Y_1)/ (X_2 -X_1)`

Note:

In m is denoted as gradient of straight lines.

` m = (Y_2 -Y_1) / (X_2 -X_1)`

Applications of Gradient of Straight Lines

In gradient of straight lines is important one it build the house roof to enabled water rain inside the roof .In a particular area aeroplane ascends has travel to sky on the top of the sky
aeroplane is take off by using the gradient method on the another area safely land. Cricket, football etc is made with gradient straight lines.

Example Problems for Gradient of Straight Lines

Problem 1:-

Find P(– 4, 5) and Q(4, 17) in gradient of straight lines passes through the joint points .

Solution:

Let (`x_1,y_1` ) = (-4,5) and (`x_2,y_2` ) = (4,17)

m = `(y_2 - y_1)/(x_2 - x_1)`

= `(17-5)/(4-4)`

= `12/8`

= 1.5

So, PQ is 1.5 in gradient of straight lines.

Note:

The slope of a lines upward and value x is increases through the positive line is called as gradient of a straight lines.

Problem 2:-

Find A(6, 0) and B(0, 3) in gradient of straight lines passes through the joint points .

Solution:

Let (`x_1,y_1` ) = (6,0) and (`x_2,y_2` ) = (0,3)

m = `(y_2 - y_1)/(x_2 - x_1)`

= `(3-0)/(0-6)`

= `3/-6`

= `-1/2`

So, the gradient of the line AB is `-1/2`

Note:

The slope of a lines downward and value x is increases through the negative line is called as gradient of a straight lines.

Problem 3:-

A horse gallops for 35 minutes and covers a distance of 25 km, as shown in the diagram to describe the gradient of the line and its meaning.

Solution:

Let (`t_1,d_1` ) = (0,0) and (`t_2,d_2` ) = (35,25)

m = `(d_2 - d_1)/(t_2 - t_1)`

= `(35-0)/(25-0)`

= `35/25`

= `7/5`

So, the gradient of the line is `7/5` km/min.In the above example, show the gradient of the distance-time graph gives the speed (in kilometers per minute) and the distance covered by the horse
can be represented by the equation:

d = `7/5` t (Distance = Speed x Time)

Problem 4:-

In documents has been transporting to courier department it shown in the line diagram to describe the gradient of straight lines.

Solution:

Let (`d_1, C_1` ) = (0,5) and (`d_2,c_2` ) = (5,24)

m = `(c_2 - c_1)/(d_2 - d_1)`

= `(24-5)/(5-0)`

= `19/5`

= 3.8

So, the gradient of the line is 3.8. This means that the cost of transporting documents is $3.8 per km plus a fixed charge of $5, that is it costs $5 for the courier to arrive and $5 for every
kilometer travelled to deliver the documents.