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Aniket Patra Grade: Upto college level
        

If m is fixed and c is an arbitrary constant.then the equation y=mx+c represents a family of straight line.How is this possible?If c changes then slope also changes.Then hoe is it possible,Please interpret the result geometrically.

6 years ago

Answers : (1)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear aniket,


In the straight line equation: y = mx + c:




'x' and 'y' are the coordinates of the points that satisfy the function and so lie on the straight line graph.


'm' is the gradient of the straight line graph, and


'c' is the 'y intercept' of the straight line graph.


 




About 'gradient'...



'Gradient' is a number that represents the steepness of a straight line. A horizontal line has gradient zero. A 45º line has gradient 1, and a vertical line has infinite gradient.


This diagram shows some different lines and their gradients (the 'm' values):







The sign of the gradient is important. Positive gradients (like those in the diagram above) mean that the line is slopinguphill as you go left to right. If a line slopes downhill going left to right, then it has a negative gradient, as shown below:







Gradient is an exact quantity. This is how it is defined mathematically...



1. Start with any two points on a straight line, (x1,y1) and (x,y):





2. Starting at (x1,y1) on the left, then moving to the right along the line to (x,y), we measure the change in 'x', and the change in 'y'.



The change in 'x'  =  x - x1 while the change in 'y' =   y - y1



Gradient 'm' is defined as follows:





The Greek capital letter 'Delta' : D is often used in mathematics to mean a change in some quantity. So the gradient definition can be written like this:





To measure the change in 'x' and 'y' between the two points in the graph above, we draw a right triangle beneath the graph. The length of the vertical leg is: (y-y1), and the horizontal leg has length (x-x1)...





 So the gradient of this line is: 




Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.


All the best.


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Askiitians Expert


Sagar Singh


B.Tech, IIT Delhi


 




6 years ago
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