Simple gradient/graph question.

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In summary, the points A and B have coordinates (6,-1) and (2,5) respectively and the gradient of the line AB is -3/2. To find the equation of the line, you can use the formula y-y1 = m(x-x1), where m is the gradient and (x1, y1) is one of the points on the line. In this case, using (2,5) as (x1,y1), we get the equation 3x+2y=16. To find the equation of the perpendicular line that crosses through point B, we can use the fact that the gradient of perpendicular lines are negative reciprocals. Thus, the equation of the perpendicular line is
  • #1
DeanBH
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1. The points A and b have Coords (6,-1) and (2,5) respectively

a.Show gradient is - 3/2

-1-5 = -6
6-2 = 4

= -3/2

Hence find the equation of the line AB with answer in form Ax + By = C

A B C are integers.

- now this isn't a homework question I am just doing some extra work in my own time. I have never been taught this, but when looking through past papers i have seen that it has actualy come up, can someone explain to me how exactly you do these. I have attempted it, and know the answer because i have the mark scheme, but i don't 100% understand it.


i know you use y-5 and y+1 somewhere, which is the inverse of the Y coordinates in the 2 points I am given. and i know that you somehow times the X by the gradient with those coordinates inverted but I am not sure how it works.

answer = 3x+2y = 16 can anyone explain this?
 
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  • #2
You got the gradient by (1) subtracting the two y values, (2) subtracting the two x values, (3) dividing the first by the second.

And, of course, you could do that for any two points on the line and get the same thing. Suppose you pick an arbitrary point on the line and call it (x, y). Now use (x,y) and (2, 5) to calculate the gradient. You would have (y- 5)/(x- 2) and that must give you the same thing as (-1-5)/(6-2)= -3/2. That is, you must have (y- 5)/(x-2)= -3/2. "Cross multiplying", that is, multiplying both sides by 2 and (x-2), we get 2(y- 5)= -3(x- 2). Multiply those out, move the "x" term to the left, and the constants to the right and see what you get.
 
  • #3
2y-10=-3x+6
2y+3x=16

hurraynext question is to find equation of perpendicular line that crosses through point b.

is that (y-5)/(X-2) = 2/3

=3y-15 = 2x-4
=3y-2x=11?
or is something wrong
 
Last edited:

1. What is a simple gradient/graph question?

A simple gradient/graph question is a type of math problem that involves finding the slope or rate of change of a line on a graph. It usually requires you to calculate the rise over run, or the change in y over the change in x, to determine the slope of the line.

2. How do I find the slope of a line on a graph?

To find the slope of a line on a graph, you need to choose two points on the line and calculate the difference in their y-values (rise) and x-values (run). Then, divide the rise by the run to get the slope. This can also be represented as the change in y over the change in x, or Δy/Δx.

3. What is the difference between positive and negative slope?

A positive slope means that the line on the graph is increasing as you move from left to right, while a negative slope means that the line is decreasing. This is determined by the direction of the line and can also be represented by the sign of the slope value. A positive slope will have a positive slope value, while a negative slope will have a negative slope value.

4. How do I use the slope-intercept form to graph a line?

The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept (the point where the line crosses the y-axis). To graph a line using this form, plot the y-intercept on the y-axis and then use the slope to find other points on the line. From the y-intercept, move up or down by the rise, and then move left or right by the run to find another point. Connect these points to graph the line.

5. What are some real-world applications of simple gradient/graph questions?

Simple gradient/graph questions are used in many fields, including physics, engineering, economics, and geography. Some examples of real-world applications include calculating velocity and acceleration in physics, determining the optimal production level for a business in economics, and finding the slope of a mountain in geography.

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