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# The A to Z of mathematics a basic guide - Sidebotham T.H.

Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
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Example. Find the average gradient of the viaduct through Otira Gorge from the following data (see figure d). The roadway is 452 meters long. At the start of the viaduct, point A it is 744 m above sea level, and at the end, point B is 797 m above sea level.
B
53 m C
(d)
Solution. The height of point A above sea level is 744 m and the height of B is 797 m, so the length of BC is 797 — 744 = 53 m. To obtain the gradient of AB the length of AC is required. Write
AB2 = BC2 + AC2 4522 = 532 + AC2 AC2 = 204,304 - 2809 AC = -y/201,495 AC = 449 m
Now
AC
53 “ 449
x 100 = 11.8% (to 1 dp)
449 1
Coordinate Geometry Suppose two points A and B have coordinates A(xi, yi) and B(x2, y2)- Then the gradient of the line through the points A and B is given by the formula
xi -x2
Pythagoras’ theorem in triangle ABC Substituting AB = 452 and BC = 53 Rearranging equation, and squaring 452 and 53 Subtracting, and then taking the square root To nearest whole number
Example. Find the gradient of the line through the two points A (2, — 1) and £(-5,3).
Solution. A diagram is not necessary, because there is no need to decide whether or not the gradient is positive or negative, since the formula will take care of it. By comparing A and B with the formula we can see that X\ = 2, yx = — 1 and x2 = —5, y2 =3. Write
yi — y2
m =---------------- The formula applies equally well with the order reversed,
X2 -Xi
-1-3
m = ——— Substituting values into the formula
m = — ^ Simplifying
The gradient of the line through AB is m = —
References: Average Speed, Canceling, Gradient-Intercept Form, Graphs, Parallel, Perpendicular Lines, Trigonometry.
This entry refers to a method of drawing straight-line graphs without using a table of values. The equations of straight-line graphs can be written in the form y = mx + c, and this is called the gradient-intercept form of the straight-line graph. The graph of a straight line is often called a linear graph.
In the equation y = mx + c, the gradient of the straight-line graph is m and its y intercept is c. When we draw the graph of a straight line we first need to make sure that its equation is written in the form y = mx + c in order to identify the values of m and of c. The graph can be drawn if the values of m and c are known. The whole process is explained in the following examples.
Example 1. Draw the graph of the straight line whose equation is y = 2x — 3.
Solution. Compare the equation y = 2x — 3 with the formula y = mx + c. We can see that the y intercept is c = — 3, and m = 2 (see figure a). The gradient should be expressed as a fraction m = 2/1 so that the gradient can be drawn. Always start with fixing the point c = — 3 on the y-axis, which tells us that the graph crosses this axis at the point y = — 3. From this point on the y-axis we draw a straight line with a gradient of m = 2/1. Then we continue the straight line with a series of gradients m = 2/1 until we have an accurate graph of the straight line y = 2x — 3. The line should be extended at each end to show that it continues.
224
(a)
Example 2. Draw the graph of the straight line x + 2y = 4.
Solution. The equation of the straight line must first be rearranged until it is in the form y = mx + c. Write
lx + 2y = 4 Writing x as la:
2y = — la: + 4 Subtracting la: from both sides of the equation
y = — \x + 2 Dividing both sides of equation by 2
m = — | and c = 2 Comparing with the formula y = mx + c.
Start by fixing the point y = 2 on the y-axis, and from this point draw a straight line with a gradient of m = — | (see figure b). A line with a negative gradient slopes to the left, \. Then continue the straight line with another gradient of m = — | until you have an accurate graph of the straight line x + 2y = 4.
(b)