coordinat lines, and point 2

8/9/2019 Coordinat Lines, And Point 2

http://slidepdf.com/reader/full/coordinat-lines-and-point-2 1/10

2010

SMAN 1 BLITAR

Yahya Rizki Darmawan

[COORDINATES, POINTS, AND LINES]



1. The distance between two points

The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry.Here's how we get from the one to the other:

Suppose you're given the two points ( – 2, 1) and (1,

5), and they want you to find out how far apart they

are. The points look like this:

You can draw in the lines that form a right-angledtriangle, using these points as two of the corners:

It's easy to find the lengths of the horizontal and

vertical sides of the right triangle: just subtract the x-

values and the y-values:

Then use the Pythagorean Theorem to find the length of the third side (which is the hypotenuse of theright triangle):

c2

= a2

+ b2

...so:



This format always holds true. Given two points, you can always plot them, draw the righttriangle, and then find the length of the hypotenuse. The length of the hypotenuse is the distancebetween the two points. Since this format always works, it can be turned into a formula:

Distance Formula: Given the two points ( x1, y1) and ( x2, y2), the distance between these

points is given by the formula:

Don't let the subscripts scare you. They only indicate that there is a "first" point and a "second"point; that is, that you have two points. Whichever one you call "first" or "second" is up to you. The

distance will be the same, regardless.

2. The mid-point of a line segment

The point on a line segment dividing it into two segments of equal length. Themidpoint of a line segment is easy to locate by first constructing a lens using circular arcs,

then connecting the cusps of the lens. The point where the cusp-connecting line intersects the

segment is then the midpoint (Pedoe 1995, p. xii). It is more challenging to locate the

midpoint using only a compass (i.e., a Mascheroni construction).

For the line segment in the plane determined by and , the

midpoint can be calculated as

(1)

Similarly, for the line segment in space determined by and, the midpoint can be calculated as

http://mathworld.wolfram.com/LineSegment.html

http://mathworld.wolfram.com/Lens.html


http://mathworld.wolfram.com/Intersection.html

http://mathworld.wolfram.com/Compass.html

http://mathworld.wolfram.com/MascheroniConstruction.html

http://mathworld.wolfram.com/MascheroniConstruction.html

http://mathworld.wolfram.com/Compass.html

http://mathworld.wolfram.com/Intersection.html



http://mathworld.wolfram.com/LineSegment.html



3. The gradient of a line segment

The gradient of a line is simply a measure of how steep it is. One can think of the

gradient as the amount (and direction) of changed experienced in the y-axis per unit change

in the positive x-direction. This is put more simply as:

The letter "m" denotes the gradient, and it is conventional to use "m". To formalise

this one can use the delta symbol which denotes "change in", giving:

Now one merely has to calculate each of these changes. Using the same idea of the

construction of a right-angled triangle that has been used previously, one can calculate the

gradient of the hypotenuse.

Example

Consider the points (2, 2), and (4, 4); calculate the gradient of the line, a line segment

of which is bounded by these points.

Initially one should take a line horizontally, and another vertically to form a right-

angled triangle whose hypotenuse is the line which one wants to calculate the gradient of. As

previously one can simply see the intersection of these lines in the image below.

From this, one can calculate both the change in the y-direction, and the change in the

x-direction. In this example, one will simply do the following:

http://www.thestudentroom.co.uk/wiki/Image:Gradient-Generalisation.png



Now one can simply apply this new information to the original idea of gradient:

Hence the gradient is 1.

Generalisation

This concept is easy to generalise. Assume (as previously) that there are two points,

, and . One must find the change in the y-direction, and also the change

in the x-direction. The intersection of the lines which one has drawn (see Figure 002) is the

point which one must reference in order to calculate the changes in the y-direction and the

changes in the x-direction, as it has the same x-coordinate of one of the coordinates (

), and the same y coordinates as the other coordinate ( ). Hence one can

say that:

Hence, the gradient of the line is calculated thus:

.

Note that it is possible for the gradient to be negative, this occurs when line slopes

"downward", not "upward" (Figure 001, and Figure 002 show examples of positive gradients,a line that slopes "upwards"). It is also possible to have an undefined gradient. If one has a

vertical line (a line that is parallel to the y-axis) it has an undefined gradient.

4. The equation of a straight line

It is possible to express any straight line as an equation of the form:

Where m is the gradient, and c is the y-axis intercept.

Finding the equation of a straight line is only a slightly longer process than finding the

gradient.

Example

Consider the points (2, 3), and (4, 5); calculate the equation of the line, a line segment

of which is bounded by these points.

One considers first that as this is a straight line, the gradient and y-axis intercept must

be calculated. To calculate the gradient, one does the following (see the previous section for amore in-depth explanation of this process):



Therefore the equation of the line will be in the form of:

Now one must consider the y-axis intercept. This is something that is relatively easy

to calculate. Consider that "c" is a constant value, and one is aware that the points (2, 3), and

(4, 5) both are on the line. If one is aware of a set of coordinates that are on the line, it

follows that the x-coordinate will produce the y-coordinate when the function (or equation) is

applied to it. Hence, one can say:

Hence the equation is:

A good check is to use the second coordinate to ensure that it satisfies the equation:

Hence the equation must be correct.

Generalisation

This generalisation refers to a line which has two points , and

which lie upon it.

It is relatively simple to generalise this equation as one already has a generalisation

for the gradient:

One must, therefore, consider the y-axis intercept. In the numerical example the y-

axis intercept was calculated by substituting known values of y, and x into the equation, in

order to reduce it down to a simple algebraic expression, from which one can calculate the

value of the y-axis intercept.



Now substitute:

One might wish to revert back to the use of "m" to denote the gradient, such that one

obtains a simple expression for "c":

This is a simple way to calculate "c", and thus complete the equation of a line.

It may be asked that you obtain m, or c from an equation in the form:

If this is the case it is important to remember that "c" is not the y-intercept, one must

rearrange, to obtain:

This is because the reference to "c" in:

Is due to the form of the equation, and not the positioning of the variable, or the letterwhich denotes it ("d" could represent the y-intercept, or any other symbol).

5. The point of intersection of two lines

Two lines will intersect when they have equivalent y-coordinates and x-coordinates. This

occurs at one place at most with linear equations, but calculating the coordinates of the

intercept is simply an application of simultaneous equations.

Example

1. The lines , and are plotted; find the coordinates of the point of

interception.

One merely has to find the place where the y-coordinates of both are equal, and their x-

coordinates are equal. Consider that the y-coordinate is obtainable though a function of the

variable "x", (that function is given as the equation of the line), hence, if the y-coordinates are

equal, and the x-coordinates are equal (an interception), it follows that one can equate the

functions that produced the y-coordinates. Thus:



One must be careful to ensure that one gives the answer in the correct form (in this case it is

the coordinates, so one might wish to use the general representation of coordinates in order to

give an answer).

Hence the coordinates of interception are (0, 1).

Generalisation

A generalisation is not particularly helpful in this context as one can encounter many

different styles of equation, and it is very simple to solve the simultaneous equations in order

to find the coordinates of interception on the Cartesian plane. However, it is useful for some

people.

Suppose the following are plotted:

Hence one can equate the two:

It is not recommended that a student tries to remember this, one would be better to simply

learn how to solve simultaneous linear equations, and use the skills in this context.

6. The gradients of perpendicular lines

Finding the gradient of a line which is perpendicular to a known line is certainly helpful, and

also relatively easy.

Generalisation

Suppose that one has the same situation as previously discussed for the calculation of the

gradient of a line (this being a construction of a right-angled triangle). If the line is rotated to

make it perpendicular, this is an anti-clockwise rotation, and has the following effect on the

vector which describes the line from the centre of rotation to a given point:



Now consider that if one is to take any point of rotation it will have the desired effect on the

line (as one is rotating an infinite line upon which the right-angled triangle can be constructed

in any place).

Suppose that the lowest part of the line to be the vertex of the triangle is the centre of rotation

(see Figure 002). Now:

Hence:

Also:

Now consider the effects of the rotation (to make a perpendicular line) on the right-angled

triangle:

Hence, "-y" is now , and "x" is now . Hence the gradient of the perpendicular is:

Now use the notation of "m":

The original line:

Hence, with reference to the perpendicular line:

This can be applied to any line.



Example

1. Find the gradient of the perpendicular to the line .

This is a simple task. First one must find the gradient of the original line. This is evidently 1

as it is expressed in the form of , and the coefficient of is 1.

Now one simply uses the formula derived above, thus:

One can apply this method to calculate the equation of the perpendicular at a given point. It

must be understood that one can only state the gradient of the perpendicular, and not the

equation if one does not have any information of the exact intersection of the perpendicular

with the original line.

coordinat lines, and point 2

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