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Revision:OCR Core 1 - Coordinates, points, and lines

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 1 - Coordinates, points, and lines

1 1. Coordinates, points, and lines
2 Also See
3 Comments

1. Coordinates, points, and lines

The distance between two points

Consider the line shown in the image below.

Figure 001

How would one calculate the distance between two points on this line? It is a simple problem of Pythagoras' Theorem.

Example

1. Calculate the distance between the points (2, 2), and (4, 4).

This is a simple problem, and it involves the line that has been displayed previously. Consider the line segment bounded by the points (2, 2), and (4, 4). This line segment is the line whose distance needs to be calculated.

Now, one might consider the line as the hypotenuse of a right-angled triangle. The logic behind this decision is simply that one can easily calculate the lengths of the other two sides, and hence apply Pythagoras' Theorem, in order to obtain the desired length.

In this example one would consider the lines y = 2 , and x = 4 . The reason for choosing these lines is that one wishes to construct a right angled triangle whose hypotenuse is the length that one wishes to calculate.

One knows that these lines will intersect at (4, 2), and hence one is able to calculate the lengths of the two sides. One side is the line which goes through (4, 2), and (4, 4), having a length of 2; and the other line goes through the points (2, 2), and (4, 2), having a length of two.

Now apply Pythagoras' Theorem to these values:

$a^{2} + b^{2} = c^{2}$

$2^{2} + 2^{2} = c^{2}$

$8 = c^{2}$

$c = \sqrt{8}$

Obviously, this example goes into much more depth than one would go into on a question normally, however it demonstrates the reasoning behind the method, and allows for greater understanding. This technique is a useful and easy technique that can be generalised in a simple manner.

Generalisation

Consider the points $x_{1} ,\ y_{1}$ , and $\left( x_{2} ,\ y_{2} \right)$ ; calculate the distance between these points.

One needs to obtain the lengths of the sides (other than the hypotenuse) of the triangle (as seen in the numerical example on Page 003). To do this, simply do the following: $a = y_{2} - y_{1}$ , and $b = x_{2} - x_{1}$ (note that the variables a, and b do not necessarily have to correspond to a specific side, they must merely refer to one of the two shorter sides, and this side must not be the one which is referred to by the other variable).

Now one has all the information that one needs to conduct Pythagoras' Theorem:

$a^{2} + b^{2} = c^{2}$

$\left( y_{2} - y_{1} \right) ^{2} + \left( x_{2} - x_{1} \right) ^{2} = c^{2}$

$c = \sqrt{ \left( y_{2} - y_{1} \right) ^{2} + \left( x_{2} - x_{1} \right) ^{2} }$

This is therefore the method that one can use in order to calculate the distance between the two points.

The mid-point of a line segment

It is sometimes desirable to calculate the coordinates of the point which is the mid-point of a line segment. The method for the calculation of the mid-point of a line segment is somewhat similar to the one covered previously (finding the length of a line segment), in that this method relies on the construction of a right-angled triangle with the line segment that one is considering as the hypotenuse.

Example

1. Consider the line in Figure 001. Consider the line segment from (2, 2) to (4, 4). Calculate the mid-point of this line segment.

This is a simple question which relies on the same principles as the previous sub-topic did. Construct a right-angled triangle whose hypotenuse is the line segment which one is dealing with. The mid-point of the line segment will be halfway between the coordinates which one has, but if one considers a line dropped from the mid-point, and another line which is taken horizontally from it, one sees that these lines bisect the shorter sides of the right-angled triangle. Hence it is now as simple as finding the mid-points of two lines (one horizontal, and the other vertical).

The line from (4, 4) which one will draw vertically, and the one from (2, 2) which one will draw horizontally, will meet at (4, 2). Now one calculates the lengths of these lines:

4 - 2 = 2

Hence the lengths of both of the lines is 2. One can now deduce that the mid-point of these lines will be 1 unit from either end (as this is half the length of both of them). Now to find the coordinates of the shorter sides of the right-angled triangle, one can simply add the 1 unit to the coordinates of the lines (note that one should add the length to the lower of the coordinates so as to produce a mid-point). This gives:

$(2, 2) + \left( \begin{array}{ccc} 1 \\ 0 \end{array} \right) = (3, 2)$

$(4, 2) + \left( \begin{array}{ccc} 0 \\ 1 \end{array} \right) = (4, 3)$

One now has the coordinates of the mid-point for the hypotenuse. Simply take the y-coordinate of the mid-point for the vertical line, and the x-coordinate for the mid-point of the horizontal line. One now can say that the mid-point of the hypotenuse is: (3, 3).

Generalisation

Consider the points $\left( x_{1} ,\ y_{1} \right)$ , and $\left( x_{2} ,\ y_{2} \right)$ ; calculate the mid-point of the line segment between them.

It is a good idea to understand that the arithmetic mean of two real numbers is equivalent to the number which is half way between them. To show that this is true, one has to have two numbers:

$a \in \mathbb{R}$

$b \in \mathbb{R}$

b = a + c

The arithmetic mean is:

$\frac{a + (a + c)}{2} = \frac{2a + c}{2}$

$\frac{2a + c}{2} = a + \frac{c}{2}$

As "c" denotes the difference between a and b, it follows that the mid-point is half of c from a and b. The above explains that the arithmetic mean calculates this mid-point.

Applying this to the line is simple. One merely would like the arithmetic mean of the x-coordinates as the x-coordinate of the mid-point, and the arithmetic mean of the y-coordinates as the y-coordinate of the mid-point of the line. Hence, in general, one can obtain the coordinates of the mid-point of a line segment thus:

$\left( \frac{x_{1} + x_{2}}{2},\ \frac{y_{1} + y_{2}}{2} \right)$ .

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:

$m = \frac{Change\ in\ y}{Change\ in\ x}$

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:

$m = \frac{\Delta y}{\Delta x}$

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.

Figure 002

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:

$4 - 2 = Change\ in\ y = 2$

$4 - 2 = Change\ in\ x = 2$

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

$m = \frac{\Delta y}{\Delta x} = \frac{2}{2} = 1$

Hence the gradient is 1.

Generalisation

This concept is easy to generalise. Assume (as previously) that there are two points, $\left( x_{1} ,\ y_{1} \right)$ , and $\left( x_{2} ,\ y_{2} \right)$ . 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 ( $\left( x_{2} ,\ y_{2} \right)$ ), and the same y coordinates as the other coordinate ( $\left( x_{1} ,\ y_{1} \right)$ ). Hence one can say that:

$\Delta y = y_{2} - y_{1}$

$\Delta x = x_{2} - x_{1}$

Hence, the gradient of the line is calculated thus:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

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.

The equation of a straight line

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

y = mx + c

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 a more in-depth explanation of this process):

$m = \frac{\Delta y}{\Delta x} = \frac{5 - 3}{4 - 2} = \frac{2}{2} = 1$

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

y = x + c

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:

y = x + c

3 = 2 + c

c = 1

Hence the equation is:

y = x + 1

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

y = x + 1

5 = 4 + 1

Hence the equation must be correct.

Generalisation

This generalisation refers to a line which has two points $\left( x_{1} ,\ y_{1} \right)$ , and $\left( x_{2} ,\ y_{2} \right)$ which lie upon it.

It is relatively simple to generalise this equation as one already has a generalisation for the gradient:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

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.

y = mx + c

$y = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}x + c$

Now substitute:

$y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}x_{1} + c$

$y_{2} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}x_{2} + c$

One might wish to revert back to the use of "m" to denote the gradient, such that one obtains a simple expression for "c":

$c = y_{1} - mx_{1}$

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:

ax + by + c = 0

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

$y = \frac{-ax - c}{b}$

This is because the reference to "c" in:

y = mx + c

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

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 y = x + 1 , and y = 1 - x 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:

x + 1 = 1 - x

2x = 0

x = 0

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).

y = x + 1 = 0 + 1 = 1

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:

$y = m_{1}x + c_{1}$

$y = m_{2}x + c_{2}$

Hence one can equate the two:

$m_{1}x + c_{1} = m_{2}x + c_{2}$

$x \left( m_{1} - m_{2} \right) = c_{2} - c_{1}$

$x = \frac{c_{2} - c_{1}}{m_{1} - m_{2}}$

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.

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:

$\left( \begin{array}{ccc} x \\ y \end{array} \right) \to \left( \begin{array}{ccc} -y \\ x \end{array} \right)$

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:

$Let\ t = \Delta y$

$Let\ r = \Delta x$

Hence:

$m = \frac{t}{s}$

Also:

$Let\ P_{1}\ denote\ the\ lowest\ vertex\ in\ the\ triangle\ which\ is\ on\ the\ line.$

$Let\ P_{2}\ denote\ the\ highest\ vertex\ in\ the\ triangle\ which\ is\ on\ the\ line.$

$Let\ P_{3}\ denote\ the\ vertex\ whose\ angle\ is\ a\ right-angle.$

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

$\left( \begin{array}{ccc} r \\ 0 \end{array} \right) \to \left( \begin{array}{ccc} 0 \\ r \end{array} \right)$

$\left( \begin{array}{ccc} r \\ t \end{array} \right) \to \left( \begin{array}{ccc} -t \\ r \end{array} \right)$

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

$\frac{\Delta y}{\Delta x} = \frac{x}{-y}$

Now use the notation of "m":

The original line:

$m = \frac{y}{x}$

Hence, with reference to the perpendicular line:

$\frac{x}{-y} = - \frac{1}{m}$

This can be applied to any line.

Example

1. Find the gradient of the perpendicular to the line y = x + 2 .

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 y = mx + c , and the coefficient of is 1.

Now one simply uses the formula derived above, thus:

$- \frac{1}{m} = - \frac{1}{1} = -1$

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.