What are coordinates

In classical mathematics, analytic geometry, also known as coordinate geometry, or Cartesian geometry, is the study of geometry using a coordinate system and the principles of algebra and analysis. This contrasts with the synthetic approach of Euclidean geometry, which treats certain geometric notions as primitive, and uses deductive reasoning based on axioms and theorems to derive truth. Analytic geometry is widely used in physics and engineering, and is the foundation of most modern fields of geometry, including algebraic, differential, discrete, and computational geometry.

Usually the Cartesian coordinate system is applied to manipulate equations for planes, straight lines, and squares, often in two and sometimes in three dimensions. Geometrically, one studies the Euclidean plane (2 dimensions) and Euclidean space (3 dimensions). As taught in school books, analytic geometry can be explained more simply: it is concerned with defining and representing geometrical shapes in a numerical way and extracting numerical information from shapes' numerical definitions and representations. The numerical output, however, might also be a vector or a shape. That the algebra of the real numbers can be employed to yield results about the linear continuum of geometry relies on the Cantor–Dedekind axiom.

Coordinates

In analytic geometry, the plane is given a coordinate system, by which every point has a pair of real number coordinates. The most common coordinate system to use is the Cartesian coordinate system, where each point has an x-coordinate representing its horizontal position, and a y-coordinate representing its vertical position. These are typically written as an ordered pair (x, y). This system can also be used for three-dimensional geometry, where every point in Euclidean space is represented by an ordered triple of coordinates (x, y, z).

Other coordinate systems are possible. On the plane the most common alternative is polar coordinates, where every point is represented by its radius r from the origin and its angle θ. In three dimensions, common alternative coordinate systems include cylindrical coordinates and spherical coordinates

Equations of curves

In analytic geometry, any equation involving the coordinates specifies a subset of the plane, namely the solution set for the equation. For example, the equation y = x corresponds to the set of all the points on the plane whose x-coordinate and y-coordinate are equal. These points form a line, and y = x is said to be the equation for this line. In general, linear equations involving x and y specify lines, quadratic equations specify conic sections, and more complicated equations describe more complicated figures.

Usually, a single equation corresponds to a curve on the plane. This is not always the case: the trivial equation x = x specifies the entire plane, and the equation x2 + y2 = 0 specifies only the single point (0, 0). In three dimensions, a single equation usually gives a surface, and a curve must be specified as the intersection of two surfaces (see below), or as a system of parametric equations. The equation x2 + y2 = r2 is the equation for any circle with a radius of r.

Distance and angle

n analytic geometry, geometric notions such as distance and angle measure are defined using formulas. These definitions are designed to be consistent with the underlying Euclidean geometry. For example, using Cartesian coordinates on the plane, the distance between two points (x1, y1) and (x2, y2) is defined by the formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},\!$

which can be viewed as a version of the Pythagorean theorem. Similarly, the angle that a line makes with the horizontal can be defined by the formula

$\theta = \arctan(m)\!$

where m is the slope of the line.

The distance formula on the plane follows from the Pythagorean theorem.

Section of a line

In Analytical Geometry a section of a line can be given by the formula where (c,d)&(e,f) are the endpoints of the line & m:n is the ratio of division

S(a,b)=(nc+me/m+n, nd+mf/m+n)

Transformations

Transformations are applied to parent functions to turn it into a new function with similar characteristics. For example, the parent function y=1/x has a horizontal and a vertical asymptote, and occupies the first and third quadrant, and all of its transformed forms have one horizontal and vertical asymptote,and occupies either the 1st and 3rd or 2nd and 4th quadrant. In general, if y = f(x), then it can be transformed into y = af(b(x − k)) + h. In the new transformed function, a is the factor that vertically stretches the function if it is greater than 1 or vertically compresses the function if it is less than 1, and for negative a values, the function is reflected in the x-axis. The b value compresses the graph of the function horizontally if greater than 1 and stretches the function horizontally if less than 1, and like a, reflects the function in the y-axis when it is negative. The k and h values introduce translations, h, vertical, and k horizontal. Positive h and k values mean the function is translated to the positive end of its axis and negative meaning translation towards the negative end.

Transformations can be applied to any geometric equation whether or not the equation represents a function. Transformations can be considered as individual transactions or in combinations.

Suppose that R(x,y) is a relation in the xy plane. For example

x2 + y2 -1= 0

is the relation that describes the unit circle. The graph of R(x,y) is changed by standard transformations as follows:

Changing x to x-h moves the graph to the right h units.

Changing y to y-k moves the graph up k units.

Changing x to x/b stretches the graph horizontally by a factor of b. (think of the x as being dilated)

Changing y to y/a stretches the graph vertically.

Changing x to xcosA+ ysinA and changing y to -xsinA + ycosA rotates the graph by an angle A.

There are other standard transformation not typically studied in elementary analytic geometry because the transformations change the shape of objects in ways not usually considered. Skewing is an example of a transformation not usually considered. For more information, consult the Wikipedia article on affine transformations.

Intersections

While this discussion is limited to the xy-plane, it can easily be extended to higher dimensions. For two geometric objects P and Q represented by the relations P(x,y) and Q(x,y) the intersection is the collection of all points (x,y) which are in both relations. For example, P might be the circle with radius 1 and center (0,0): P = {(x,y) | x2+y2=1} and Q might be the circle with radius 1 and center (1,0): Q = {(x,y) | (x-1)2+y2=1}. The intersection of these two circles is the collection of points which make both equations true. Does the point (0,0) make both equations true? Using (0,0) for (x,y), the equation for Q becomes (0-1)2+02=1 or (-1)2=1 which is true, so (0,0) is in the relation Q. On the other hand, still using (0,0) for (x,y) the equation for P becomes (0)2+02=1 or 0=1 which is false. (0,0) is not in P so it is not in the intersection.

The intersection of P and Q can be found by solving the simultaneous equations:

x²+y² = 1

(x-1)²+y² = 1

Explanation for the coordinate geometry of a straight line--

Every straight line can be represented algebraically in the form y = mx + c , where

m = represents the gradient of a line (its slope, steepness)

c = represents the y -intercept (a point where the line crosses the y axis)

Furthermore, there are several ways in which you can describe a straight line algebraically

Equation of a line=

Line gradient m , the intercept on y -axis is c

,	Line gradient m , passes through the origin

,	Line gradient 1, makes an angle of 45 0 with the x -axis, and the intercept on the y axis is c
,	Line is parallel to the x axis, through (0, k)
	Line is the x axis
.	Line is parallel to the y axis, through ( k , 0)
,	Line is the y axis
	, General form of the equation of straight line

The gradient measures the steepness of the line.

It is defined as

, or

When the gradient is 1, the line makes a 45 0 angle with either axes. If the gradient is 0, the line is parallel to the x axis.

Equation of a straight line given the gradient and a point=

On the coordinate plane, the slant of a line is called the slope. Slope is the ratio of the change in the y-value over the change in the x-value, also called rise over run.
Given any two points on a line, you can calculate the slope of the line by using this formula:

slope =

f the point is given by its coordinates

, and the gradient of a line is given as m , you can deduce the equation of that line.

You are using the formula for gradient,

, to derive a formula for the line itself. The coordinates of the point can be substituted, while the y 2 and x 2 need to remain (without the superscript numbers).

Then simply substitute the given values into

The equation of a line given two points=

When you have this kind of problem, you take that, as both points belong to the same line, the gradients at both points will be the same.

It makes sense therefore to say that

All you need to do in this case will be to substitute coordinates you have for the given points

and

Parallel and perpendicular lines-

When two lines are parallel, their gradient is the same:

When two lines are perpendicular, their product equals -1:

The line length

The length of the line segment joining two points will relate to their coordinates. Have a good look at the diagram

The length joining the point A and C can be found by using Pythagoras' Theorem:

Mid-point of a line

Mid-point of the line can be found by using the same principle

So the point between A and C will have the coordinates

If you know the midpoint, you can easily find the perpendicular bisector of a given line. This new line will go through the midpoint of the given line, and it will be perpendicular to it.

Pairs of Straight Lines

Any two lines through the Origin may be written as $y = mx$ and $y = tx$ where $m$ and $t$ are their gradients. So $(y - mx)(y - tx) = 0$ giving $y - mx$ or $y - tx = 0$ must represent the pair.

The general form of this equation is given by:

$ax^2+2hxy+by^2=0$

This equation must represent a pair of straight lines, real or imaginary, through the origin. These can be written as

$b\left(\frac{y}{x} \right)^2 + 2h\left(\frac{y}{x} \right) + a = 0$

Since $\displaystyle \frac{y}{x}$ is the gradient of a line through the origin
the roots of this equation must be the gradients of the lines $m$ and $t$ .
Therefore $\displystyle m + t= - \frac{2h}{t}\f$ and $\displaystyle m t = \frac{a}{b}$

Angles Between Lines

Suppose that the lines $y = mx$ and $y = tx$ are represented by the following equation:

$ax^2 + 2hxy + by^2 = 0$

If the angle between them is $\theta$ then:

$\tan \theta = \frac{m - t}{1 - mt}= \frac{\sqrt{(m + t)^2 - 4mt}}{1 + mt}$

Hence

$tan\;\theta = \frac{\sqrt{4h^2/b^2 - 4a/b}}{1 + a/b}$

therefore

$\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$

N.B. The lines will be parallel if the values of this fraction become infinite. i.e. $a + b = 0$

To Find The Equation Of The Angle Bisectors

An angle bisector divides the angle into two angles with equal measures. An angle only has one bisector. Each point of an angle bisector is equidistant from the sides of the angle.

As before suppose that the lines $y = mx$ and $y = tx$ are represented by:

$ax^2 + 2hxy + by^2 = 0$

The equation of the angle bisectors will be:

$\frac{y - mx}{\sqrt{1 + m^2}} = \pm \frac{y - tx}{\sqrt{1 + t^2}}$

$\therefore\;\;\;\;\;(1 + t^2)(y - mx)^2 = (1 + m^2)(y - tx)^2$

$x^2(m^2 - t^2) - 2xy(m + mt^2\;-t\;-tm^2) + y^2(t^2 - m^2) = 0$

Since $m$ is not equal to $t$ , divide the above equation by $(m - t)$

$x^2(m + t) - 2xy(1 - mt) - y^2(m + t) = 0$

Substituting for $(m+t)$ and $mt$ :

$x^2(- \frac{2h}{b}) - 2xy(1 - \frac{a}{b}) - y^2(-\frac{2h}{b}) = 0$

$(x^2 - y^2)(- 2h) = 2xy(b - a)$

Therefore the required equation is

$\frac{x^2 - y^2}{xy} = \frac{a - b}{h}$

To Find The Equation Of The Pair Of Lines Joining The Points Of Intersection Of The Following Two Lines, To The Origin:

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

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$lx + my + n = 0$

From the linear equation express 1 as a linear function of $x$ and $y$ . i.e.:

$1\;= - \frac{(lx + my)}{n}$

Use this to build up every term of the quadratic equation to the second degree and we get:

$ax^2 + 2hxy + by^2 + (2gx + 2fy)\left(- \frac{lx + my}{n} \right) + c\left(- \frac{lx + my}{n} \right)^2 = 0$

Every term here is of the second degree and since any point which satisfies both:

$- \frac{(lx + my)}{n} = 1$

and

$2hxy + by^2 + 2gx + 2fy + c = 0$

must also satisfy this new equation, it must represent the required pair of lines.

To Find The Condition That The General Equation Of The Second Degree Should Represent A Pair Of Straight Lines..

So far we have considered only pairs of straight lines through the origin.

The equation of the pair of lines $ax + by + c = 0$ and $lx + my + n = 0$ is obviously given by the equation:

$(ax + by + c)(lx + my + n) = 0$

And it is worth noting that the equation:

$a(x - \alpha )^2 + 2h(x - \alpha )(y - \beta ) + b(y - \beta )^2 = 0$

represents a pair of straight lines through the point $(\alpha, \beta )$ and parallel to the pair given by:

$ax^2 + 2hxy + by^2 = 0$

The general equation in the second degree:

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

will represent a pair of straight lines if it factorizes. Expanding the equation as a quadratic in x we get:

$ax^2 + 2x(hy + g) + (by^2 + 2fy\;+c) = 0$

When we solve for $x$ we will get an expression containing a square root. If the equation represents a pair of lines $x$ must be expressible as one or other of two linear expressions in $x$ and $y$ and so this square root must be rational. $(hy + g)^2 - a(by^2 + 2fy+c)$ must be a perfect square.

The condition for this is given by: $(hy - af)^2 = (h^2 - ab)(g^2 - ac)$

Which simplifies to become:

$af^2 + bg^2 + ch^2 = 2fgh + abc$

The Equation Of A Circle

(x - a)² + (y - b)² = r²
where
a is the x co-ordinate of the centre of the circle
b is the y co-ordinate of the centre of the circle
r is the radius of the circle

In terms of showing method and avoiding error
especially with directed numbers
the following format can be very useful
[x - (a)]² + [y - (b)]² = (r)²

(x + 2)² + (y - 3)² = 5²

[x + 2]² + [y - 3]² = 5²

[x - (-2)]² + [y - (3)]² = (5)²

a = (-2)

-2 is the x co-ordinate
of the centre of the circle

b = (3)

3 is the y co-ordinate
of the centre of the circle

r = (5)

5 is the radius
of the circle

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Welcome to Mathmatics Universe: For Class V To XII

Introduction to Coordinate Geometry

What are coordinates

Coordinates

Equations of curves

Distance and angle

Section of a line

Transformations

Explanation for the coordinate geometry of a straight line--

Pairs of Straight Lines

Any two lines through the Origin may be written as $y = mx$ and $y = tx$ where $m$ and $t$ are their gradients. So $(y - mx)(y - tx) = 0$ giving $y - mx$ or $y - tx = 0$ must represent the pair.

The general form of this equation is given by:

Angles Between Lines

Suppose that the lines $y = mx$ and $y = tx$ are represented by the following equation:

To Find The Equation Of The Angle Bisectors

An angle bisector divides the angle into two angles with equal measures. An angle only has one bisector. Each point of an angle bisector is equidistant from the sides of the angle.

As before suppose that the lines $y = mx$ and $y = tx$ are represented by:

To Find The Equation Of The Pair Of Lines Joining The Points Of Intersection Of The Following Two Lines, To The Origin:

To Find The Condition That The General Equation Of The Second Degree Should Represent A Pair Of Straight Lines..

So far we have considered only pairs of straight lines through the origin.

The equation of the pair of lines $ax + by + c = 0$ and $lx + my + n = 0$ is obviously given by the equation:

The Equation Of A Circle

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Popular posts from this blog

List of trigonometric identities

Basic arithmetics math formulas

Introduction to Coordinate Geometry

What are coordinates

Coordinates

Equations of curves

Distance and angle

Section of a line

Transformations

Explanation for the coordinate geometry of a straight line--

Pairs of Straight Lines

Any two lines through the Origin may be written as and where and are their gradients. So giving or must represent the pair.

The general form of this equation is given by:

Angles Between Lines

Suppose that the lines and are represented by the following equation:

To Find The Equation Of The Angle Bisectors

An angle bisector divides the angle into two angles with equal measures. An angle only has one bisector. Each point of an angle bisector is equidistant from the sides of the angle.

As before suppose that the lines and are represented by:

To Find The Equation Of The Pair Of Lines Joining The Points Of Intersection Of The Following Two Lines, To The Origin:

To Find The Condition That The General Equation Of The Second Degree Should Represent A Pair Of Straight Lines..

So far we have considered only pairs of straight lines through the origin.

The equation of the pair of lines and is obviously given by the equation:

The Equation Of A Circle

Comments

Post a Comment

Popular posts from this blog

List of trigonometric identities

Basic arithmetics math formulas

Any two lines through the Origin may be written as $y = mx$ and $y = tx$ where $m$ and $t$ are their gradients. So $(y - mx)(y - tx) = 0$ giving $y - mx$ or $y - tx = 0$ must represent the pair.

Suppose that the lines $y = mx$ and $y = tx$ are represented by the following equation:

As before suppose that the lines $y = mx$ and $y = tx$ are represented by:

The equation of the pair of lines $ax + by + c = 0$ and $lx + my + n = 0$ is obviously given by the equation: