Analytical Geometry: Cartesian Plane & Coordinates

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Analytic geometry is a branch of Algebra wherein the position of a point on the plane can be determined by means of an Ordered Pair of Numbers, known as Coordinates.

  • Analytical geometry is also known by the name of Coordinate geometry or Cartesian geometry.
  • Analytic geometry can also be known as a type of contradiction to synthetic geometry, wherein no coordinates or formulas are used.
  • Analytical Geometry is the traditional method of developing logical thinking and problem-solving abilities.
  • Straight lines are defined as collections of points that satisfy linear equations in analytic geometry.
  • Descartes and Fermat's invention, Analytical Geometry, is a mathematical subject that solves problems using algebraic symbolism and methods.
  • It establishes the relationship between algebraic equations and geometric curves. The term "Coordinate Geometry" is also used to represent Analytical Geometry.

Read Also: Section Formula in Coordinate Geometry

Key Terms: Analytical Geometry, Coordinates, Cartesian Plane, Differential Geometry, Computational Geometry, Rotational Axes, Polar Coordinates


Analytical Geometry of Two and Three Dimensions

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Descartes and Fermat invented Analytic Geometry. It is also known as coordinate geometry. It is a branch of mathematics in which algebraic symbolism and methods are used to represent and solve geometry problems.

  • Analytic geometry creates a link between geometric curves and algebraic expressions and equations. It is frequently referred to as an algebraic branch used to model geometric objects.
  • It aids in the reformulation of algebraic problems in geometry, as well as vice versa, where the methods and principles of both fields can be used to solve problems in the other.
  • Many modern branches of geometry, such as algebraic, differential, discrete, and computational geometry, are built on the foundation of analytical three-dimensional geometry.
  • Many fields, including physics and engineering, as well as aviation, rocketry, space research, and spaceflight, require analytical geometry.
  • Midpoints and distance, parallel and perpendicular lines on the coordinate plane, dividing line segments, distance between a line and a point, and other topics are covered in trigonometry and analytic geometry.
Branches of Geometry

Branches of Geometry

Read More: Midpoint Formula


What is Analytical Geometry?

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Points are defined as ordered pairs of numbers such as (x, y) in analytic geometry, while straight lines are defined as sets of points that satisfy linear equations. Geometric axioms are derivable theorems in analytic geometry.

Here's an example: for any two distinct points, say (x1, y1) and (x2, y2), there is a single line axe + by + c = 0 that connects them. A linear system of two equations can be used to find the coefficients (a, b, c) (up to a constant factor):

ax1 + by1 + c = 0, ax2 + by2 + c = 0

Coordinate geometry employs both two-dimensional and three-dimensional geometry. The following are the most frequently used terms in analytic geometry:

  • Planes
  • Coordinates

Planes

To have a clear understanding of analytical geometry in three or two dimensions, as well as the importance and applications of analytic geometry, we must first define a plane. A plane is a flat surface that extends in both directions indefinitely. Analytic Geometry uses the coordinates of a point in the X and Y planes to help locate any point on this plane.

Coordinates

Coordinates are two ordered pairs that define the location of any given point in a plane. The box below will assist us in better understanding.

Type A B C
1 - - -
2 - x -
3 - - -

The letter x is located in column B and row 2 of this grid. As a result, the coordinates of this box, x, are B and 2.

Coordinates of a Point

Coordinates can be defined as an address, that helps to determine a point in space. In case of a two-dimensional space, the coordinates of a point are (x, y). Thus, it can be further broken down into:

  • Abscissa: It can be expressed as the x value in the point (x, y). From the origin, it is also the distance of this point along the x-axis.
  • Ordinate: It can be expressed as the y value in the point (x, y). It is the perpendicular distance of the point from the x-axis that is parallel to the y-axis.

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Types of Coordinates

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Different types of coordinates exist in three-dimensional analytical geometry. The following are the most important:

Cartesian Coordinates

The most well-known coordinate system is the Cartesian coordinate system, in which each point has an x-coordinate and y-coordinate that represent its horizontal and vertical positions, respectively.

Polar Coordinates

Polar coordinates are a coordinate system in which each point in a plane is denoted by the distance 'r' from the origin and the angle from the polar axis.

Coordinates in Cylindrical Form

All points in cylindrical coordinates are expressed by their height, radius from the z-axis, and angle projected on the xy-plane with respect to the horizontal axis. The letters h, r, and angle represent the height, radius, and angle, respectively.

Spherical Coordinates

The point in space is denoted in this type of coordinates by its distance from the origin (ρ), the angle projected on the xy-plane with respect to the horizontal axis (θ), and another angle with respect to the z-axis (φ).

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Cartesian Plane

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A Cartesian plane is a graph with two axes, one for the x-axis and one for the y-axis. These two axes are parallel to each other. The origin (O) is located in the exact centre of the graph, at the point where the two axes intersect.

  • On the x-axis, numbers to the right of zero are positive, and whole numbers to the left of zero are negative.
  • For the y-axis, numbers less than zero are negative, and whole numbers greater than zero are positive.
  • To locate a point, we need two numbers in the order of writing the location of the X-axis first and the Y-axis second.
  • Both will reveal the plane's single and unique position.
  • You must strictly adhere to the order of the points on the plane, i.e. the x coordinate is always the first of the pair. ((x, y))

Equation of a Line

The different forms of the equation of a line are as follows.

  • Point-Slope Form: In the equation of a line, it requires a point on the line and the slope of the given line. The point on the given line can be expressed as (x1, y1), while the slope of the line is m. The point-slope form of the equation of a line can be defined as: (y – y1) = m(x – x1).
  • Two-Point Form: The Point-slope form (m = (y2 – y1)/(x2 – x1), here, is substituted to form the two-point form of the equation of a line. The equation of two-point form can thus be expressed as: \((y -y_1) = \frac{(y_2 - y_1)}{(x_2 - x_1)}(x - x_1) \).

Equation of a Line

Equation of a Line

  • Slope-intercept form: The slope-intercept form of a line can be expressed as y = mx + c. Herein, m is known as the slope of the line, while 'c' is the y-intercept of the line. The line here bisects the y-axis at the point (0, c). Here, c is the distance of this point on the y-axis from the origin.
  • Intercept form: The equation of a line in intercept form can be determined with the x-intercept 'a' and the y-intercept 'b'. The given line then bisects the x-axis at point (a, 0), and the y-axis at point(0, b),. Herein, a and b are the distances of the points from the origin. Thus, \(\frac{x}{a} + \frac{y}{b} = 1 \).
  • Normal form: The normal form of the equation is on the basis of the perpendicular to the line, passing via the origin. The normal form equation in the line is xcosθ + ysinθ = p.

Analytical Geometry Formulas

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The Geometry formulas are extremely useful for finding measurements of geometric figures. Graphs and coordinates are two of analytic geometry's many applications. Analytic geometry is used in engineering and science to study the rate of change in varying quantities and to demonstrate the relationship between the quantities involved. 

Here are the analytic geometry formulas.

Distance Formula

Let A and B be two points with coordinates (x1,y1) and (x2,y2), respectively.

As a result, the distance between two points can be calculated as-

⇒ \(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Formula for Midpoint Theorem

Let A and B be two points in a plane that are connected to form a line and have coordinates (x1,y1) and (x2,y2), respectively. If M(x,y) is the midpoint of the line connecting points A and B, then its formula is:

\(M(x, y) =\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \)

Angle Formula

Angle Calculator

Let m1 and m2 be the slopes of two lines, and be the angle formed by the two lines, A and B. This is represented as follows:

Tan = m1-m2/1 + m1m2

Conic Section

The standard form of equations for a variety of conics are:

  • Circle: x2+y2= a2
  • Ellipse: x2/a2 + y2/b2 = 1
  • Hyperbola: x2/a2 – y2/b2 = 1
  • Parabola: y2= 4ax when a>0

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Applications of Analytical Geometry

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Engineering and physics are two fields that make extensive use of analytic geometry. It is also widely used in fields such as space science, rocket science, aviation, and space flights, among others. Many things are now possible because of analytical geometry, including the following:

  • We can determine whether the given lines are parallel or perpendicular.
  • We can find the line segment's midpoint, equation, and slope.
  • We can calculate the distance between the two points.
  • We can also calculate the perimeter and area of the polygon formed by the plane's points.
  • It defines the ellipse, curve, and circle equations.

Analytical Geometry: Translation and Rotation of Axes

Translation of Axes

The coordinate axes with origin O come with coordinates (x, y). The origin is transferred to a new origin O' located at the point (h, k) as per the old coordinate axes. The new coordinate axes are converted in a way that the new axes are parallel to the old ones. The coordinates of a point thus changes from (x, y) to (x' + h, y' + k). 

Rotation of Axes

The coordinate axes ox and oy have been rotated anticlockwise by an angle θ, in order to get new axes ox' and oy'. The coordinates of the point with reference to the old axes are (x, y). Upon rotation, the coordinates of the new axes become (x', y'). We can obtain the old coordinates by substituting (x', y') as (xCosθy -Sinθ, xSinθ + ycosθ).

Read More: Perimeter and Area of a Circle


Things to Remember

  • Analytical geometry is the study of geometric properties and relationships between points, lines and angles in the Cartesian plane. 
  • Gradient (m) describes the slope or steepness of the line joining two points. The gradient of a line is determined by the ratio of vertical change to horizontal change.
  • Analytical geometry comprises basic formulas of coordinate geometry, equations of line and curves, translation and rotation of axes, and three-dimensional geometry concepts.
  • Cartesian plane divides the plane space into two dimensions and is useful to easily locate the points. It is also referred to as the coordinate plane. 
  • Coordinate axes in analytical geometry can be translated by moving the axes such that the new axes are parallel to the old axes.

Read Also: Area of Ellipse


Previous Year Questions


Sample Questions

Ques: What is the formula for the angle between two lines having direction ratios? (1 Mark)

Ans: The angle between two lines that have direction ratios a1,b1,c1 and a2,b2,c respectively is  \(Cosθ =\left |\dfrac{a_1.a_2 +b_1.b_2+c_1.c_2}{\sqrt{a_1^2 + b_1^2+c_1^2}.\sqrt{a_2^2 + b_2^2+c_2^2}}\right | \)

Ques: Find the distance between the points (2,3) and (4,1) (-5,7)(-1,3). (1 Mark)

Ans: Using the distance formula,

  1. d = (x2x1)2 + (y2y1)2

\(\sqrt{}\)(4-2)2+(1-3)2  = 2\(\sqrt{}\)2.

  1. D = (x2x1)2 + (y2y1)2

\(\sqrt{}\) (-1+5)2+(3-7)2 = 4√ 2

Ques: Determine the equation of a line in analytical geometry which has an x-intercept of 5 units, and a y-intercept of 6 units. (2 Marks)

Ans: As per the given intercepts, it can be said that:

  • For x-axis, a = 5
  • For y-axis, b = 6

Thus, the equation of straight line is x/a + y/b = 1.

Hence, x/5 + y/6 = 1

⇒ 6x + 5y = 30

Thus, the equation of the line is 6x + 5y = 30.

Ques: Find the value of x if the points A(x, 2), B(-3, 4) and C(7, -5) are collinear. (2 Marks)

Ans: When the points are collinear,

x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0

x(-4 – (-5)) + (-3)(-5 – 2) + 7(2 – (-4)) = 0

x(1) + 21 + 42 = 0

x + 63 = 0

∴ x = -63

Ques: ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Calculate the length of one of its diagonals. (2 Marks)

Ans: AB = 4 units

BC = 3 units

AC2 = AB2 + BC2 …[Pythagoras’ theorem]

= (4)2 + (3)2

= 16 + 9 = 25

∴ AC = 5 cm

Ques: If the midpoint of the line segment joining the points A (3, 4) and B (k, 6) is P (x, y) and x + y – 10 = 0, find the value of k. (2 Marks)

Ans: Midpoint of the line segment joining A (3, 4) and B (k, 6) = 3+k/2, 4+6/2

= 3+k/2, 5

Or 3+k/2, 5 = (x,y)

Therefore 3+k/2 = x, 5 = y

Since x + y – 10 = 0, we have:

(3+k/2) + 5 – 10 = 0

i.e., 3 + k = 10

Therefore, k = 7

Ques: How do you calculate the distance between two points A and B whose coordinates are (5, -3) and (2, 1), respectively?
Given that, the coordinates are as follows:
A = (-5, -3) (x1, y1)
(2, 1) = B (x2,y2) (3 Marks)

Ans: The formula for calculating the distance between two points is as follows:

  • Distance,d = \(\sqrt{}\)(x2−x1)2+(y2−y1)2(x2−x1)2+(y2−y1)2
  • d = \(\sqrt{}\)(2−5)2+(1−(−3))2(2−5)2+(1−(−3))2
  • d =\(\sqrt{}\)(−3)2+(4)2(−3)2+(4)2
  • d =\(\sqrt{}\)9+169+16
  • d =\(\sqrt{}\)(25)
  • d = 5

As a result, the distance between points A and B is 5.

Ques: Determine the slope of the line that passes through A(5, -3) and intersects the y-axis at 7. (3 Marks)

Ans: The given point is A = (5, -3)

If the line intersects the y-axis, we know that, x2 = 0

Thus, (x2, y2) = (0, 7)

Slope Formula:

m = (y2 –y1)/ (x2 -x1)

Now, substitute the values

m = (7- (-3))/(0 – 5)

m = 10/-5

m = -2

The slope of line comes out to be -2.

Ques: Determine an x-y relationship such that the point P(x, y) is equidistant from the points A (2, 5) and B. (-3, 7). (3 Marks)

Ans: Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7).

∴ AP = BP …[Given]

AP2 = BP2 …[Squaring both sides]

(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2

⇒ x2 – 4x + 4 + y2 – 10y + 25

⇒ x2 + 6x + 9 + y2 – 14y + 49

⇒ -4x – 10y – 6x + 14y = 9 +49 – 4 – 25

⇒ -10x + 4y = 29

∴ 10x + 29 = 4y is the required relation.

Ques: Determine if the points (1,5), (2,3) and (−2,−11) are collinear. (5 Marks)

Ans: Given that,

Let the three points be (1,5),(2,3)(1,5),(2,3) and (−2,−11)(−2,−11)

To determine if the given points are collinear

Let A(1,5) B(2,3) C(−2,−11) be the vertices of the given triangle.

AB = \(\sqrt{}\)(2-1)2 + (3-5)2

= \(\sqrt{}\)12 + (-2)2

= \(\sqrt{}\)1+4 = \(\sqrt{}\)5

BC = \(\sqrt{}\)(-2-2)2 + (-11-3)2

BC= \(\sqrt{}\)(-4)2 + (-14)2

BC = \(\sqrt{}\)16+196

BC = \(\sqrt{}\)212

AB = \(\sqrt{}\)4 x 53 = 2√53

AC = \(\sqrt{}\)(-2-1)2 + (-11-5)2

AC = \(\sqrt{}\)(-3)2 + (-16)2

AC = \(\sqrt{}\)9 + 256 = √265

Hence A, B and C are not collinear.


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CBSE CLASS XII Related Questions

1.

Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta  &-sin\alpha \\   -sin\beta&cos\beta  &0 \\   sin\alpha cos\beta&sin\alpha\sin\beta  &cos\alpha  \end{vmatrix}\)

      2.

      Evaluate\(\begin{vmatrix} 1 & x & y\\ 1 & x+y & y\\1&x&x+y \end{vmatrix}\)

          3.
          Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).

              4.
              2, b, c are in A.P. and the range of determinant $\begin{vmatrix}1&1&1\\ 2&b&c\\ 4&b^{2}&c^{2}\end{vmatrix}$ is [2, 16]. Then find range of c is

                • [6, 8]
                • [0, 4]
                • [8, 12]
                • [4, 6]

                5.

                The cost of 4 kg onion,3 kg wheat and 2kg rice is Rs 60.The cost of 2 kg onion,4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

                    6.
                    If A'= \(\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 &1 \end{bmatrix}\)\(\begin{bmatrix}  -1 & 2 & 1 \\ 1 &2 & 3\end{bmatrix}\) , then verify that 
                    (i) \((A+B)'=A'+B' \)
                    (ii) \((A-B)'=A'-B'\)

                        CBSE CLASS XII Previous Year Papers

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