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Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Solving Geometric Problems using Coordinate Methods (2D) Solving Geometric Problems using Coordinate Methods (3D) Applications in Physics, Engineering, and other Fields


Applications of Coordinate Geometry



Solving Geometric Problems using Coordinate Methods (2D)

Coordinate geometry, also known as analytic geometry, provides a powerful framework for solving problems that involve geometric figures by translating them into algebraic terms. By assigning coordinates to points and representing lines, curves, and other geometric entities with equations, we can use the tools of algebra and calculus to analyze and solve geometric problems.

General Approach using Coordinate Methods

Solving a geometric problem using coordinate geometry typically involves the following steps:

  1. Introduce a Coordinate System: Place the given geometric figure on a Cartesian coordinate plane. Assign coordinates to the relevant points (vertices, specific points mentioned in the problem). Choose the position and orientation of the axes strategically to simplify the coordinates and subsequent calculations (e.g., place a vertex at the origin, align a side with a coordinate axis).

  2. Represent Geometric Elements Algebraically: Write the equations for the lines, curves, or other geometric entities involved based on the chosen coordinate system and the assigned coordinates. For example, find the equation of a line segment, a circle, etc.

  3. Translate Geometric Conditions into Algebraic Equations: Express the given geometric conditions or properties algebraically using standard coordinate geometry formulas and relationships:

    • Distances/Lengths: Use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
    • Midpoints: Use the midpoint formula for a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
    • Slopes: Use the slope formula for a line through $(x_1, y_1)$ and $(x_2, y_2)$: $m = \frac{y_2-y_1}{x_2-x_1}$.
    • Parallel Lines: Slopes are equal ($m_1 = m_2$).
    • Perpendicular Lines: Product of slopes is -1 ($m_1 m_2 = -1$) for non-vertical lines.
    • Collinear Points: Three points are collinear if the slope of the segment between the first two equals the slope of the segment between the second and third, OR if the area of the triangle formed by them is zero.
    • Section of a Segment: Use the section formula for internal or external division.
    • Intersection Points: Find the coordinates of the intersection by solving the system of equations of the intersecting lines or curves.
    • Angles: Use the formula for the angle between two lines in terms of their slopes.
    • Area of Polygon: Use the coordinate formula for the area of a triangle or the Shoelace formula for a general polygon.
  4. Solve the Algebraic Problem: Perform algebraic manipulations (solving equations, simplifying expressions, etc.) to prove the required geometric statement or calculate the unknown geometric quantity.

  5. Interpret the Result: Translate the algebraic solution or conclusion back into a geometric statement or result regarding the original figure.


Examples of Applications in 2D

1. Proving Properties of Geometric Figures

Coordinate methods are often used to prove geometric theorems algebraically. Choosing coordinates wisely is key to simplifying the proof.

Example 1. Prove that the diagonals of a parallelogram bisect each other.

Proof using Coordinates:

Step 1: Represent the Geometry Algebraically

Let ABCD be a parallelogram. We place the parallelogram on the coordinate plane in a convenient position. Let vertex A be at the origin $(0, 0)$. Let side AB lie along the positive x-axis. Let the length of side AB be $a$. Then the coordinates of B are $(a, 0)$.

Let the coordinates of vertex D be $(b, c)$. Since ABCD is a parallelogram, the side DC is parallel and equal in length to AB. Also, AD is parallel and equal in length to BC.

Using the property that $\vec{AB} = \vec{DC}$ (vectors are equal), where $\vec{AB} = (a-0, 0-0) = (a, 0)$.

Let the coordinates of C be $(x_C, y_C)$. Then $\vec{DC} = (x_C - b, y_C - c)$.

Equating the components of the vectors: $x_C - b = a \implies x_C = a+b$, and $y_C - c = 0 \implies y_C = c$.

So, the coordinates of the vertices of the parallelogram are: $A(0, 0)$, $B(a, 0)$, $C(a+b, c)$, and $D(b, c)$.

Parallelogram ABCD with coordinates

Step 2: Translate Geometric Condition into Algebraic Equations

The property to prove is that the diagonals of a parallelogram bisect each other. Diagonals bisect each other if and only if they share the same midpoint. The diagonals of parallelogram ABCD are AC and BD.

We will find the coordinates of the midpoint of diagonal AC using the midpoint formula for A$(0, 0)$ and C$(a+b, c)$:

$M_{AC} = \left( \frac{0 + (a+b)}{2}, \frac{0 + c}{2} \right)$

... (i)

$M_{AC} = \left( \frac{a+b}{2}, \frac{c}{2} \right)$

... (ii)

Now find the coordinates of the midpoint of diagonal BD using the midpoint formula for B$(a, 0)$ and D$(b, c)$:

$M_{BD} = \left( \frac{a + b}{2}, \frac{0 + c}{2} \right)$

... (iii)

$M_{BD} = \left( \frac{a+b}{2}, \frac{c}{2} \right)$

... (iv)

Step 3: Solve Algebraically (Compare Results)

Comparing the coordinates of $M_{AC}$ from equation (ii) and $M_{BD}$ from equation (iv), we see that the x-coordinates are equal $\left(\frac{a+b}{2} = \frac{a+b}{2}\right)$ and the y-coordinates are equal $\left(\frac{c}{2} = \frac{c}{2}\right)$.

Thus, the midpoint of diagonal AC is the same point as the midpoint of diagonal BD.

Step 4: Interpret the Algebraic Result Geometrically

Since both diagonals of the parallelogram ABCD share the same midpoint, this proves that the diagonals bisect each other.

2. Finding Loci

Finding the equation of a locus involves translating the geometric condition that defines the locus into an algebraic equation involving the coordinates of a general point on the locus. This was discussed in detail in the section on Locus and its Equation.

Example: Finding the equation of the perpendicular bisector of a line segment (which is the locus of points equidistant from the two endpoints).

3. Finding Equations of Geometric Elements

Coordinate geometry allows us to find the equations of lines related to geometric figures, such as the equations of the sides, medians, altitudes, angle bisectors, or perpendicular bisectors of a triangle, given its vertices. These involve using the formulas for slope, midpoint, distance, and the various forms of the equation of a line.

4. Calculating Areas

The area of a triangle or any polygon can be calculated directly from the coordinates of its vertices using specific formulas (e.g., the area of a triangle using the determinant or Shoelace formula, or by dividing the polygon into triangles). This avoids the need to find side lengths and heights.

5. Classifying Figures

Given a set of points that form a geometric figure (like a triangle or quadrilateral), coordinate methods allow us to classify the specific type of figure. For example, we can determine if a triangle is isosceles, equilateral, or right-angled by calculating its side lengths using the distance formula and checking the Pythagorean theorem. We can classify a quadrilateral as a parallelogram, rectangle, rhombus, or square by calculating side lengths, diagonal lengths, and checking for conditions like opposite sides equal, all sides equal, diagonals equal, etc.

Coordinate geometry provides a systematic and algebraic approach to these and many other geometric problems, making it a fundamental tool in mathematics.



Solving Geometric Problems using Coordinate Methods (3D)

Just as coordinate geometry provides algebraic methods for solving 2D geometric problems, it extends to three dimensions to handle geometric figures and relationships in space. While visualizing 3D geometry can be challenging, the algebraic machinery remains systematic and powerful, relying on 3D coordinates and formulas.

Approach and Tools in 3D

The general strategy for solving a geometric problem using 3D coordinate methods is analogous to the 2D approach:

  1. Introduce a 3D Coordinate System: Place the geometric figure within a 3D Cartesian coordinate system. Assign coordinates $(x, y, z)$ to relevant points. Choose the placement of the origin and the orientation of the axes strategically to simplify coordinates and calculations (e.g., placing a vertex at the origin, aligning an edge or a face with coordinate axes or planes).
  2. Represent Geometric Elements Algebraically: Use the assigned 3D coordinates to represent lines, planes, and other geometric entities algebraically (e.g., equations of lines in 3D, equations of planes). Note that line and plane equations in 3D are more complex than in 2D and often involve vectors, which are related to coordinate geometry.
  3. Translate Geometric Conditions into Algebraic Equations: Express the given geometric conditions or properties algebraically using 3D coordinate geometry formulas:
    • Distances/Lengths: Use the 3D distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
    • Midpoints: Use the 3D midpoint formula: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.
    • Collinearity: Check if $|AB| + |BC| = |AC|$ using the 3D distance formula, or use the concept of direction ratios being proportional for the segments AB and BC.
    • Section of a Segment: Use the 3D section formula for internal or external division.
    • Parallelism/Perpendicularity of Lines/Planes: These conditions often involve direction ratios/vectors and normal vectors. For lines with direction vectors $\vec{v_1}$ and $\vec{v_2}$, parallel if $\vec{v_1} = k \vec{v_2}$, perpendicular if $\vec{v_1} \cdot \vec{v_2} = 0$. For planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$, parallel if $\vec{n_1} = k \vec{n_2}$, perpendicular if $\vec{n_1} \cdot \vec{n_2} = 0$.
    • Angles: Use dot product of vectors to find the angle between lines or planes.
    • Intersections: Solve systems of equations representing geometric entities. Intersection of two planes is a line; intersection of a line and a plane is a point; intersection of three planes (generally) is a point.
  4. Solve the Algebraic Problem: Perform necessary algebraic calculations (solving systems of equations, simplifying expressions) to reach the desired result.
  5. Interpret the Result: Translate the algebraic findings back into geometric conclusions about the original figure.

Examples of Applications in 3D

1. Classifying Figures in Space

Using the 3D distance formula, we can calculate the lengths of edges and diagonals of polyhedra defined by vertex coordinates to classify their types. For example, showing that a figure is a square in 3D requires proving all four sides are equal and the two diagonals are equal using the 3D distance formula.

2. Collinearity of Points

As shown in the 3D distance formula section, the collinearity of three points A, B, and C in space can be checked by verifying if the sum of two distances equals the third distance, e.g., $|AB| + |BC| = |AC|$.

3. Finding Division Ratios by Planes or Axes

Given a line segment and a coordinate plane (or axis), we can find the ratio in which the plane (or axis) divides the segment by using the section formula and the equation of the plane (or axis).

Example 1. Find the ratio in which the YZ-plane divides the line segment formed by joining the points P(-2, 4, 7) and Q(3, -5, 8).

Answer:

Given the points $P(x_1, y_1, z_1) = (-2, 4, 7)$ and $Q(x_2, y_2, z_2) = (3, -5, 8)$.

The YZ-plane is the plane where the x-coordinate is zero. The equation of the YZ-plane is $\mathbf{x=0}$.

Let the YZ-plane divide the line segment PQ at a point $R(x, y, z)$ in the ratio $k:1$. Since the plane intersects the segment, the division is internal, so $k$ will be a positive value.

Using the section formula for internal division in 3D, the coordinates of R are:

$R(x, y, z) = \left( \frac{k \cdot x_2 + 1 \cdot x_1}{k + 1}, \frac{k \cdot y_2 + 1 \cdot y_1}{k + 1}, \frac{k \cdot z_2 + 1 \cdot z_1}{k + 1} \right)$

... (i)

Substitute the coordinates of P and Q ($x_1=-2, y_1=4, z_1=7$ and $x_2=3, y_2=-5, z_2=8$) into equation (i):

$R(x, y, z) = \left( \frac{k(3) + 1(-2)}{k + 1}, \frac{k(-5) + 1(4)}{k + 1}, \frac{k(8) + 1(7)}{k + 1} \right)$

... (ii)

$R = \left( \frac{3k - 2}{k + 1}, \frac{-5k + 4}{k + 1}, \frac{8k + 7}{k + 1} \right)$

... (iii)

Since the point $R$ lies on the YZ-plane, its x-coordinate must be equal to 0.

Equate the x-coordinate of R from equation (iii) to 0:

$\frac{3k - 2}{k + 1} = 0$

(R is on YZ-plane)

For a fraction to be equal to zero, the numerator must be zero (provided the denominator is non-zero. Here $k \neq -1$, which corresponds to P, so if R is distinct from P, $k \neq -1$).

$3k - 2 = 0$

... (iv)

Solve equation (iv) for $k$:

$3k = 2$

$k = \frac{2}{3}$

... (v)

The ratio of division is $k:1$. Substitute the value of $k$ from equation (v):

Ratio $= \frac{2}{3} : 1$.

To express this ratio with integers, multiply both parts by 3:

Ratio $= \left(\frac{2}{3} \times 3\right) : (1 \times 3) = 2 : 3$.

Since the value of $k$ ($2/3$) is positive, the division is indeed internal.

The YZ-plane divides the line segment joining P(-2, 4, 7) and Q(3, -5, 8) internally in the ratio $\mathbf{2:3}$.

We can find the coordinates of the intersection point R by substituting $k=2/3$ into equation (iii), but it's not required by the question.

4. Finding Centroids

The centroid of a triangle or a tetrahedron in 3D space is calculated as the average of the coordinates of its vertices. This formula is derived using the 3D section formula and the properties of medians/lines connecting vertices to centroids of opposite faces.

5. Finding Points Equidistant

Problems involving finding points equidistant from given points in 3D space can be solved using the 3D distance formula. Setting up equations based on equal distances (squared distances are often easier) leads to algebraic equations involving the unknown coordinates of the equidistant point. If the point is constrained to lie on an axis or plane, some of its coordinates will be fixed at zero.

Coordinate geometry in 3D provides a systematic approach to these and more complex problems involving lines, planes, spheres, and other surfaces in space.



Applications in Physics, Engineering, and other Fields

Coordinate geometry is not merely a theoretical branch of mathematics; it is a powerful tool with extensive practical applications across numerous scientific, technical, and even artistic domains. By providing a systematic way to represent geometric objects and spatial relationships algebraically, it enables quantitative analysis and problem-solving in real-world contexts.

Physics

Physics fundamentally describes the universe in terms of space, time, and motion, making coordinate systems indispensable.


Engineering

Engineering disciplines rely heavily on mathematical modeling of physical systems and designs, where coordinate geometry is a core tool.


Computer Science

Fields within computer science that deal with spatial data or visual output are heavily dependent on coordinate geometry.


Navigation and Geography

Locating positions and mapping the Earth involve coordinate systems.


Other Fields

Coordinate geometry's utility extends to various other areas:

In summary, coordinate geometry provides the mathematical language and tools to describe and solve problems involving spatial relationships, shapes, sizes, positions, and movement. Its concepts are foundational and permeate numerous fields, enabling precise analysis and practical applications.