Coordinate Geometry

Coordinate Geometry is a mathematical branch that merges algebra and geometry. It utilizes coordinates to represent and analyze geometric shapes and figures. This blend allows us to study geometric properties, relationships, and spatial configurations using algebraic equations. The beauty of coordinate geometry lies in its ability to describe geometric shapes numerically and enable us to perform calculations such as finding distances, midpoints, and angles.

1.0About Coordinate Geometry

Coordinate Geometry, also known as Cartesian Geometry, was developed by the French mathematician René Descartes. By representing geometric points using a pair of numerical values called coordinates, Descartes revolutionized the way we approach geometric problems. Each point in the coordinate plane is defined by an ordered pair of numbers, (x, y), representing its horizontal (x-axis) and vertical (y-axis) distances from the origin.

2.0Basic Coordinate Geometry Concepts

1. The Cartesian Plane: The basic building block of coordinate geometry is the Cartesian plane, which consists of 2 perpendicular axes, the x-axis and the y-axis, intersecting at the origin (0, 0). This plane is divided into four quadrants, each having distinct signs for the x and y coordinates.

2. Coordinates of a Point: A point in 2D space is represented as (x, y), where x is the perpendicular distance from the y-axis, and y-coordinate represents the perpendicular distance from the x-axis.
3. Distance Formula: The distance between 2 points, A (x₁, y₁) and B (x₂, y₂), is expressed by: 

$\text{ Distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

4. Midpoint Formula: The midpoint of a line segment joining 2 points, A (x₁, y₁) and B (x₂, y₂), is calculated as: 

$Midpoint=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

5. Slope of a Line: The slope, or gradient, of a line passing through 2 points, (x₁, y₁) and (x₂, y₂), is given by:

$\text{ Slope }=\frac{y_2-y_1}{x_2-x_1}$

6. Equation of a Line: The equation of a straight line in the slope-intercept form is:

y = mx + c, where m denotes the slope and c signifies the y-intercept.

3.0Exploring 3D Coordinate Geometry 

When we move from two dimensions to three dimensions, we introduce a new axis, the z-axis, perpendicular to both the x and y axes. This gives rise to 3D Coordinate Geometry, where a point is represented as (x, y, z).

1. Three-Dimensional Coordinate Geometry: A point in 3D space is identified by three coordinates, (x, y, z), where:

• x is the perpendicular distance from the yz-plane.
• y is the perpendicular distance from the xz-plane.
• z is the perpendicular distance from the xy-plane.

2. Distance Formula in 3D: The distance between 2 points, A (x₁, y₁, z₁) and B (x₂, y₂, z₂), is expressed by:

Distance = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

3. Midpoint Formula in 3D: The midpoint of a line segment joining two points, C (x₁, y₁, z₁) and D (x₂, y₂, z₂), is:

Midpoint = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$

4.0Applications of Coordinate Geometry

Coordinate Geometry is widely used in various fields such as physics, engineering, computer graphics, and even in everyday problem-solving scenarios. For instance, it helps in determining the trajectory of a moving object, optimizing routes in navigation systems, and creating realistic 3D models in computer graphics.

3D Coordinate Geometry finds applications in understanding the spatial distribution of celestial bodies in astronomy, designing mechanical parts in engineering, and modeling geographical landscapes.

5.0Solved Examples on Coordinate Geometry 

Example 1: Find the distance between points A (1, 2) and B (4, 6).

Solution:

Using the distance formula:

$\text{ Distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$\text{ Distance }=\sqrt{(4-1{)}^2+(6-2{)}^2}$

$\text{ Distance }=\sqrt{3^2+4^2}$

$\text{ Distance }=\sqrt{9+16}$

$\text{ Distance }=\sqrt{25}$

$\text{ Distance }=5$

Hence the distance between points A (1, 2) and B (4, 6) is 5.

Example 2: Determine the midpoint of the line segment joining points C (–2, 1) and D (2, 3).

Solution: 

Using the midpoint formula:

$\text{ Midpoint }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$\text{ Midpoint }=\left(\frac{-2+2}{2},\frac{1+3}{2}\right)$

$\text{ Midpoint }=(0,2)$

Hence, the midpoint of the line segment joining points C (–2, 1) and D (2, 3) is (0, 2).

Example 3: Find the distance between points A (3, 4) and B (–1, 7).

Solution: 

Using the distance formula:

$\text{ Distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Substitute the given values A (3, 4) = (x₁, y₁) and B (–1, 7) = (x₂, y₂):

$\text{ Distance }=\sqrt{(-1-3{)}^2+(7-4{)}^2}$

$\text{ Distance }=\sqrt{(-4{)}^2+(3{)}^2}$

$\text{ Distance }=\sqrt{16+9}$

$\text{ Distance }=\sqrt{25}=5$

Thus, the distance between points A and B is 5 units.

Example 4: Find the midpoint of the line segment joining points P (2, –3) and Q (6, 5).

Solution: 

Using the midpoint formula:

$\text{ Midpoint }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Substitute P (2, –3) = (x₁, y₁) and Q (6, 5) = (x₂, y₂):

$\text{ Midpoint }=\left(\frac{2+6}{2},\frac{-3+5}{2}\right)$

$\text{ Midpoint }=(4,1)$

Hence, the midpoint of the line segment PQ is (4, 1).

Example 5: Determine the slope and the equation of a line passing through points C (–1, 2) and D (3, –4).

Solution:  

The slope of a line passing through two points C (x₁, y₁) and D (x₂, y₂) is given by:

$\text{ Slope }=m=\frac{y_2-y_1}{x_2-x_1}$

Substitute C (–1, 2) and D (3, –4):

$\text{ Slope }=\frac{-4-2}{3-(-1)}$

$\text{ Slope }=\frac{-6}{4}=-\frac{3}{2}$

Thus, the slope of the line is $-\frac{3}{2}.$

To find the equation of the line, we use the point-slope form:

$y-y_1=m\left(x-x_1\right)$

Substitute $m=-\frac{3}{2}$ and (x₁, y₁) = (–1, 2):

$y-2=-\frac{3}{2}(x+1)$

$y-2=-\frac{3}{2}x-\frac{3}{2}$

$y=-\frac{3}{2}x+\frac{1}{2}$

Hence, the equation of the line is:

$y=-\frac{3}{2}x+\frac{1}{2}$

Example 6: Find the distance between points A (1, 2, 3) and B (4, 6, 8) in three-dimensional space.

Solution:  

Using the distance formula in 3D:

$\text{ Distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Substitute A (1, 2, 3) and B (4, 6, 8):

$\text{ Distance }=\sqrt{(4-1{)}^2+(6-2{)}^2+(8-3{)}^2}$

$\text{ Distance }=\sqrt{3^2+4^2+5^2}$

$\text{ Distance }=\sqrt{9+16+25}$

$\text{ Distance }=\sqrt{9+16+25}$

$\text{ Distance }=\sqrt{50}=5\sqrt{2}$

Thus, the distance between points A and B is $5\sqrt{2}$ units.

6.0Practice Questions on Coordinate Geometry

1. Find the distance between the following pairs of points:  

a) A (2, 3) and B (5, 7)   

b) C (–3, –4) and D (4, 0) 

2. Determine the midpoint of the line segment joining the points:  

a) P (1, 2) and Q (3, –2)  

b) R (–4, 5) and S (0, 1)

3. Find the equation of the line passing through the given points:  

a) A (2, -3) and B (–1, 4)   

b) C (0, 0) and D (5, 10) 

4. Determine the slope of the line that passes through the following points:  

a) M (–2, 3) and N (4, –1)   

b) P (–5, 7) and Q (1, –3) 

5. Determine the distance between the points A(1, 2, 3) and B(4, –1, 5).

7.0Sample Questions on Coordinate Geometry

1. What is the Distance Formula in Coordinate Geometry?

Ans: 

The distance between two points A (x₁, y₁) and B(x₂, y₂) in 2D is given by:

$\text{ Distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

In 3D space, the distance between two points A (x₁, y₁, z₁) and B (x₂, y₂, z₂) is:

$\text{ Distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

2. What is the Midpoint Formula?

Ans: The midpoint of a line segment joining points A(x₁, y₁)  and B(x₂, y₂) in 2D is given by:

$\text{ Midpoint }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

In 3D, the midpoint of a line segment joining A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is:

$\text{ Midpoint }=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

3. What is the Equation of a Line in 2D and 3D?

Ans:  

• In 2D: The equation of a line passing through a point (x₁, y₁) with slope m is given by:

$y-y_1=m\left(x-x_1\right)$

• In 3D: The equation of a line passing through a point (x₁, y₁, z₁) and parallel to vector d = (a, b, c) is:

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

4. How do you find the Equation of a Circle in Coordinate Geometry?

Ans:

The standard equation of a circle with center (h, k) and radius r is:

$(x-h{)}^2+(y-k{)}^2=r^2$

If the circle is centered at the origin (0, 0), the equation simplifies to:

$x^2+y^2=r^2$

5. What is the General Form of a Plane in 3D Coordinate Geometry?

Ans:

The general equation of a plane in 3D is given by:

$ax+by+cz+d=0$

where (a, b, c) is the normal vector of the plane, and d is a constant.

6. What are the Collinearity and Coplanarity Conditions in Coordinate Geometry?

Ans: 

• Collinearity: Three or more points are said to be collinear if they lie on the same straight line. For points A (x₁, y₁), B (x₂, y₂), and C (x₃, y₃) in 2D, they are collinear if the slopes of AB and BC are equal.
• Coplanarity: Four or more points are said to be coplanar if they lie on the same plane in 3D space. A necessary condition for points A, B, C, and D to be coplanar is that the scalar triple product of vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$,  and $\overline{AD}$ is zero.