What is co-ordinate geometry? I think you all are here to enhance your knowledge about the topic of coordinate geometry. So, in today’s article, I will tell you about what is coordinate geometry, the History of coordinate geometry and many more. So, stay tuned.

As we all know 90% of students love mathematics a lot. Am I right, ok ok fine don’t speak wrong words I was just joking. I know very well that most of the students hate Mathematics. If I tell about myself, I too at first gets always irritated to solve maths problems, but then I started to explore things about Maths then I came to know how interesting and logical it is.

* Explaining the position of a point in two-dimension or we can say the branch of Mathematics which combines the use of Algebra and Geometry together is called coordinate geometry.**Coordinate geometry is a branch of geometry where the position of the points on the plane is defined with the help of an ordered pair of numbers also known as coordinates*

I will try my best to make the topic easy and interesting for all of you. I will tell more about Co-ordinate geometry and different types of skills used to solve the problems and history of the topic in the upcoming paragraphs. So, let’s get started.

*What is Co-ordinate Geometry?*

*What is Co-ordinate Geometry?*

*History-:*

*History-:*

The method of describing the location of points in this way was proposed by the French mathematician René Descartes (1596 – 1650). (Pronounced “day CART”). He proposed further that curves and lines could be described by equations using this technique, thus being the first to link algebra and geometry. In honor of his work, the coordinates of a point are often referred to as its Cartesian coordinates, and the coordinate plane as the Cartesian Coordinate Plane.

*Rectangular Co-ordinate System*-:

*Rectangular Co-ordinate System*

The coordinate axes divide the coordinate plane into four quadrants. The position of point P which is at a distance of units from the y-axis and at a distance of b units from the x-axis is shown as P (a, b) where a is called the abscissa of the point or x-coordinate and b is called ordinate of the point or y-coordinate. (a, b) is called ordered pair.

** Remember**-: In 1st Quadrant : x-coordinate (+), y-coordinate (+). In 2nd Quadrant : x-coordinate (-), y-coordinate (+). In 3rd Quadrant : x-coordinate (-), y-coordinate (-). In 4th Quadrant : x-coordinate (+), y-coordinate (-).

*Note-:*

- The coordinates of the origin are (0, 0).
- y-coordinate of all the points on x-axis is zero for example (2, 0), (-3, 0) etc , are the points on x-axis.
- x-coordinate of all the points on the y-axis is zero for example (0, 5), (0, -3) are the points on the y-axis.

*Plotting of Point*-:

*Plotting of Point*

To plot a given point (-3, 5), first note the signs coordinates and find quadrant in which the point lies. Draw two perpendicular lines as x-axis and y-axis and mark O as origin as the point of intersection. To plot (-3, 5), x-coordinate is -3 and y-coordinate is 5.

The point lies in 2nd quadrant. Label this point as A (-3, 5).

*Inclination/Slope (or gradient) of a Straight Line-:*

*Inclination/Slope (or gradient) of a Straight Line-:*

** Slope of a line**-: The slope, Inclination, and gradient are the same. The

**slope**or

**gradient**of a line is a number that describes both the

*direction*and the

*steepness*of the line. The slope is often denoted by the letter

*‘m*‘.

- If a line makes an angle θ with the positive direction of an x-axis, then tan θ is called the slope of the line. It is written as m = tan θ and the line is said to be inclined at an angle of θ with the x-axis.
- When the line is || ( parallel ) to x-axis its inclination is zero or slope = m = tan 0
^{0}= 0. - When the line is || to y-axis its inclination is 90
^{0}, or slope = m =tan 90^{0}which is not defined. - https://stormypassion.com/trigonometry-best-used-formulas/ open the link to gain interest in trigonometry.

** Note**-:

- x = 0 is the equation of y-axis.
- x=a is the equation of a line || to y-axis and at a distance of ‘a’ units from it.
- y = 0 is the equation of x-axis.
- y=b is the equation of a line || to x-axis at a distance of ‘b’ units from it.

*Graph of Linear Equation*-:

*Graph of Linear Equation*

The graph of linear equation is always a straight line.

** Type 1**-: When linear equation is in the form of y= mx.

** Type 2-:** When the linear equation is of the form y= mx + c, where c is a rational number but not zero ( numbers which are in the form of p/q, where q is not equal to 0 are called rational numbers).

c = intercept on y-axis

OA = intercept on x-axis

OB = c = intercept on y-axis. c is +ve, if above the origin and -ve, if below the origin.

*Note-:*

- A vertical line has no intercept.
- The graph of y = mx always passes through (0, 0).

To draw the graph of a given line, we give two or three values to x and find the corresponding three values of y and write it in the form of a table. Plot these points from the table on the graph paper and join these points to get a straight line.

Consider an equation -: x + 2y = 4

Table 1 depicts that if x is 0 than y is 2 and if x is 2 than y is 1. A (0, 2) and B (2, 1).

Table 2 depicts the information that if x is 4 than y is 0 and if x is 0 than y is 2. C (4, 0) and D(0, 2).

So if we plot the graph of the equation x + 2y = 4 by using these two tables

In equation i) ** x** is the independent variable and

**is dependent as value of**

*y***depends on**

*y*

*x*In equation ii) ** y** is independent variable and

**is dependent as value of**

*x***depends on the solution of**

*x***.**

*y*So according to this graph, we can clearly see that the graph of the equation will remain the same. You can find the values by taking differents points at your convenience but the end result of the graph will be the same.

*Dependent and Independent Variables-:*

*Dependent and Independent Variables-:*

The general equation of a straight line in two variables x and y is ax + by +c = 0, where a is not equal to 0, b is not equal to 0. It can be expressed in terms of y and x as below.

y = ( ^{-a}/_{b} )x – ^{c}/_{b} _ _ _ _ i ) and x= ( ^{b}/_{a })y – ^{c}/_{a} _ _ _ _ii )

In equation i ), we can find the value of y if the value of x is given so y is called the dependent variable and x is called the independent variable.

In equation ii), we can find the value of x if value of y is given so x is called dependent variable, y is called independent variable.

*Some examples for you lets do some questions*-:

*Some examples for you lets do some questions*

*Plot the points associated with the pairs P(3, -4), Q(- 3, 4), R(- 5, – 3), S(0, 5), E(- 5, 0) and F(5, 3).*

** Sol**-: Swipe left to see the graph

** 2**.

*Find the values of x and y, if**i) (2x + 1, 3y – 5) = (5, 4)*

*ii) (3x – 4y, 5x – 3y) = (10, 24)*

** Sol-:** (

**) As two ordered pairs are equal**

*i*∴ their corresponding components are equal

(i.e., their first components are equal and their second components are also equal)

∴ 2x + 1 = 5 and 3y – 5 = 4

=> x = 2 and 3y = 9

=> x = 2, y = 3

*3. Express the equation 3x – 2y +12 = 0 in the form, so that*

*i) x is independent variable and y is dependent variable*

*ii)**y is independent variable and x is dependent variable*

** Sol-: **i) 3x – 2y + 12

=> 2y = 3x + 12

=> y = (^{3}/_{2})x + 6

Here, ** x** is independent variable and

**is dependent variable.**

*y*ii) 3x – 2y + 12 = 0

=> 3x = 2y – 12

=> x = (^{2}/_{3})y – 4

Here, ** y** is independent variable and

**is dependent variable.**

*x**4. In each of the following, find the coordinates of the point whose abscissa is the solution of first equation and ordinate is the solution of second equation.*

*i) 2 – 3x + 7 = 0, 4y +3 = 8 – y*

*ii) ( ^{5x}/_{3}) – 8 = x, ^{(13.5 – 3y)}/_{5} = ^{(y+1)}/_{2}*

** Sol-: i) **2 – 3x + 7 = 0

=> 3x = 2 + 7 = 9

=> x = ^{9}/_{3} = 3

4y + 3 = 8 – y

4y + y = 8 – 3 = 5

=> y = ^{5}/_{5} = 1

Co-ordinates are (3, 1)

*ii)*^{ 5x}/_{3} – 8 = x

=> ^{5x}/_{3} – x = 8

=> ^{(5x – 3x)}/_{3} = 8

=> ^{2x}/_{3} = 8

=> x = ^{24}/_{2} = 12

^{(13.5 –} ^{3y)}/_{5} = ^{(y – 1)}/_{2} (cross multiply)

27 – 6y = 5y + 5

=> 27 – 5 = 11y

=> 11y = 22

=> y = ^{22}/_{11} = 2

*5. Three vertices of a square are P(2, 3), Q(-3, 3) and R(-3, -2). Plot these points on a graph paper and hence use it to find the co-ordinates of the fourth vertex. Also, find the area of the square.*

** Sol-:** Plot the points P(2, 3), Q(-3, 3), and R(-3, -2) on the graph paper as usual. Join the points to complete the square PQRS as shown in the figure.

Now, we read the coordinates of the point S from the graph paper.

Clearly S is (2, -2)

Side of square = 3 + 2 = 5 units

∴ Area of square PQRS = (side)^{2} = 5^{2}

= 25 sq. units.

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Nice one?

Thnku

Explained really well. Well done ✅

Thnku

Best explanation ?? Really understood well Keep it up

Thnku