Coordinate geometry is very useful to precisely locate any particle in two or three dimensional space.
We all use Google map.when we have to search for a place. Have you ever thought how Google gets to know the exact location of that place.The answer to this question is very simple google use the coordinates of that place to exactly navigate to that one.
Coordinates of a point in Two dimension
Cartesian coordinate system is widely used to locate a particle in two dimensional or Three dimensional Plane.
In Cartesian coordinate system the coordinate of origin are defined as (0,0). Origin is always taken as reference point to get the coordinates of any other point in this system.
The notation to write the coordinates of a point P is (x,y). Where x is the distance from origin in x direction and y is the distance from origin in y direction.
The Cartesian coordinate system is Divided into four parts. i.e four quadrants
1 st Quadrant
In the first quadrant, For every point the value of x and y is always Postive.
Example: (3,1) ,(7,8) ,(3,7),(3,0) all these points lie in first quadrant
General Notation (x,y) = (+,+)
2nd Quadrant
In the second quadrant, For every point the value of x is Negative and for y is Postive.
Example: (-2,1) ,(-7,0) ,(-3,5),(-1,9) all these points lie in second quadrant.
General Notation (x,y) = (-,+)
3rd Quadrant
In the third quadrant, For every point the value of x and y are Negative.
Example : (-2,-2) ,(-2,-5) ,(-7,-2),(-2,-8) all these points lie in third quadrant.
General Notation (x,y) = (-,-)
4th Quadrant
In the 4th quadrant, For every point the value of x is positive and for y is Negative.
Example : (1,-2) ,(5,-5) ,(6,-7),(6,-1) all these points lie in third quadrant.
General Notation (x,y) = (+,-)
For instance if a point A has coordinates (3,4) which means it is 3 unit away from origin in +x-axis and 4 unit away from origin in +Y-axis.
Distance Formula
The distance between two points \( A(x_1,x_2) \) and \( B(x_2,y_2) \) is given by the following formula.
\(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Example 1 : The coordinates of the points P and Q are (3,2) and (8,5). Find the Distance between PQ.
Solution: \(x_1 = 3 , y_1=2\)
\( x_2 = 8 , y_2=5 \)
PQ = \(\sqrt{(8-3)^2 + (5-2)^2}\)
= \(\sqrt{25 + 9}\)
= √34
Example 2 : The distance between two points is 5 and Coordinates of point M and N are (2,4) and (5,x).
Solution: \( 5 = \sqrt{(5-2)^2 + (x-4)^2} \)
Squaring Both sides
\( 25 = 9 + (x-4)^2 \)
\( \sqrt{16} = x - 4 \)
x = 4+4
= 8
Section Formula
The coordinates of a point that divides the line PQ into m:n is :
\( X = \frac{m x_1 + n x_2}{m + n} \)
\( Y = \frac{m y_1 + n y_2}{m + n} \)
This formula also can be written in diffrent form.Let the Point divides the line joining P and Q in \( k: 1 \) then the formula converts like this
\( X = \frac{k x_2 + x_1}{k + 1}, \quad Y = \frac{k y_2 + y_1}{k + 1} \)
where \( k = m/n \)
If the Point divides the line PQ in 1:1 then the point must be mid point and its coordinates are given by :
\( X = \frac{x_1 + x_2}{2} \)
\( Y = \frac{y_1 + y_2}{2} \)
Example 1 : Two points A(7,8) and B(5,2) form a line.find the coordinates of the the point that divides it into 4:2.
Ans : Let the coordinates of that point be (X,Y)
X = \( \frac{4 \times 5 + 2 \times 7}{4 + 2} \)
X= 34/6
Y =\( \frac{4 \times 2 + 2 \times 8}{4 + 2} \)
= 4
Hence the coordinates of the point that divides the line joining A(7,8) and B(5,2) is:
(34/6,4)
Outer Section formula
When the dividing point is outside the line in that case
Let the coordinates of that point be (X,Y) then
\( X = \frac{m x_2 - n x_1}{m - n} \)
\( Y = \frac{m y_2 - n y_1}{m - n} \)
The coordinates of that point are:
\( C(X, Y) = \left( \frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n} \right) \)
Example : If a point P divides the line joining A(2, 4) and B(6, 8) externally in the ratio 1 : 2, find the coordinates of point P using the outer section formula.
Ans: X = \( \frac{1 \times 6 - 2 \times 2}{1 - 2} = -2 \)
Y = \( \frac{1 \times 8 - 2 \times 4}{1 - 2} = 0 \)
The point of the coordinates P( -2,0).
Special Note : No external point can divide a line into 1:1.
Area of the Triangle
The area of a triangle can also be calculated using the method of Coordinate geometry.Hence it is given by:\(\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|\)
Condition for Collinearity
The condition for collinearity of three points is that the area of the triangle formed by them is zero.
\( \Delta \) = \( 0\)
Coordinate Geometry Worksheets