Class 10 Maths notes for Coordinate geometry

• Introduction
• Distance formula
• Section Formula
• Area of triangle
• How to Solve the  line segment bisection ,trisection and four-section problem's
• How to Prove three points are collinear
• How to solve general Problems of Area in Coordinate geometry

Introduction

• We require two perpendicular axes to locate a point in the plane. One of them is horizontal and other is Vertical.The plane is called Cartesian plane and axis are called the coordinates axis
• The horizontal axis is called x-axis and Vertical axis is called Y-axis
• The point of intersection of axis is called origin.
• The distance of a point from y axis is called x –coordinate or abscissa and the distance of the point from x –axis is called y – coordinate or Ordinate
• The x-coordinate and y –coordinate of the point in the plane is written as (x, y) for point and is called the coordinates of the point
• Above all in details can be read in Class IX Maths Coordinate Geometry notes

Distance formula

Section Formula

Area of Triangle ABC

How to Solve the  line segment bisection ,trisection and four-section problem's

1. You will be given coordinates of the two point A , B
2. If the problem is to find bisection, then you can simply found the mid point using
3. If the problem is to find trisection(three equal parts of the line ).Let us assume the point are P and Q, then AP=PQ=QB
   Now P divides the line AB into 1:2 part
   While Q divides the line AB into 2:1 part
   So we can use section formula to get the coordinate of point P and Q
   
4. If the problem is to find four equal parts .Let us assume the point are P ,Q And R  such that AP=PQ=QR=RB
   Now P divides the line AB into 1:3 part
   Q divides the line AB into 1:1 part
   R divides the line AB into 3:1 part
   So we can use section formula to get the coordinate of point P  ,Q and R
   

How to Prove three points are collinear

How to solve general Problems of Area in Coordinate geometry

Solved Examples

Question 1
The points (4, 5), (7, 6) and (6, 3) are collinear. True or false
Solution
False. Since the area of the triangle formed by the points is 4 sq. units, the points are not collinear

Question 2
If the mid-point 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.
Solution
Mid-point of the line segment joining A (3, 4) and B (k, 6) = $\frac {(3+k)}{2}$, 5
Now mid -point is P(x,y)
$x=\frac {(3+k)}{2}$ or 3+k=2x
y=5
Since $x + y - 10 = 0$, we have
$\frac {3+k}{2} +5 - 10 = 0$
3 + k = 10
or k=7

Question 3
ABCD is a parallelogram with vertices A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃). Find the coordinates of the fourth vertex D in terms of x₁,x₂,x₃,y₁,y₂,y₃
Solution
Let the coordinates of D be (x, y). We know that diagonals of a parallelogram bisect each other.
Therefore, mid-point of AC = mid-point of BD
$[ \frac{(x_1 + x_3)}{2} ,\frac{(y_1 + y_3)}{2}]=[ \frac{(x + x_2)}{2} ,\frac {(y + y_2)}{2}]$
i.e $x_1 + x_3 = x_2 + x$ and $y_1 + y_3 = y_2 + y$
$x= x_1 + x_3 - x_2$ and $y=y_1+ y_3 -y_2$
So coordinates are $(x_1 + x_3 - x_2, y_1+ y_3 - y_2)$

Question 4<
The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
Solution


$A=\frac {1}{2} [a(c+a -a-b) + b(a+b-b-c) + c( b+c -c-a)]$
$A=\frac {1}{2} [ac-bc + ba-bc + bc-ac)]$
A=0

Crossword Puzzle

Across
1. the distance of point(3,4) from origin
3. x-coordinate is called
5. the y -axis is a .....line
6. The (-6,-2) lies in .....quadrant
7. the x-axis is a ......line
9. The point of intersection of axis
10. The figure formed by joining the point (0,0), (2,0),(2,2) and (0,2)
Down
2. the coordinates (3,0) (-3,0) and $(0,3 \sqrt {3})$ formed an ...... triangle
4. The three point are said to be ....if the area of triangle formed by them is zero
8. y-coordinate is called
Check your Answers

1)five
2)equilateral
3)abscissa
4)collinear
5)vertical
6)fourth
7)horizontal
8)Ordinate
9) origin
10) square