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
(1) Find the distance of point (3,4) and ( 4,3) from Origin?
Solution
Distance from Origin is given by
$D=\sqrt {x^2 + y^2}$
for (3,4),
$D=\sqrt {3^2 + 4^2} =5$
for (4,3),
$D=\sqrt {4^2 + 3^2} =5$
Practice Questions
(1) Arrange these points in ascending order of distance from origin
(a) (1,2)
(b) (5,5)
(c) (-1, 3)
(d) (1/2, 1)
(e) (.5 ,.5)
(f) (-1,1)
(2) Find the distance between the points
(a) (1,2) and (3,4)
(b) (5,6) and ( 1,1)
(c) (9,3) and ( 3,1)
Section Formula
A point P(x,y) which divide the line segment AB in the ratio m_{1} and m_{1} is given by
The mid point P is given by
Area of Triangle ABC
Area of triangle ABC of coordinates A(x_{1},y_{1}) , B(x_{2},y_{2}) and C(x_{3},y_{3}) is given by
For point A,B and C to be collinear, The value of A should be zero
How to Solve the line segment bisection ,trisection and four-section problem's
You will be given coordinates of the two point A , B
If the problem is to find bisection, then you can simply found the mid point using
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
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
1) We need to assume that if they are not collinear,they should be able to form triangle.
We will calculate the area of the triangle,if it comes zero, that no triangle can be found and they are collinear
How to solve general Problems of Area in Coordinate geometry
Area of Triangle
Three vertices will be given,you can calculate the area directly using formula
Area of Square
Two vertices will be given, we can calculate either side or diagonal depending on vertices given and apply the square area formula
Area Of rhombus
Given: all the vertices coordinates
Two ways
1) Divide the rhombus into two triangle. Calculate the area of both the triangle and sum it
2) Calculate the diagonal and apply the Area formula
Area of parallelogram
Three vertices are sufficent to find the area of parallelogram
Calculate the area of the traingle formed by the three verticles and double it to calculate the area of parallelogram
Area of quadilateral
Given: all the vertices coordinates
Divide into two triangle. Calculate the area seperately and sum it
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 (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}).
Find the coordinates of the fourth vertex D in terms of x_{1},x_{2},x_{3},y_{1},y_{2},y_{3} 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
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
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