- 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
- The Origin has zero distance from both x-axis and y-axis so that its abscissa and ordinate both are zero. So the coordinate of the origin is (0, 0)
- A point on the x –axis has zero distance from x-axis so coordinate of any point on the x-axis will be (x, 0)
- A point on the y –axis has zero distance from y-axis so coordinate of any point on the y-axis will be (0, y)
- The axes divide the Cartesian plane in to four parts. These Four parts are called the quadrants
- The coordinates of the points in the four quadrants will have sign according to the below table

Quadrant |
x-coordinate |
y-coordinate |

Ist Quadrant |
+ |
+ |

IInd quadrant |
- |
+ |

IIIrd quadrant |
- |
- |

IVth quadrant |
+ |
- |

Distance between the points AB is given by

Distance of Point A from Origin

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)

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),

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

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)

A point P(x,y) which divide the line segment AB in the ratio m

The mid point P is given by

For point A,B and C to be collinear, The value of A should be zero

- 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

We will calculate the area of the triangle,if it comes zero, that no triangle can be found and they are collinear

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 |

The points (4, 5), (7, 6) and (6, 3) are collinear. True or false

False. Since the area of the triangle formed by the points is 4 sq. units, the points are not collinear

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.

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

ABCD is a parallelogram with vertices A (x

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 $(x_1 + x_3 - x_2, y_1+ y_3 - y_2)$

The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is

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

$A=\frac {1}{2} [ac-bc + ba-bc + bc-ac)]$

A=0

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)

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

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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