 Class 10 Maths notes for Coordinate geometry

Flash Back from Class IX Coordinate geometry notes

1. We require two perpendicular axes to locate a point in the plane. One of them is horizontal and other is Vertical
2. The plane is called Cartesian plane and axis are called the coordinates axis
3. The horizontal axis is called x-axis and Vertical axis is called Y-axis
4. The point of intersection of axis is called origin.
5. 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
6. 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
7. 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)
8. 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)
9. 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)
10. The axes divide the Cartesian plane in to four parts. These Four parts are called the quadrants
11. 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 formula 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)

Section Formula A point P(x,y) which divide the line segment AB in the ratio m1 and m1 is given by The mid point P is given by Area of Triangle ABC

Area of triangle ABC of coordinates A(x1,y1) , B(x2,y2) and C(x3,y3) 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

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

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 (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

Quiz Time

Question 1 Point on y axis and x axis has coordinates: ?
A) (a,0) & (0,b)
B) (0,a) & (0,b)
C) (a,b) & (a,b)
D) (0,a) & (b,0)
Question 2 The perimeter of a PQR triangle with vertices P(0, 4), Q(0, 0) and R(3, 0) is: ?
A) 6
B) 7
C) 11
D) 12
Question 3 Which point is at minimum distance from origin
A) (2, -3)
B) (2,2)
C) (6,8)
D) (6,6)
Question 4 The end points of diameter of circle are (0, 0) & (24, 7). The radius of the circle is:
A) 12.5
B) 15.5
C) 12
D) 10
Question 5 The mid point of point A(4,5) and B(-8,7) lies in quadrant?
A) I
B) II
C)III
D) IV
Question 6 The point on the x-axis which is equidistant from (-2,5) and (2, -3) is?
A)(0,3)
B) (0,0)
C)(5,0)
D) (-2,0)

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

Reference Books for class 10

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

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

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20

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