Chapter 7: Coordinate Geometry — Class 10 Mathematics NCERT Solutions | CBSE
Chapter 7: Coordinate Geometry – NCERT Class 10 Mathematics Solutions | CBSE
Chapter 7. COORDINATES GEOMETRY
Chapter 7 . Coordinate Geometry |
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Exercise 7.1 complete solution Exercise 7.2 complete solution Exercise 7.3 complete solution Exercise 7.4 (Optional) complete solution |
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Distance formula : 1. The distance between |
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Section formula : i.e. , i.e., 5. If |
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Area of a Triangle : 6. The area of the triangle formed by the points 7. The points |
and 
is given by
from the origin 
is given by 
which divides the line segment joining the points 
and 
internally in the ratio 
are given by 
and 
and 
are the vertices of 
, then the coordinates of the centroid is 
EXERCISE 7.1
1. Find the distance between the following pairs of points :
(i) 
(ii) 
(iii) 
Solution : (i)
Let, 
and 
are two points .
Using distance formula , we have
Here, 
units
(ii)
Let, 
and 
are two points .
Using distance formula , we have
Here, 
units
(iii)
Let, 
and 
are two points .
Using distance formula , we have 
Here, 
units
2. Find the distance between the points 
and 
. Can you now find the distance between the two towns A and B discussed in Section 7.2 .
Solution : Let, 
and 
are two points .
Here, 
Using distance formula , we have
units
3. Determine if the points 
and 
are collinear .
Solution : Let, 
and 
are three points respectively .
Using distance formula , we have
units
units
units
So , 
.
Therefore, the points and are not collinear .
4. Check whether 
and 
are the vertices of an isosceles triangle .
Solution : Let 
and 
are the vertices of any triangle respectively .
Using distance formula , we have
units
units
units
So, 
Therefore , the points 
and 
are the vertices of an isosceles triangle .
5. In a classroom, 4 friends are seated at the points A , B , C and D as shown in Fig. 7.8 . Champa and Chemeli walk into the class and after observing for a few minutes Champa asks Chameli ‘‘ Don’t you think ABCD is a square ? Chameli disagrees . Using distance formula , find which of them is correct .
Solution: Given, the coordinates of the four friends are A(3,4) , B(6,7) , C(9,4) and D(6,1) respectively .
Using distance formula, we have
units
units
units
units
units
and 
units
So, AB = BC = CD = AD and AC = BD
Therefore, ABCD is a square . Chapma is correct .
6. Name the type of quadrilateral formed, if any , by the following points, and give reasons for your answers :
(i) 
(ii) 
(iii) 
Solution: (i)
Let, 
and 
are the vertices of the quadrilateral respectively.
Using distance formula , we have
units
units
units
units
units
units
So, 
and 
Therefore , 
and 
are vertices of the square .
(ii)
Solution: Let, 
and 
are the vertices of the quadrilateral respectively.
Using distance formula , we have
units
units
units
units
So, 
Therefore , and are not of the vertices of quadrilateral .
(iii)
Solution: Let, 
and 
are the vertices of the quadrilateral respectively.
Using distance formula , we have
units
units
units
units
units
units
So, 
, 
and 
Therefore , and are vertices of the parallelogram .
7. Find the point on the 
-axis which is equidistant from 
and 
.
Solution : Let, 
is equidistant from the points A(2 , – 5 ) and B (– 2 ,9) .
Given , 
-axis , i.e., 
.
A/Q , 
Therefore , the coordinate of the point P is 
.
8. Find the values of 
for which the distance between the points 
and
is 10 units .
Solution : We have , 
[ using distance formula]
or 
Thus, the value of are – 9 and 3 .
9. If 
is equidistant from 
and 
, find the values of . Also find the distance QR and PR .
Solution : Since is equidistant from and 
.
A/Q, 
The distance of and 
is
units
The distance of and 
is
units
The distance of and is
units
The distance of and is
units
10. Find a relation between 
and 
such that the point
is equidistant from the point 
and 
.
Solution : Given ,the point 
is equidistant from the point 
and 
.
A/Q, 
[Squaring both side]
EXERCISE 7.2
1. Find the coordinates of the point which divides the join of 
and 
in the ratio 
.
Solution: Here, 
, 
, 
Let, the coordinate of the point is 
.
Using section formula , we have
and 
Therefore, the coordinates of the point is (1 , 3) .
2. Find the coordinates of the points of trisection of the line segment joining 
and 
.
Solution: let the coordinates of the points are 
and 
.
For point P : Here, 
, 
, 
Using section formula , we have
and 
For point Q : Here, , , 
and 
Therefore, the coordinates of the points are 
.
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD , lines have been drawn with chalk powder at a distance of 1 m each . 100 flowers pots have been placed at a distance of 1 m from each other along AD , as shown in Fig. 7.12 . Niharika runs 
th the distance AD , on the 
line and posts a green flag . Preet runs 
th the distance AD on the eighth line and posts a red flag . What is the distance between both the flags ? If Rashmi has to post a blue flag exactly halfway between the line segment jointing the two flags, where should she post her flag ?
Solution: Given, ABCD is a rectangular school ground , then the distance of the side AD = 1m × 100 = 100 m .
The distance of AD run by Niharika on the second line 
Therefore, the coordinate of the point is (2 , 25) .
Again , the distance of AD run by Preet on the eighth line 
Therefore, the coordinate of the point is (8 , 20) .
Using distance formula, we have
the distance between both the flags 
units
Since, Rashmi has to post a blue flag exactly halfway between the line segment jointing the two flags ,i.e., Rashmi is equidistant distance from Niharika and Preet . let the coordinate of Rashmi is 
Using section formula , we have
and 
Therefore, the position of Rashmi has to post a blue flag on the 5th line at a distance of 22.5 m .
4. Find the ratio in which the line segment joining the points 
and 
is divided by 
.
Solution: Let , the ratio be 
.
Here, 
, 
, 
Using section formula , we have
and
Now,
Therefore, the ratio is 2 : 7 .
5. Find the ratio in which the line segment joining 
and
is divided by the 
-axis . Also , find the coordinates of the point of division .
Solution: let , the ratio be and the coordinate is 
.
Here, 
, 
Using section formula , we have
and
Now,
Again, 
Therefore, the ratio is 1 : 1 and the coordinate is 
.
6. If 
and 
are the vertices of a parallelogram taken in order, find and 
.
Solution: Since, 
and 
are the vertices of a parallelogram.
We know that , the diagonals of a parallelogram bisect each other .
A/Q, The coordinate of the mid-point of the diagonal AC = The coordinate of the mid-point of the diagonal BD .
and 
or 
Therefore, the value of 
and 
.
7. Find the coordinates of a point A , where AB is the diameter of a circle whose centre is 
and 
is 
.
Solution: Here, AP = BP = Radius . So, 
Let the coordinates of a point 
.
Here, 
, 
Using the section, we have
and
Therefore, the coordinate of the point 
.
8. If A and B are 
and 
, respectively, find the coordinates of P such that 
and P lies on the line segment AB .
Solution: let the coordinates of the point P is .
and 
Here, 
,
, 
Using section formula , we have
and 
Therefore, the coordinate of the point P is 
.
9. Find the coordinates of the points which divide the line segment joining 
and 
into four equal parts .
Solution: Let the coordinate of the points are 
and 
.
Here, 
, 
For point P : 
Using section formula , We have
and 
The coordinate of the point P is 
.
For point Q : Here, 
and 
The coordinate of the point Q is (0 , 5) .
For point R : Here, 
and 
The coordinate of the point R is 
.
10. Find the area of a rhombus if its vertices are 
and 
taken in order .
[ Hint : Area of a rhombus 
(product of its diagonals)]
Solution: let the vertices of the rhombus are 
and 
respectively .
Using distance formula , we have,
units
And 
units
The area of a rhombus ABCD 
sq. units
EXERCISE 7.3
1. Find the area of the triangle whose vertices are :
(i) 
(ii) 
Solution: (i) 
Let A(2,3) , B(– 1 , 0) and C(2 ,– 4) are the vertices of the triangle ABC respectively .
Here, 
We know that , Area of 
Sq. units.
(ii) 
Let A(– 5 ,– 1) , B(3 ,– 5) and C(5 , 2) are the vertices of the triangle ABC respectively .
Here, 
We know that , Area of
Sq. units
2. In each of the following find the value of ‘ k ’ , for which the points are collinear :
(i) 
(ii) 
Solution: (i) 
Here, 
We know that , 
Therefore, the value of k is 4 .
(ii) 
Here, 
We know that ,
Therefore, the value of k is 3 .
3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are 
and
.Find the ratio of this area to the area of the given triangle .
Solution: Let A(0 ,– 1) , B(2 , 1) and C(0 , 3) are the vertices of the triangle ABC respectively .
Here, 
We know that , Area of
sq. units
Let D , E and F are the mid-point of the sides AB , BC and AC of the triangle ABC respectively .
So, the coordinate of the point D is 
The coordinate of the point E is 
The coordinate of the point F is 
Therefore, D(1 , 0) , E(1 , 2) and C(0 , 1) are the vertices of the triangle DEF respectively .
Here, 
Area of 
sq. unit
4. Find the area of the quadrilateral whose vertices, taken in order are 
and
.
Solution: Let, A(– 4 ,– 2) , B(– 3 , – 5) , C(3 ,– 2) and D(2,3) are the vertices of the triangle ABC respectively and join diagonal AC .Then ,we find two triangles ABC and ADC .
For triangle ABC : The vertices of the triangle ABC are A(– 4 ,– 2) , B(– 3 , – 5) and C(3 ,– 2) respectively.
Here, 
We know that , Area of
For triangle ADC : The vertices of the triangle ABC are A(– 4 ,– 2) , D(2 , 3) and C(3 ,– 2) respectively.
Here, 
We know that , Area of
Sq. units (positive value)
Therefore, Area of ABCD = Area of ABC + Area of ADC 
Sq.units .
5. You have studied in Class IX , (Chapter 9 , Example 3) , that a median of a triangle divides it into two triangles of equal areas . Verify this result for 
whose vertices are 
and C
.
Solution: let AD is the median of a triangle ABC . The vertices of the triangle ABC are and C . We find two triangles are ABD and ACD .
The coordinate of the point D is 
For triangle ABD : The vertices of the triangle ABD are A( 4 ,– 6) , B(3 ,– 2) and D(4 ,0) respectively.
Here, 
We know that , Area of 
sq. units
For triangle ACD : The vertices of the triangle ABD are A( 4 ,– 6) , B(5 , 2) and D(4 ,0) respectively.
Here, 
, 
We know that , Area of 
Therefore, area of the ABD = Area of the ACD . Verified
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