NCERT solutions for class 10 maths chapter 7 Coordinate Geometry: A system of mathematics in which the points on a plane are described using an ordered pair of numbers is called coordinate geometry. In solutions of NCERT class 10 maths chapter 7 coordinate geometry, you will get a detailed explanation to each and every question present in the chapter including the optional exercises. Coordinate Geometry is one of the important chapters for higher mathematics in upcoming classes. You will use at least some part of coordinate geometry in the field whether you are going to study commerce, engineering, maths, medical, arts, etc.
CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry is designed to give you a step by step solution for a particular question. In Coordinate Geometry, you will study the algebraic forms of geometric figures. NCERT solutions for class 10 maths chapter 7 Coordinate Geometry will be helpful for doing homework as well as while preparing for the board examinations. Apart from this chapter, if you want NCERT solutions for other classes and subjects then you can download these solutions by clicking on the above link.
Q1 (i) Find the distance between the following pairs of points : (2, 3), (4, 1)
Given points: (2, 3), (4, 1)
Distance between the points will be:
Q1 (ii) Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)
Given points: (– 5, 7), (– 1, 3)
Distance between the points will be:
Q1 (iii) Find the distance between the following pairs of points :(a, b), (– a, – b)
Given points: (a, b), (– a, – b)
Distance between the points will be:
Given points: (0, 0) and (36, 15)
Distance between the points will be:
The distance between the two towns A and B is, thus 39 km for given towns location
and .
Q3 Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
Let the points (1, 5), (2, 3) and (– 2, – 11) be representing the vertices A, B, and C of the given triangle respectively.
Therefore,
Since these are not satisfied.
As these cases are not satisfied.
Hence the points are not collinear.
Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
The distance between two points is given by:
So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangle A, B, and C respectively.
Therefore, AB = BC
Here two sides are equal in length.
Therefore, ABC is an isosceles triangle.
The coordinates of the points:
are the positions of 4 friends.
The distance between two points is given by:
Hence,
And the lengths of diagonals:
So, here it can be seen that all sides of quadrilateral ABCD are of the same lengths and diagonals are also having the same length.
Therefore, quadrilateral ABCD is a square and Champa is saying right.
Let the given points be representing the vertices A, B, C, and D of the given quadrilateral respectively.
The distance formula:
Finding the length of the diagonals:
It is clear that all sides are of the same lengths and also the diagonals have the same lengths.
Hence, the given quadrilateral is a square.
Q6 (ii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)
Let the given points be representing the vertices A, B, C, and D of the given quadrilateral respectively.
The distance formula:
All the sides of the given quadrilateral have different lengths.
Therefore, it is only a general quadrilateral and not a specific one like square, rectangle, etc.
Q6 (iii) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)
Let the given points be representing the vertices A, B, C, and D of the given quadrilateral respectively.
The distance formula:
And the diagonals:
Here we can observe that the opposite sides of this quadrilateral are of the same length.
However, the diagonals are of different lengths.
Therefore, the given points are the vertices of a parallelogram.
Q7 Find the point on the xaxis which is equidistant from (2, –5) and (–2, 9).
Let the point which is equidistant from be as it lies on Xaxis.
Then, we have
Distance AX:
and Distance BX
According to the question, these distances are equal length.
Hence we have,
Solving this to get the required coordinates.
Squaring both sides we get,
Or,
Hence the point is .
Q8 Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Given the distance between the points and is 10 units.
The distance formula :
So, given
After squaring both sides
Therefore, the values are .
Answer:
Given is equidistant from and .
Then, the distances .
Distance
Distance
Squaring both sides, we get
The points are:
CASE I: when R is
The distances QR and PR.
CASE II: when R is
The distances QR and PR.
Let the point is equidistant from and .
Then, the distances
and
Squaring both sides: we obtain
Thus, the relation is between x and y.
Q1 Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Let the coordinates of point which divides the line segment joining the points and , internally, in the ratio then,
Section formula:
Substituting the values in the formula:
Here,
Hence the coordinate is .
Q2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Let the trisection of the line segment and have the points and
Then,
Section formula:
By observation point, P divides AB internally in the ratio .
Hence,
Substituting the values in the equation we get;
And by observation point Q, divides AB internally in the ratio
Hence,
Substituting the values in the equation above, we get
Hence, the points of trisections are and
Niharika posted the green flag at the distance P, i.e.,
from the starting point of line.
Therefore, the coordinates of this point are
Similarly, Preet posted red flag at of the distance Q i.e.,
from the starting point of line.
Therefore, the coordinates of this point Q are .
The distance is given by,
and the point at which Rashmi should post her Blue Flag is the midpoint of the line joining these points. Let this point be .
Then, by Section Formula,
Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.
Let the ratio be :
Then, By section formula:
Given point
Hence, the point divides the line AB in the ratio .
Let the point on the xaxis be and it divides it in the ratio .
Then, we have
Section formula:
Hence, the value of k will be:
Therefore, the xaxis divides the line in the ratio and the point will be,
Putting the value of in section formula.
Let the given points .
Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonal AC and BD also divides these diagonals.
Therefore, O is the midpoint of AC and BD.
The coordinates of the point O when it is midpoint of AC.
The coordinates of the point O when it is midpoint of BD.
Since both coordinates are of same point O.
Therefore,
and
Or,
As the centre point will be the midpoint of the diameter AB.
Then, the coordinates of point A will be .
Given point .
Therefore,
.
Therefore, the coordinates of A are
From the figure:
As
hence the ratio is 3:4,
Now, from the section formula, we can find the coordinates of Point P.
Section Formula:
From the figure:
Points C, D, and E divide the line segment AB into four equal parts.
Now, from the section formula, we can find the coordinates of Point C, D, and E.
Section Formula:
Here point D divides the line segment AB into two equal parts hence
Now, point C divides the line segment AD into two equal parts hence
Also, point E divides the line segment DB into two equal parts hence
From the figure:
Let the vertices of the rhombus are:
Area of the rhombus ABCD is given by;
Hence we have to find the lengths of the diagonals AC and BD of the rhombus.
The distance formula:
Length of the diagonal AC:
Length of the diagonal BD:
Thus, the area will be,
As we know, the area of a triangle with vertices (x1,x2) ,(y1, y2) and (z1 z2 ) is given by :
So Area of a triangle whose vertices are(2, 3), (–1, 0)and (2, – 4) is
Hence, the area of the triangle is 10.5 per unit square.
Q1 (ii) Find the area of the triangle whose vertices are (–5, –1), (3, –5), (5, 2)
From the figure:
Area of the triangle is given by:
Substituting the values in the above equation, we obtain
The points (7, –2), (5, 1), (3, k) are collinear if the area of the triangle formed by the points will be zero.
Area of the triangle is given by:
Substituting the values in the above equation, we obtain
Hence, the points are collinear for k=4.
The points (8,1), (k, 4), (2,5) are collinear if the area of the triangle formed by these points will be zero.
Area of the triangle is given by:
Substituting the values in the above equation, we obtain
Hence, the points are collinear for k = 3.
From the figure:
The coordinates of the point P, Q, and R are:
Point P is the midpoint of side AB, hence the coordinates of P are :
Point Q is the midpoint of side AC, hence the coordinates of Q are :
Point R is the midpoint of side BC, hence the coordinates of R are :
Hence, the area of the triangle formed by the midpoints PQR will be,
And the area formed by the triangle ABC will be:
Thus, the ratio of Area of to the Area of will be .
From the figure:
The coordinates are
Divide the quadrilateral into 2 parts of triangles.
Then the area will be,
Area of the triangle formed by ABC will be,
Area of the triangle formed by ADC will be,
Therefore, the area of the quadrilateral will be:
Alternatively,
The points A and C are in the same ordinates.
Hence, the length of base AC will be
Therefore,
Area of triangle ABC:
Area of triangle ADC:
Therefore, the area will be,
From the figure:
The coordinates of midpoint M of side BC is:
Now, calculating the areas of the triangle ABM and ACM :
Area of triangle, ABM:
Area of triangle, ACM:
However, the area cannot be negative, Therefore, area of is 3 square units.
Clearly, the median AM divided the in two equal areas.
Let the line divide the line segment AB in the ratio at point C.
Then, the coordinates of point C will be:
Point C will also satisfy the given line equation , hence we have
Therefore, the ratio in which the line divides the line segment joining the points and is internally.
Q2 Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
If the points are collinear then, the area formed by these points will be zero.
The area of the triangle is given by,
Substituting the values in the above equation, we have
Or,
Hence, the required relation between x and y is .
Q3 Find the center of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
From the figure:
Let the center point be .
Then the radii of the circle are equal.
The distance OA:
The distance OB:
The distance OC:
Equating the radii of the same circle.
When equating,
Squaring both sides and applying
or
...................................(1)
When equating,
Squaring both sides and applying
...................................(2)
Now, adding the equations (1) and (2), we get
.
From equation (1), we get
Therefore, the centre of the circle is .
From the figure:
We know that the sides of a square are equal to each other.
Therefore, AB = BC
So,
Squaring both sides, we obtain
Now, doing
We get
Hence .
Applying the Pythagoras theorem to find out the value of y.
Answer:
Taking A as origin then, the coordinates of P, Q, and R can be found by observation:
Coordinates of point P is
Coordinates of point Q is
Coordinates of point R is
The area of the triangle, in this case, will be:
Taking C as origin, then CB will be xaxis and CD be yaxis.
The coordinates fo the vertices P, Q, and R are: respectively.
The area of the triangle, in this case, will be:
It can be observed that in both cases the area is the same so, it means that the area of any figure does not depend on the reference which you have taken.
From the figure:
Given ratio:
Therefore, D and E are two points on side AB and AC respectively, such that they divide side AB an AC in the ratio of .
Section formula:
Then, coordinates of point D:
Coordinates of point E:
Then, the area of a triangle:
Substituting the values in the above equation,
Hence the ratio between the areas of and is
From the figure:
Let AD be the median of the triangle
Then, D is the midpoint of BC
Coordinates of Point D:
From the figure,
The point P divides the median AD in the ratio, AP: PD = 2: 1
Hence using the section formula,
From the figure,
The point Q divides the median BE in the ratio, BQ : QE = 2 : 1
Hence using the section formula,
The point R divides the median CF in the ratio, CR: RF = 2: 1
Hence using the section formula,
Q7 (iv) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of D ABC. What do you observe?
We observed that the coordinates of P, Q, and R are the same. Therefore, all these are representing the same point on the plane. i.e., the centroid of the triangle.
From the figure,
Let the median be AD which divides the side BC into two equal parts.
Therefore, D is the midpoint of side BC.
Coordinates of D:
Let the centroid of this triangle be O.
Then, point O divides the side AD in a ratio 2:1.
Coordinates of O:
From the figure:
P is the midpoint of side AB.
Therefore, the coordinates of P are,
Similarly, the coordinates of Q, R and S are: respectively.
The distance between the points P and Q:
and the distance between the points Q and R:
Distance between points R and S:
Distance between points S and P:
Distance between points P and R the diagonal length:
Distance between points Q and S the diagonal length:
Hence, it can be observed that all sides have equal lengths. However, the diagonals are of different lengths.
Therefore, PQRS is a rhombus.
Chapter No. 
Chapter Name 
Chapter 1 
CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers 
Chapter 2 

Chapter 3 
Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4 
CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations 
Chapter 5 
NCERT solutions for class 10 chapter 5 Arithmetic Progressions 
Chapter 6 

Chapter 7 
CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry 
Chapter 8 
NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 
Chapter 9 
Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10 

Chapter 11 

Chapter 12 
Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13 
CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14 

Chapter 15 
First of all, go through the concepts and examples given in the textbook to understand the basics of the chapter.
Memorize the formulae to find out some directly asked questions.
After doing the above two activities you can come to the practice exercises.
During the practice of exercises, you can take the help of NCERT solutions for class 10 maths chapter 7 Coordinate Geometry.
After doing all the abovementioned things, you can come to practice the previous year's question to boost your preparation.
Keep working hard & happy learning!