In coordinate geometry, points, lines, curves, and other geometric objects are represented using coordinates on a graph. The coordinates consist of two numbers, usually denoted by (x, y), which represent the horizontal and vertical distances of a point from the origin of the coordinate system.
The coordinate system consists of two perpendicular number lines called the xaxis and yaxis. The point where the two axes intersect is called the origin, and it is usually denoted by (0, 0). The xaxis is the horizontal axis, and the yaxis is the vertical axis.
Using the coordinates of two points, it is possible to determine the distance between them, as well as the slope and equation of the line that passes through them. Similarly, the equation of a circle, parabola, ellipse, or hyperbola can be found using algebraic techniques.
Coordinate geometry has applications in various fields, including physics, engineering, and computer graphics. It is also used in navigation and mapping, as well as in geometry and trigonometry.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
There are total four exercises in Coordinate Geometry Class 10 which contain 33 problems. Distance between two points in class 10 board are important model questions on their coordinates, area of a triangle, using dividing line (ratio formula).
Topics and Sub Topics in Class 10 Maths Chapter 7 Coordinate Geometry:
Section Name  Topic Name 
7  Coordinate Geometry 
7.1  Introduction 
7.2  Distance Formula 
7.3  Section Formula 
7.4  Area of a Triangle 
7.5  Summary 
NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1
The NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 are part of the NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1.
Board  CBSE 
Textbook  NCERT 
Class  Class 10 
Subject  Maths 
Chapter  Chapter 7 
Chapter Name  Coordinate Geometry 
Exercise  Ex 7.1 
Number of Questions Solved  10 
Category  NCERT Solutions 
Define Coordinate Geometry
Coordinate geometry is a branch of mathematics that deals with the study of geometric figures using algebraic techniques. It involves the use of a coordinate system to represent points, lines, curves, and other geometric objects. In this system, each point is represented by an ordered pair of numbers, usually denoted by (x, y), which represent the horizontal and vertical distances of a point from the origin of the coordinate system. The coordinate system consists of two perpendicular number lines called the xaxis and yaxis, which intersect at a point called the origin. Using algebraic techniques, it is possible to analyze the properties of geometrical objects such as distance between points, slope and equation of lines, and equation of circles, parabolas, ellipses, or hyperbolas. Coordinate geometry has applications in various fields including physics, engineering, and computer graphics.
Use of Coordinate Geometry in daily life
Coordinate geometry has many practical applications in daily life. Some examples include:
 Navigation: The GPS (Global Positioning System) used for navigation relies on coordinate geometry. By using the coordinates of a destination and the current location, the GPS system can calculate the distance and direction to travel to reach the destination.
 Construction: Architects and builders use coordinate geometry to design and construct buildings, roads, and bridges. They use coordinates to determine the location and orientation of structural elements, such as walls, columns, and beams.
 Mapping: Maps and cartography rely heavily on coordinate geometry. The coordinates of a location are used to plot it on a map and determine the distance and direction to other locations.
 Robotics: Robots use coordinate geometry to navigate and perform tasks. They use the coordinates of objects in their environment to locate them and perform operations on them.
 Computer graphics: Coordinate geometry is used extensively in computer graphics to create and manipulate images. The coordinates of the pixels that make up an image are used to position and transform them to create various effects.
Tips and tricks Coordinate Geometry
Here are some tips and tricks for coordinate geometry:
 Use the distance formula to find the distance between two points. The distance formula is derived from the Pythagorean theorem and is given by: distance = √[(x2 – x1)² + (y2 – y1)²]
 To find the midpoint of a line segment, use the midpoint formula: midpoint = [(x1 + x2)/2, (y1 + y2)/2]
 To find the slope of a line passing through two points, use the slope formula: slope = (y2 – y1)/(x2 – x1)
 To determine if two lines are parallel or perpendicular, use their slopes. Two lines are parallel if their slopes are equal. Two lines are perpendicular if the product of their slopes is 1.
 To find the equation of a line passing through a given point and parallel to another line, use the pointslope form of the equation of a line: y – y1 = m(x – x1), where m is the slope of the given line and (x1, y1) is the given point.
 To find the equation of a circle, use the standard form of the equation: (x – h)² + (y – k)² = r², where (h, k) is the center of the circle and r is the radius.
 To find the equation of a parabola, use the vertex form of the equation: y – k = a(x – h)², where (h, k) is the vertex of the parabola and a is a constant that determines the shape of the parabola.
 To find the equation of an ellipse or hyperbola, use the standard form of the equation: (x – h)²/a² + (y – k)²/b² = 1 or (y – k)²/a² – (x – h)²/b² = 1, where (h, k) is the center of the ellipse or hyperbola, and a and b are the semimajor and semiminor axes, respectively.
These tips and tricks can help you to solve problems in coordinate geometry more efficiently and accurately.
Coordinate Geometry Exerciser Questions Solution
Ex 7.1 Class 10 Maths Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (5, 7), (1, 3)
(iii) (a, b), (a, b)
(i) The distance between (2, 3) and (4, 1) is given by the distance formula:
Distance = √[(4 – 2)² + (1 – 3)²] = √[2² + (2)²] = √8 = 2√2
Therefore, the distance between (2, 3) and (4, 1) is 2√2 units.
(ii) The distance between (5, 7) and (1, 3) is given by the distance formula:
Distance = √[(1 – (5))² + (3 – 7)²] = √[4² + (4)²] = √32 = 4√2
Therefore, the distance between (5, 7) and (1, 3) is 4√2 units.
(iii) The distance between (a, b) and (a, b) is given by the distance formula:
Distance = √[(a – a)² + (b – (b))²] = √[4a² + 4b²] = 2√(a² + b²)
Therefore, the distance between (a, b) and (a, b) is 2√(a² + b²) units.
Ex 7.1 Class 10 Maths Question 2.
Find the distance between the points (0, 0) and (36, 15).
The distance between the points (0, 0) and (36, 15) is given by the distance formula:
Distance = √[(36 – 0)² + (15 – 0)²] = √[36² + 15²] = √(1296 + 225) = √1521 = 39
Therefore, the distance between the points (0, 0) and (36, 15) is 39 units.
Ex 7.1 Class 10 Maths Question 3.
Determine if the points (1, 5), (2, 3) and (2, 11) are collinear.
We can determine if the points (1, 5), (2, 3) and (2, 11) are collinear by checking if the slope between any two points is the same as the slope between the other two points.
The slope between the points (1, 5) and (2, 3) is:
m1 = (3 – 5)/(2 – 1) = 2
The slope between the points (2, 3) and (2, 11) is:
m2 = (11 – 3)/(2 – 2) = 14/4 = 7/2
The slope between the points (1, 5) and (2, 11) is:
m3 = (11 – 5)/(2 – 1) = 16/3 = 16/3
Since the slope between any two pairs of points is not equal, the points (1, 5), (2, 3) and (2, 11) are not collinear.
Ex 7.1 Class 10 Maths Question 4.
Check whether (5, 2), (6, 4) and (7, 2) are the vertices of an isosceles triangle.
To check whether the points (5, 2), (6, 4) and (7, 2) are the vertices of an isosceles triangle, we need to check if any two sides of the triangle have equal length. We can use the distance formula to calculate the length of each side of the triangle:
 Length of side 1: (5, 2) to (6, 4) d1 = √[(6 – 5)² + (4 – (2))²] = √[1² + 6²] = √37
 Length of side 2: (6, 4) to (7, 2) d2 = √[(7 – 6)² + (2 – 4)²] = √[1² + (6)²] = √37
 Length of side 3: (7, 2) to (5, 2) d3 = √[(5 – 7)² + (2 – (2))²] = √[2²] = 2
Since only one pair of sides have equal length (d1 = d2 = √37), the triangle with vertices (5, 2), (6, 4) and (7, 2) is not an isosceles triangle.
Ex 7.1 Class 10 Maths Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in the given figure. Champa and Chameli 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:
Ex 7.1 Class 10 Maths Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) (1, 2), (1, 0), (1, 2), (3, 0)
(ii) (3, 5), (3, 1), (0, 3), (1, 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) (1, 2), (1, 0), (1, 2), (3, 0):
The given points are (1, 2), (1, 0), (1, 2) and (3, 0). Plotting these points on the coordinate plane, we can see that they form a rhombus.
The reason for this is that the opposite sides of the quadrilateral are parallel and the adjacent sides are perpendicular to each other. Also, all four sides are equal in length.
(ii) (3, 5), (3, 1), (0, 3), (1, 4):
The given points are (3, 5), (3, 1), (0, 3) and (1, 4). Plotting these points on the coordinate plane, we can see that they do not form any wellknown type of quadrilateral.
(iii) (4, 5), (7, 6), (4, 3), (1, 2):
The given points are (4, 5), (7, 6), (4, 3) and (1, 2). Plotting these points on the coordinate plane, we can see that they form a trapezium.
The reason for this is that one pair of opposite sides of the quadrilateral are parallel, but the other pair of opposite sides are not parallel.
Ex 7.1 Class 10 Maths Question 7.
Find the point on the xaxis which is equidistant from (2, 5) and (2, 9).
Solution:
Let the point on the xaxis be (x, 0). Since this point is equidistant from (2, 5) and (2, 9), we can use the distance formula to write:
√[(x – 2)² + (5 – 0)²] = √[(2 – x)² + (9 – 0)²]
Simplifying this equation, we get:
(x – 2)² + 25 = (x + 2)² + 81
Expanding the squares and simplifying further, we get:
x² – 4x + 4 + 25 = x² + 4x + 4 + 81
Simplifying this equation, we get:
8x = 58
Therefore, x = 58/8 = 7.25
Hence, the point on the xaxis which is equidistant from (2, 5) and (2, 9) is (7.25, 0).
Ex 7.1 Class 10 Maths Question 8.
Find the values of y for which the distance between the points P (2, 3) and Q (10, y) is 10 units.
Solution:
We can use the distance formula to find the distance between points P (2, 3) and Q (10, y):
√[(10 – 2)² + (y – (3))²] = 10
Simplifying this equation, we get:
√[64 + (y + 3)²] = 10
Squaring both sides of the equation, we get:
64 + (y + 3)² = 100
Subtracting 64 from both sides, we get:
(y + 3)² = 36
Taking the square root of both sides, we get:
y + 3 = ±6
Solving for y, we get:
y = 3 ± 6
Therefore, the values of y for which the distance between the points P (2, 3) and Q (10, y) is 10 units are y = 9 and y = 3
Ex 7.1 Class 10 Maths Question 9.
If Q (0, 1) is equidistant from P (5, 3), and R (x, 6), find the values of x. Also, find the distances QR and PR.
Solution:
Since Q (0, 1) is equidistant from P (5, 3) and R (x, 6), we can use the distance formula to set up two equations:
d(QP) = d(QR) √[(0 – 5)² + (1 – (3))²] = √[(0 – x)² + (1 – 6)²]
Simplifying this equation, we get:
√34 = √(x² + 25)
Squaring both sides of the equation, we get:
34 = x² + 25
Solving for x, we get:
x² = 9
x = ±3
Therefore, the possible values of x are x = 3 and x = 3.
To find the distances QR and PR, we can use the distance formula:
QR = √[(0 – x)² + (1 – 6)²] = √[x² + 25] PR = √[(5 – x)² + (3 – 6)²] = √[(x – 5)² + 81]
If x = 3, then QR = √34 and PR = 5. If x = 3, then QR = √34 and PR = 7.
Therefore, the distances QR and PR are both equal to √34, and the possible values of x are x = 3 and x = 3
Ex 7.1 Class 10 Maths Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (3, 4).
Solution:
Let (x, y) be the point which is equidistant from (3, 6) and (3, 4).
Using the distance formula, we have:
√[(x – 3)² + (y – 6)²] = √[(x + 3)² + (y – 4)²]
Squaring both sides of the equation, we get:
(x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
Expanding the terms, we get:
x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16
Simplifying the equation, we get:
12x + 24 = 4y
Dividing both sides by 4, we get:
3x + 6 = y
Therefore, the relation between x and y such that the point (x, y) is equidistant from (3, 6) and (3, 4) is y = 3x + 6.
Class 10 Maths Coordinate Geometry Mind Maths
Coordinate Geometry Examples
Here are some examples of problems in coordinate geometry:
 Find the distance between the points (3, 4) and (2, 1).
Using the distance formula: distance = √[(x2 – x1)² + (y2 – y1)²] distance = √[(2 – 3)² + (1 – 4)²] distance = √[25 + 9] distance = √34 Answer: The distance between the points (3, 4) and (2, 1) is √34 units.
 Find the midpoint of the line segment with endpoints (2, 3) and (6, 9).
Using the midpoint formula: midpoint = [(x1 + x2)/2, (y1 + y2)/2] midpoint = [(2 + 6)/2, (3 + 9)/2] midpoint = [4, 6] Answer: The midpoint of the line segment with endpoints (2, 3) and (6, 9) is (4, 6).
 Find the slope of the line passing through the points (3, 2) and (5, 6).
Using the slope formula: slope = (y2 – y1)/(x2 – x1) slope = (6 – 2)/(5 – (3)) slope = 4/8 slope = 1/2 Answer: The slope of the line passing through the points (3, 2) and (5, 6) is 1/2.
 Find the equation of the line passing through the point (4, 3) and parallel to the line y = 2x + 1.
Using the pointslope form of the equation of a line: y – y1 = m(x – x1), where m is the slope of the given line and (x1, y1) is the given point.
Since the given line is parallel to the new line, they have the same slope. Therefore, the slope of the new line is also 2.
Substituting the values: y – (3) = 2(x – 4) y + 3 = 2x – 8 y = 2x – 11 Answer: The equation of the line passing through the point (4, 3) and parallel to the line y = 2x + 1 is y = 2x – 11.
 Find the equation of the circle with center (2, 5) and radius 3.
Using the standard form of the equation of a circle: (x – h)² + (y – k)² = r², where (h, k) is the center of the circle and r is the radius.
Substituting the values: (x – 2)² + (y – (5))² = 3² (x – 2)² + (y + 5)² = 9 Answer: The equation of the circle with center (2, 5) and radius 3 is (x – 2)² + (y + 5)² = 9.
Coordinate of a Point in XY – Plane
The coordinates of a point in the XYplane are a pair of numbers (x,y) that represent the horizontal and vertical distances of the point from the origin (0,0), respectively.
The xcoordinate is the horizontal distance from the origin to the point, and it is represented by the first number in the ordered pair (x,y). The ycoordinate is the vertical distance from the origin to the point, and it is represented by the second number in the ordered pair (x,y).
For example, consider the point P with coordinates (3,4) in the XYplane. The xcoordinate of P is 3, which means that the point is located 3 units to the right of the origin. The ycoordinate of P is 4, which means that the point is located 4 units above the origin.
Note that the order of the coordinates is important, as (x,y) and (y,x) represent different points in the plane.
Signconventions in the XYPlane
Signconventions in the XYplane refer to the convention used to assign signs to the coordinates of points in the plane. The convention used is based on the Cartesian coordinate system, where the Xaxis is horizontal and the Yaxis is vertical, and the origin (0,0) is the point where they intersect.
The sign of the Xcoordinate depends on whether the point is to the left or right of the Yaxis. Points to the right of the Yaxis have positive Xcoordinates, while points to the left have negative Xcoordinates.
The sign of the Ycoordinate depends on whether the point is above or below the Xaxis. Points above the Xaxis have positive Ycoordinates, while points below have negative Ycoordinates.
For example, consider the point P with coordinates (3, 4) in the XYplane. The Xcoordinate is negative, which means that the point is located to the left of the Yaxis. The Ycoordinate is positive, which means that the point is located above the Xaxis.
Similarly, if we have a point Q with coordinates (5, 2), the Xcoordinate is positive, which means that the point is located to the right of the Yaxis. The Ycoordinate is negative, which means that the point is located below the Xaxis.
By convention, the positive Xaxis points to the right, and the positive Yaxis points upward, as shown below:

 (+y)


(x)  (+x) 
 (y)

Distance Formula
The distance formula is used to find the distance between two points in the XYplane. It is derived from the Pythagorean theorem and can be applied to any two points (x1,y1) and (x2,y2) in the plane.
The distance formula is given by:
d = √[(x2 – x1)^2 + (y2 – y1)^2]
Where d is the distance between the two points (x1,y1) and (x2,y2).
To use the formula, simply substitute the coordinates of the two points into the formula and simplify the expression. The result will be the distance between the two points.
For example, let’s find the distance between the points P(2,3) and Q(4,6):
d = √[(4 – 2)^2 + (6 – 3)^2]
d = √[2^2 + 3^2]
d = √(4 + 9)
d = √13
Therefore, the distance between the points P(2,3) and Q(4,6) is √13 units.
Section Formula
The section formula is a formula used to find the coordinates of a point that lies on a line segment joining two given points in the XYplane. The formula gives the coordinates of the point that divides the line segment into two parts in a given ratio.
Suppose we have two points A(x1,y1) and B(x2,y2) in the XYplane, and we want to find the coordinates of a point P that lies on the line segment AB, such that the ratio of the distance of P from A to the distance of P from B is given by m:n, where m and n are positive integers.
Then, the coordinates of the point P can be found using the section formula:
Px = [(nx2) + (mx1)] / (m + n)
Py = [(ny2) + (my1)] / (m + n)
where Px and Py are the coordinates of the point P, and m and n are the given ratio of the distances.
Note that if m+n=1, then the point P lies on the line segment AB and divides it in the ratio m:n.
For example, let’s find the coordinates of a point P that divides the line segment joining the points A(2,4) and B(6,8) in the ratio 2:3.
Using the section formula, we have:
Px = [(36) + (22)] / (2 + 3) = 4
Py = [(38) + (24)] / (2 + 3) = 6
Therefore, the coordinates of the point P are (4,6).
Area of a Triangle
The formula to find the area of a triangle in the XYplane is given by:
Area = 1/2 * base * height
where the base and height are the perpendicular distance between any two sides of the triangle.
There are different ways to find the base and height of a triangle, depending on the information given. Here are three common methods:
 If the coordinates of the vertices of the triangle are given, we can use the distance formula to find the length of each side of the triangle, and then use the formula for the area of a triangle.
 If the equation of the three sides of the triangle is given, we can use the intersection point of two sides to find the height, and then use the formula for the area of a triangle.
 If the coordinates of the vertices and the lengths of two sides of the triangle are given, we can use the formula for the area of a triangle given by Heron’s formula:
Area = √[s(sa)(sb)(sc)]
where s is the semiperimeter of the triangle, given by:
s = (a + b + c)/2
and a, b, and c are the lengths of the sides of the triangle.
For example, let’s find the area of a triangle with vertices A(2,4), B(6,8) and C(4,10):
 Using the distance formula, we find that AB = BC = 2√2 and AC = 2√5. Then, using the formula for the area of a triangle, we get:
Base = AB = BC = 2√2
Height = the perpendicular distance from vertex A to the line containing BC.
Using the equation of the line BC, we can find that its slope is 1, so the slope of the perpendicular line from A is 1. The equation of the line containing the perpendicular is y = x + 6, and the intersection point of this line with BC is (4,6). So, the height of the triangle is the distance between A and (4,6), which is 2√2. Therefore, the area of the triangle is:
Area = 1/2 * base * height = 1/2 * 2√2 * 2√2 = 2
 If the equation of the sides of the triangle are given as y = x, y = x + 14 and x = 2, then we can see that the intersection point of y = x and y = x + 14 is (7,7). So, the height of the triangle is the perpendicular distance from vertex C to the line containing AB, which is the line x = 2. Therefore, the height is 2. The base is the distance between A(2,4) and B(6,8), which is 2√2. So, the area of the triangle is:
Area = 1/2 * base * height = 1/2 * 2√2 * 2 = 2√2
 Using Heron’s formula, we find that the semiperimeter of the triangle is s = (AB + BC + AC)/2 = 2√2 + 2√2 + 2√5 / 2 = √2 + √5. Then, we have:
Area = √[s(sa)(sb)(sc)] = √[(√2 + √5)(√2 + √5 – 2√2)(√2 + √5 – 2√2)(√2 + √5
FAQs About NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Here are some frequently asked questions about coordinate geometry:
 What is the Cartesian plane?
The Cartesian plane, also known as the coordinate plane, is a twodimensional plane that is used to graph mathematical equations and geometric figures. It consists of two perpendicular number lines, the xaxis and yaxis, that intersect at the origin (0,0).
 What is a slope?
Slope is a measure of the steepness of a line. It is defined as the change in y divided by the change in x between any two points on the line. In other words, it is the ratio of the vertical change to the horizontal change. The slope of a line is denoted by the letter m.
 What is the distance formula?
The distance formula is a formula used to find the distance between two points in a coordinate plane. It is derived from the Pythagorean theorem and is given by:
distance = √[(x2 – x1)² + (y2 – y1)²]
where (x1, y1) and (x2, y2) are the coordinates of the two points.
 What is the midpoint formula?
The midpoint formula is a formula used to find the midpoint of a line segment in a coordinate plane. It is given by:
midpoint = [(x1 + x2)/2, (y1 + y2)/2]
where (x1, y1) and (x2, y2) are the coordinates of the endpoints of the line segment.
 What is the equation of a circle?
The equation of a circle is a formula used to graph a circle on a coordinate plane. It is given by:
(x – h)² + (y – k)² = r²
where (h, k) is the center of the circle and r is the radius.
 What is the pointslope form of the equation of a line?
The pointslope form of the equation of a line is given by:
y – y1 = m(x – x1)
where m is the slope of the line and (x1, y1) is a point on the line. This form is useful for finding the equation of a line given the slope and a point on the line.
 What is the slopeintercept form of the equation of a line?
The slopeintercept form of the equation of a line is given by:
y = mx + b
where m is the slope of the line and b is the yintercept (the point where the line intersects the yaxis). This form is useful for graphing a line or finding the equation of a line given the slope and yintercept.
 What is a system of equations?
A system of equations is a set of two or more equations that are to be solved simultaneously. In coordinate geometry, systems of equations can be used to find the intersection of two lines or the intersection of a line and a curve.
 What is a linear equation?
A linear equation is an equation of the form:
y = mx + b
where m is the slope and b is the yintercept. Linear equations describe lines with a constant slope and are used to model many realworld situations.
Here are some frequently asked questions about NCERT Solutions for Class 10 Maths Chapter 7:
 What is NCERT Solutions for Class 10 Maths Chapter 7?
NCERT Solutions for Class 10 Maths Chapter 7 is a set of solutions to the exercises given in Chapter 7 “Coordinate Geometry” of the NCERT Class 10 Mathematics textbook. These solutions provide stepbystep explanations and solutions to all the questions in the chapter.
 What topics are covered in NCERT Solutions for Class 10 Maths Chapter 7?
NCERT Solutions for Class 10 Maths Chapter 7 covers topics such as introduction to coordinate geometry, plotting a point in the coordinate plane, the distance formula, the section formula, and the area of a triangle.
 Why are NCERT Solutions for Class 10 Maths Chapter 7 important?
NCERT Solutions for Class 10 Maths Chapter 7 are important because they help students to understand and solve problems related to coordinate geometry. These solutions are designed to provide a clear understanding of the concepts and develop problemsolving skills.
 How can I access NCERT Solutions for Class 10 Maths Chapter 7?
NCERT Solutions for Class 10 Maths Chapter 7 can be accessed online on various educational websites, including the official NCERT website. These solutions are also available in the form of PDF files that can be downloaded and used offline.
 Are NCERT Solutions for Class 10 Maths Chapter 7 sufficient for exam preparation?
NCERT Solutions for Class 10 Maths Chapter 7 are designed to cover all the concepts and problems related to coordinate geometry that are covered in the Class 10 Mathematics textbook. These solutions are sufficient for understanding the concepts and solving problems related to coordinate geometry. However, students are advised to practice additional problems and take mock tests to prepare well for the exams.