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Coordinate Geometry is an integral part of Mathematics. This chapter can be found in NCERT class 10 Maths chapter 7 and it deals with geometrical figures and coordinates like the x-axis and y-axis to locate the exact position of the points in various types of planes. This chapter consists of an introduction phase into coordinate geometry and gradually makes students aware of topics like distance formula and section formula.
NCERT class 10 maths ch 7 solutions can be often challenging. Students often get confused during the class 10 chapter 7 NCERT solutions. This article is for every student to understand all the Chapter 7 Maths class 10 NCERT solutions. Students can also read Chapter 7 maths class 10 pdf solutions for a quick revision. If students want NCERT solutions for other classes and subjects, they can get the solutions by clicking on the above link. Students must refer to the NCERT Class 10 Maths books and read all the topics without fail. NCERT class 10 maths ch 7 solutions
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Distance Formulae:
For a line defined by two points A(x1, y1) and B(x2, y2), the distance between these points can be calculated using the formula:
Distance AB =
Section Formula:
When a point P divides a line AB, with A(x1, y1) and B(x2, y2) as endpoints, in a ratio of m:n, the coordinates of point P can be found using the formula:
Point P =
Midpoint Formula:
The midpoint of a line AB, defined by A(x1, y1) and B(x2, y2), can be determined using the following formula:
Midpoint P =
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Area of a Triangle:
Consider a triangle formed by points A(x1, y1), B(x2, y2), and C(x3, y3). The area of this triangle can be calculated using the formula:
Area ∆ABC =
Free download NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry PDF for CBSE Exam.
Class 10 Maths ch 7 solutions Exercise: 7.1
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 the given town location
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,
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
So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangles 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:
The distance between two points
Hence,
And the lengths of diagonals:
So, here it can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are also having the same length.
Therefore, quadrilateral ABCD is a square and Champa is saying right.
Let the given points
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
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 a 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
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 x-axis which is equidistant from (2, –5) and (–2, 9).
Let the point which is equidistant from
Then, we have,
Distance AX:
and Distance BX
According to the question, these distances are equal in length.
Hence, we have,
Squaring both sides we get,
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
The distance formula :
So, given
After squaring both sides,
Therefore, the values are
Answer:
Given
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.
Answer:
Let the point
Then, the distances
Squaring both sides: we obtain
Thus, the relation is
Class 10 Maths ch 7 solutions Exercise: 7.2
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
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
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
Niharika posted the green flag at the distance P, i.e.,
Therefore, the coordinates of this point
Similarly, Preet posted red flag at
Therefore, the coordinates of this point Q are
The distance
and the point at which Rashmi should post her Blue Flag is the mid-point 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
Let the point on the x-axis be
Then, we have
Section formula:
Hence, the value of k will be:
Therefore, the x-axis divides the line in the ratio
Putting the value of
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 mid-point of AC and BD.
The coordinates of the point O when it is the mid-point of AC.
The coordinates of the point O when it is the mid-point of BD.
Since both coordinates are of the same point O.
Therefore,
Or,
As the centre point
Then, the coordinates of point A will be
Given point
Therefore,
Therefore, the coordinates of A are
From the figure:
As
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 Points 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 be:
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,
The NCERT class 10 CoordMaths Geometry chapter includes several key features to aid students in their understanding and practice of the material provided in Class 10 Chapter 7 Example 9 and answer all exercise questions given in the NCERT textbook
Class 10 Chapter 7 solutions include diagrams and alternate methods to help students understand the concepts.
Chapter 7 Maths Class 10 pdf solutions help to familiarise students with important formulas and standards.
Class 10 Chapter 7 ncert solutions include a variety of examples to help students connect the geometry and numerical concepts to real-life scenarios.
Memorise the formula to find out some directly asked questions.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | Coordinate Geometry |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 |
Students can follow the steps provided below to use the class 10 maths ncert solutions chapter 7 in the best way possible.
Coordinate geometry is not only a useful chapter in class 10 but also for higher studies and competitive exams. If your basic concepts are clear right now, then later you will not face any difficulties solving coordinate geometry questions. These are some further benefits listed below.
Good luck and keep on reading.
In class 10 Maths Chapter 7 following topics of coordinate geometry are covered.
Let the two points in a Cartesian plane be
So, their distance =
If a point
The formula of the area of a triangle can be used to prove that three points
If the area of the triangle formed by these three points is zero(0), then the points are collinear.
Area of a triangle
Let the two points in a Cartesian plane be
Then their midpoint
Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
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After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
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Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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