NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science, Kerala Syllabus 9th Standard Physics Solutions Guide, Kerala Syllabus 9th Standard Biology Solutions Guide, NCERT Solutions for class 10 Maths in Hindi, Ekarthak Shabd in Hindi | एकार्थक शब्द की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Tatsam Tadbhav Shabd in Hindi | तत्सम तद्भव शब्द की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Shabd Vichar in Hindi | शब्द विचार की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Kriya Visheshan in Hindi | क्रिया विशेषण की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Paryayvachi Shabd in Hindi | पर्यायवाची शब्द की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Anek Shabdon Ke Liye Ek Shabd in Hindi | अनेक शब्दों के लिए एक शब्द की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Chhand in Hindi | छन्द की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Anekarthi Shabd in Hindi | एकार्थक शब्द की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Vilom Shabd in Hindi | विलोम शब्द (Antonyms) की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Samvaad Lekhn in Hindi(Dialogue Letter)-संवाद-लेखन, Vismayadibodhak in Hindi | विस्मयादिबोधक (Interjection) की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Samuchchay Bodhak in Hindi | समुच्चयबोधक (Conjunction) की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Sambandh Bodhak in Hindi | संबंधबोधक (Preposition) की परिभाषा एवं उनके भेद और उदाहरण (हिन्दी व्याकरण), Patra lekhan in Hindi – पत्र-लेखन (Letter-Writing) – Hindi Grammar, ‎हिन्दी निबंध – Essay in Hindi Writing- Hindi Nibandh. Solution: Question 17: Also, find the coordinates of the point of division. = (2(5)2) 2016. x1 = 3 , y1 = 2 If D(,) E(7,3) and  F(,) are the mid-points of sides of ΔABC,then find the area of the ΔABC. Hence, the required intersection point is (0, 4). Hence, the required value of p is ± 4. Solution: So, the coordinates of the point M of division A (- 4, – 6) and B(-1,7) are. True Here, we see that, OQ > OP Solution: Check - Coordinate Geometry - Class 10. The area of a triangle with vertices (a, b + c) , (b, c + a) and (c, a + b) is Class 10 math (India) Unit: Coordinate geometry. Also, diagonals are equal i.e., AC = BD If so, what should be his position? Hence, the fourth vertex of parallelogram is D s (x4, y4) s (0,-1). Solution: (a) We know that, the perpendicular bisector of the any line segment divides the^jjpe segment into two equal parts i.e., the perpendicular bisector of the line segment always passes through the mid-point of the line segment. Question 11: i.e.,                        (Hypotenuse)2 =( Base)2 + (Perpendicular)2 Question 15: Now, using distance formula between two points, If the distance between the points (4, p) and (1, 0) is 5, then the value of pis Question 7: (iii) Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ:QE = 2:1 and CR:RF = 2:1 The area of a triangle with vertices A(3,0), B(7, 0) and C(8, 4) is Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6). (a)AP =  AB                              (b) AP = PB                            (c)PB =  AB                      (d)AP = AB Since, the points A(k + 1,2k), B(3k, 2k + 3) and C(5k -1, 5k) are collinear. Question 5: Question 16: All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. Question 5: How many such points are there? A circle has its centre at the origin and a point P (5, 0) lies on it. Solution: Hence, the required distance of AB is 2√61. = 7.07 To know the type of triangle formed by the points Q, A and B. Question 1: He has been teaching from the past 9 years. (a) a = b                      (b) a = 2b                  (c) 2a = b                   (d) a = – b Let A ≡ (x1,y1) s (5,1), B = (x2, y2) = (- 2, – 3), C s (x3, y3) = (8,2m) Question 2: Point P(0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(-l, 1) and B(3, 3). The points A(x1, y1), B(x2 y2) and C(x3, y3) are the vertices of ΔABC. x1 = 2 , y1 = 3 Solution: CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 10 Maths chapter 7 Coordinate Geometry solved by expert teachers as per NCERT (CBSE) book guidelines. The fourth vertex D of a parallelogram ABCD whose three vertices are A(- 2, 3), B(6, 7) and C(8, 3) is Short Answer Type Questions I [2 Marks] Question 1. By distance formula. Students who are preparing for their Class 10 exams must go through Important Questions for Class 10 Math Chapter 7 Coordinate Geometry. A, B, C and D are the positions of four students as shown in figure. (b) v Distance between the points (x1, y1) and (x2, y2), Question 3: Here, we see that the sides AB = CD and BC = DA Question 18: Question 4: Now, using distance formula between two points, By section formula, CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 10 Maths chapter 7 Coordinate Geometry solved by expert teachers as per NCERT (CBSE) book guidelines. Solution: Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A (- 5, – 2) and B (4, – 2). Solution: Question 8: (a) Let the coordinate of the point which is equidistant from the three vertices 0(0, 0), A(0,2y) and B(2x, 0) is P(h,k). Then, x = Perpendicular distance from Y-axis Hence, required condition is AP =. Find the coordinates of the points of division. Solution: Hence, the required area of ΔABC is 6 sq units. Learn Science with Notes and NCERT Solutions. Teachoo is free. If the point A (2,7) lies on the perpendicular bisector of the line segment, then the point A satisfy the equation of perpendicular bisector. Hence, the points A (-1, – 2), B (4, 3), C (2, 5)and D (- 3 0)form a rectangle. Hence, both the triangles are similar. Question 20: HOTS Questions , Class 10, Math, CBSE- Coordinate Geometry. Hence, there are two points lies on the axis, which are (5, 0) and (9, 0), have 2V5 distance from the point (7, – 4). Solution: Question 13: Solution: Question 11: Hence, the required ratio is 1 : 5. The distance between the points (0, 5) and (- 5, 0) is Solution: (a) (0,13)                  (b) (0,-13)                 (c) (0,12)                  (d) (13,0) Question 11: Question 1: Given that, the vertices of triangles (b) Let the vertices of a triangle are, A ≡ (x1, y1) ≡ (a, b + c) Find the coordinates of the point R on the line segment joining the points P(- 1, 3) and Q(2, 5) such that If the area of triangle formed by the points (x1,y2), (x2, y2) and (x3, y3) is zero, then the points are collinear, Hence, L and M are the same points. We find the equation of perpendicular bisector of the line segment AS. (a)-4                           (b) -12                        (c) 12                        (d) -6 In what ratio does the X-axis divide the line segment joining the points (- 4, – 6) and (- 1, 7)? Question 10: Question 7: According to the question, Solution: x2 = 2 , y2 = 3 If P(9a -2, – b) divides line segment joining A(3a + 1,-3) and B(8a, 5) in the ratio 3 : 1, then find the values of a and b. Also, The point P(5, – 3) is one of the two points of trisection of line segment joining the points A(7, – 2) and B(1, – 5). Calculating PR Solution: Check - Coordinate Geometry - Class 10, Example 1 (b) Let the fourth vertex of parallelogram, D≡(x4 ,y4) and L, M be the middle points of AC and BD, respectively, Now,perimeter of ΔAOB=Sum of the length of all its sides = d(AO) + d(OB) + d(AB) ∴ Distance between the points (x,, y,) and (x2, y2), If the distance between the centre and any point is equal to the radius, then we say that point lie on the circle. which is not equal to the radius of the circle. To find the type of quadrilateral, we find the length of all four sides as well as two diagonals and see whatever condition of quadrilateral is satisfy by these sides as well as diagonals. x2 = 2 , y2 = 3 Solution: Solution: (i) We know that, the median bisect the line segment into two equal parts i.e., here D is the mid-point of BC. The solutions are divided into parts so that they can be easily understood by the students. Solution: Since, (a, b) is the mid-point of line segment AB. Solution: Find the value of k. Question 10: (c) If A (2,7) lies on perpendicular bisector of P(6,5) and Q (0, – 4), then AP=AQ Distance between the centre C(2a, a – 7) and the point P( 11, – 9), which lie on the circle = Radius of circle A line intersects the y-axis and X-axis at the points P and Q, respectively.