Question 72. 10p + 60 = ± 30 ⇒ 2 ± 4 = x Editing Exercise Exercise 1: England has been long famous for her dramatist, a)…………… the greatest of who was William Shakespeare. Solution: Question 45. PA2 = QA2 … [Squaring both sides Question 77. Find the value of p. (2015OD) We have A(4, 3), B(-1, y) and C(3, 4). Two of its vertices are (2, 1) and (3,-2). Find the ratio in which the y-axis divides the line segment joining the points (-4, -6) and (10, 12). = \(\frac{1}{2}\)(4 + 21 – 16) = \(\frac{9}{2}\) sq.units …(i) (2016OD) B(-3, 0), A (-3, 4), Question 2. ∴ Required ratio = 1 : 5. Given points are P(5, 4), Q(7, k) and R(9, -2). ⇒ (k – 5) (k – 1) = 0 = \(\frac{1}{2}\)[9+ 0 + 12] = 21 sq. Solution: ∴ Taking +ve sign, p = \(\frac{-60+30}{10}=\frac{-30}{10}\) = -3 ⇒ k – 2 = 0 or 2k – 1 = 0 ⇒ [3k – 3k2 + 3 – 3k + 9k2 – 15k + 3) = 0 (2012OD) Free PDF Download - Best collection of CBSE topper Notes, Important Questions, Sample papers and NCERT Solutions for CBSE Class 10 Math Coordinate Geometry. Coordinate Geometry (1 Mark Questions and Answers), IP Univ BCA, MCA, BBA Study Notes, Q Papers, CBSE Class 10/9/8 - English - Reading Comprehension (Unseen Passage) (Set-14)(#eduvictors)(#readingComprehension), CBSE Class 10 - Biology - Chapter: Life Processes - Assertion Reasoning Type Questions (#eduvictors)(#cbsenotes), CBSE Class 9 - English Grammar - Dialogue Completion, CBSE Class 10 - Chemistry - Assertion Reason Based Questions (#classs10Chemistry)(#eduvictors), English Grammar - Editing Exercise-1 (#eduvictors)(#cbsenotes). Question 78. = \(\frac{1}{2}\) [t(t + 2 – t) + (t + 2)(t – (t – 2)) +(t + 3)((t – 2) – (t + 2))] Question 4. Also find the value of y. Question 23. (6 – 3)2 + (0 – 0)2 = (x – 6)2 + (1 – 0)2 = (x – 3)2 + (y – 0)2 y – 5 = y + 3 y – 5 = -y – 3 Area of ∆ADC Similar to Question 69, Page 112. 4k2 + 4 – 8k + k2 + 10k + 25 = 4k2 + 9 + 12k + k2 – 12k + 36 You can also get complete NCERT solutions and Sample papers. (B) 3. Solution: Question 5. Solution: ⇒ 2k2 – 4k – 1k + 2 = 0 Question 7. CBSE Class 10 Maths Chapter-7 Coordinate Geometry – Free PDF Download. ⇒ -2bx = – 2ay ⇒ [(k + 1) (2k + 3 – 5k) + 3k (5k – 2k) + (5k – 1) (2k – 2k – 3)] = 0 Question 13. (2011OD) ⇒ (a + b)2 + x2 – 2(a + b)x + (b – a)2 + y2 – 2(b – a)y = (a – b)2 + x2 – 2(a – b)x + (a + b)2 + y2 – 2(a + b)y …[∵ (a – b) 2 = (b – a)2 ⇒ 2k + 29 = 45 = (4)2 + (3)2 Solution: Taking -ve sign, p = \(\frac{-60-30}{10}=\frac{-90}{10}\) = -9. ⇒ y2 – 7y + y – 7 = 0 (2013D) = \(\frac{1}{2}\) [x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)] 2k2 – 12k + 10 = 0 Find the area of the triangle whose vertices are (1, 2), (3, 7) and (5, 3). Prove that the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus ABCD. Let A (3, 3), B (6, y), C (x, 7) and D (5, 6). ⇒ -2[b – (-1)] + a(-1 – 1) + 4(1 – b) = 0 (2015D) = \(\frac{1}{2}\) [4(-2 – 0) + 3(0 + 6) + 4(-6 + 2)] ⇒ 14 + 4c + c = 4 For the triangle ABC formed by the points A(4, -6), B(3,-2) and C(5, 2), verify that median divides the triangle into two triangles of equal area. ⇒ 6k2 – 15k + 6 = 0 ∴ AP = BP …[Given Similar to Question 26, Page 103. ∴ Point P (0, 2). ⇒ (-4)2 + (y2 – 8y + 16) Question 73. ∴ b = 3, c = -2. Area of ABCD (||gm) = 2(Area of ∆ABC) Since diagonal of a ||gm divides it into two equal areas. (2011D) ⇒ x = 2 + 4 = 6 or x = 2 – 4 = -2. Find the area of the triangle ABC with A(1, 4) and mid-points of sides through A being (2, -1) and (0, -1). x(-4 – (-5)) + (-3)(-5 – 2) + 7(2 – (-4)) = 0 Question 30. (2013OD) Solution: Since, the diagonals of a parallelogram bisect each other. Solution: (2014OD) ⇒ k(k – 3) – 3(k – 3) = 0 = Solution: Question 54. If (2, -5) is the mid point of PQ, then find the coordinates of P and Question (2017OD) Solution: If P(2, 4) is equidistant from Q(7, 0) and R(x, 9), find the values of x. PA2 = PB2 … [Squaring both sides If the points A(x, y), B(3, 6) and C(-3, 4) are collinear, show that x – 3y + 15 = 0. Mid-point of AB = Coordinates of D, Question 50. If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b), prove that bx = ay. Solution: Question 43. ∴ Area of ∆ABC 9 – 8 = x or 9+ 8 = x A(x, y), B(3, 6), C(-3, 4) are collinear. (2013D) (2014OD0 If A(-3,5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area. ⇒ -2(a + b)x + 2(a – b)x = -2(a + b)y + 2(b – a)y Similar to Question 24, Page 103. (D) 5. Solution: ⇒ b = 7 + 2c AB2 = BC2 = AC2 …[Squaring throughout AD : AB = 1 : 3; AE : AC = 1 : 3 Area of ∆ABD = Area of ∆ADC = \(\frac{1}{2}\) (2+ + 2+ + 4 – 46 – 12] = \(\frac{1}{2}\) [-8] = -4 8 = 4y ⇒ y = 2 Solution: Solution: Taking +ve sign then, as we know 6 = \(\frac{1}{2}\)[(k + 1)(-3 + k) + 4(-k – 1) + 7(1 + 3)] 2k – 1 – (k + 9) = 2k + 7 – (2k – 1) (9 – x) = ± 8 …[Taking square-root on both sides Question 57. The x-coordinate of a point P is twice its y-coordinate. ⇒ bx = ay (Hence proved), Question 28. = 2x – 25 + 5y – 4y + 28 = 0 AC2 = AB2 + BC2 …[Pythagoras’ theorem Solution: Solution: ∴ x(2 – 1) + 1(1 – y) + 2(y – 2) = ±12 (2015OD) k2 + 3k + k + 3 = 0 Inte... CBSE Class 10 - Chemistry - Assertion Reason Based Questions A ssertion and Reasoning based questions are bit tricky. Answer: B ( 6, 7, 8, 9, 10, Math, CBSE- Coordinate Geometry 1! 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