coordinate geometry class 10 1 mark questions

(2014OD) Question 18. Area of ∆ADC Area of ∆ABD = Area of ∆ADC Solution: ⇒ 2k = 16 We get, k = 2, \(\frac{1}{2}\), Question 52. AD : AB = 1 : 3; AE : AC = 1 : 3 Let P(2, 4), A(5, k) and B(k, 7). (A) 2. Find the distance of the point (-3, 4) from the x-axis. ∴ AP = BP …[Given 2k = 6 ∴ k = \(\frac{6}{2}\) = 3, Question 25. ⇒ k2 + 4k + 3 = 0 …[Dividing by 3 AB = 10 units … [Given ... 1 and 4 (b) 1 and 5 ... (1, 3) (b) (3, 2) (c) (2, 1) (d) (3, 1) 10.Find the coordinate of the centroid of a triangle whose vertices are (0, 5) (7, 11) and (6, 0) 11. Question 24. ∴ OA = OB OA2 = OB2 …[Squaring both sides ∆ABC is an isosceles right angled triangle. Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) & D(2,6), are equal and bisect each other. Given: Area of ∆ = 5 sq. (2013OD) 4 Marks Questions. ar(∆ABD) = 12 sq.units The distance of the point P (2, 3) from the x-axis is. Question 35. Question 41. (2015D) = \(\frac{1}{2}\) [1(3 + 3) + 2(-3 + 2) + (-4)(-2 – 3)] = \(\frac{1}{2}\) (2+ + 2+ + 4 – 46 – 12] = \(\frac{1}{2}\) [-8] = -4 Let A (7, 10), B(-2, 5), C(3, -4) be the vertices of a triangle. Q1: What will the formula calculate the distance between two points that lie on the x-axis? ⇒ -2y – 4 = 0 ⇒ -2y = 4 MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. CBSE Class 10 Math 1 Mark Questions Coordinate Geometry. (2011D) Calculate the area of ∆ADE and compare it with area of ∆ABC. From (i) and (ii), units ⇒ k2 – 6k + 5 = 0 …[Dividing by ⇒ -2b – 2 – 2a + 4 – 4b = 0 .. Area of ∆ABD ∴ Area of ∆ABC Similar to Question 19, Page 102. Find the values of y. ⇒ 6k + 3 + 7k + 3 – 9k – 9 = 0 The three vertices of a parallelogram ABCD are A(3, 4), B(-1, -3) and C(-6, 2). 12 = (-3k + k2 – 3 + k – 4k – 4 + 28] PQ2 = PR2 … [Squaring both sides Solution: Question 65. Let A(x, y), B(1, 2), C(2, 1). ⇒ 2x(-a – b + a – b) = 2y(-a – b + b – a) Question 9. Mid-point of AB = Coordinates of D, Question 50. (k – 4 + k – 2) (k – 4 – k + 2) = 0 PA = PB … [Given ⇒ y2 – 7y + y – 7 = 0 ⇒ y2 – 6y – 7 = 0 …[Dividing both sides by 7 ⇒ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0 (C) 1. The area of a triangle is 5 sq. ∴ Area of ∆ABC = 0 ⇒ x = 2 + 4 = 6 or x = 2 – 4 = -2. Point P(x, 4) lies on the line segment joining the points A(-5, 8) and B(4, -10). units …(i) (2013D) ∴ Diagonals also bisect each other. Let P and Q be the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A. 10p + 60 = ± 30 Points A(-1, y) and B(5, 7) lie on a circle with centre 0(2, -3y). Similar to Question 56, Page 109. Let the point P be (2k, k), Q(2,-5), R(-3, 6) Also find the coordinates of the point of division. Solution: (2013OD) Find the area of the triangle ABC with A(1, 4) and mid-points of sides through A being (2, -1) and (0, -1). Area of ∆ABC Let A (3, 3), B (6, y), C (x, 7) and D (5, 6). Solution: Solution: ⇒ \(\frac{1}{2}\) × AB × CP = 12 …[Area of ∆ = 1/2 × Base × Height, Question 69. ⇒ [(k + 1) (2k + 3 – 5k) + 3k (5k – 2k) + (5k – 1) (2k – 2k – 3)] = 0 6 = \(\frac{1}{2}\)[(k + 1)(-3 + k) + 4(-k – 1) + 7(1 + 3)] (2012OD) If the points A(x, y), B(3, 6) and C(-3, 4) are collinear, show that x – 3y + 15 = 0. = \(\frac{1}{2}\)[(-3)(4 + 1) + (-2){-1 – (-1)} +4{-1 – (- 4}}] (2015OD) If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex. Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B(2 + \(\sqrt{3}\), 5) and C(2,6). (2013OD) Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). = \(\frac{56}{2}\) = 28 sq. Solution: (2011D) All types of questions are solved for all topics. Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry Coordinate Geometry Class 10 Important Questions Very Short Answer (1 Mark) Question 1. Solution: Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5). Practice Questions For Class 10 Maths Chapter 7 Coordinate Geometry. Solution: ⇒ [(a + b) – x]2 + [(b a) – y)2 = [(a – b) – x]2 + [(a + b) – y]2 = \(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = \(\frac{1}{2}\)(-8 + 32 – 30] = \(\frac{1}{2}\) [-6] = -3 = \(\frac{1}{2}\)[15 + 20] = \(\frac{35}{2}\) sq. units, Question 67. The x-coordinate of a point P is twice its y-coordinate. x(1) + 21 + 42 = 0 But area of A cannot be negative. 16 + 64 – 16y + y2 = 36 + 36 + y2 – 12y p – 7 + 40 + 27 + 9p = ±30 Find the area of the triangle whose vertices are (1, 2), (3, 7) and (5, 3). Solution: Practising CoolGyan.Org's Class 10 Maths Chapterwise Important Questions with solutions … Solution: AC = BD = 10 Solution: PA = PB … [Given Find the values of a, if the circle passes through the point (11, –9) and has a diameter 10√ 2 units. Solution: Question 37. (2015D) y = 7 or y = -1 ⇒ p2 – 4p + 4 – p2 = 0 = \(\frac{1}{2}\) [2(5 – 6) + (2 + \(\sqrt{3}\))(6 – 4) + 2(4 – 5)) Solution: Solution: Question 63. = \(\frac{1}{2}\)[4(6 – 5) + 7(5 – 2) + 4(2 – 6)) Solution: = \(\frac{1}{2}\)(4 + 21 – 16) = \(\frac{9}{2}\) sq.units …(i) ⇒ (k – 1 – 3)2 + (2 – k)2 = (k – 1 – k)2 + (2 – 5)2 2k – 6 = 0 ∴ x – 3y + 15 = 0 … [Dividing both sides by 2, Question 51. Solution: k2 + 3k + k + 3 = 0 = \(\frac{1}{2}\) × AB × DM = 12 sq.units …[Area of ∆ = 1/2 × Base × Altitude A(4, -6), B(3, -2) and C(5, 2) are the vertices of a ∆ABC and AD is its median. \(\frac{1}{2}\) [1(p – 7) + 4(7 + 3)) + (-9)(-3 – p)] = ±15 (2012OD) Solution: B(-3, 0), A (-3, 4) Question 2. Similar to Question 37, Page 105. AP = \(\sqrt{(2+2)^{2}+(2-k)^{2}}\) (AC)2 = (6 + 2)2 + (7 – 3)2 = 82 + 42 = 80 ⇒ (k – 5) (k – 1) = 0 Let A (-2, 3), B(8,3), C(6, 7). (i) Taking y = 7. Solution: Prove that the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus ABCD. x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 …[∵ Points are collinear unit Free PDF Download - Best collection of CBSE topper Notes, Important Questions, Sample papers and NCERT Solutions for CBSE Class 10 Math Coordinate Geometry. B(-3, 0), A (-3, 4), Question 2. 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(2014OD) units. Solution: \(P\left(\frac{6-8}{3+4}, \frac{-12-8}{3+4}\right) \Rightarrow\left(\frac{-2}{7}, \frac{-20}{7}\right)\). Question 79. Find the coordinates of fourth vertex. 12 = [k2 – 6k + 21] Coordinates of P are: Question 12. AB = BC = CD = DA …[All four sides are equal units, Question 68. We know that at y-axis coordinates of points are (0, y) and at x-axis (x, 0). Here you can get Class 10 Important Questions Maths based on NCERT Text book for Class X. Maths Class 10 Important Questions are very helpful to score high marks in board exams. (2015OD) ∴ \(\frac{A P}{A B}=\frac{3 K}{4 K}=\frac{3}{4}\) = \(\frac{1}{2}\) [t(t + 2 – t) + (t + 2)(t – (t – 2)) +(t + 3)((t – 2) – (t + 2))]