In fig., the vertices of AABC are A(4, 6), B(l, 5) and C(7, 2). Solution: Question 12. ∴ (x + 2)2 + (y – 3)2 = (x – 3)2 + (y – 2)2 ∴ AP = BP ⇒ AP2 = BP2 ∴ ∆PQR is an equilateral triangle. y1 = 4, y2 = -2, y3 = 16, Question 18. (d) none of these. Solution. Point P(x, 4) lies on the line segment joining the points A(-5,8) and B(4, -10). Question 10. Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear. ∴ The given three points are collinear. (ii) y = 5 So, the coordinates of Q are Here, x1 = 3, x2 = 8, y1 = 5, y2 = 10, m = 2, n = 3. Solution: Question 39. Question 4. (a) 5 units, Question 9. Adding (iv), (v) and (vi) we get (a) (√3, – √3) According to the question, AD is the median of ∆ABC, therefore D is the midpoint of BC. ⇒ (x – 2)2 + (0 + 5)2 = (x + 2)2 + (0 – 9)2 ⇒ 6x + 4y = 4x – 6y Solution: Question 61. Solution: Question 8. Show that the points (2, 1),(5, 2), (6, 4) and (3, 3) form a parallelogram. ⇒ (y +9)(y – 3) = 0 Question 10. Solution. If A(4,2), B(7,6) and C(l, 4) are the vertices of a ∆ABC and AD is its median, prove that the median AD divides ∆ABC into two triangles of equal areas Then coordinates of B are (0, 0). Multiple Choice Questions Also find the coordinates of the point of division. (d) None of these. ⇒ 6k – 3 = – k – 1 and – 8k + 10 = 6k + 6 ∴ x3 = 1, x1 = 3 and x2 = -1. Let the coordinates of A be (x, y). Solution: Question 43. ⇒ x2 + 2x + 4 + y2 – 2y + 4 = x2 + 8x + 16 + y2 + 4y + 4 Find the value of k, if the point P(2,4) is equidistant from the points A(5, k) and B(k, 7). (a) 18 units The co-ordinates of a point Mare (3, 4). Solution: Question 6. Solution: Question 75. ⇒ 15(m1 + m2) = 20m1 + 5m2 Solution. (d) (7, 0). ⇒ x = 1. Solution. Question 4. Solution: Question 20. ⇒ x2 + y2 + 2gx + 2fy + (g2 + f2 – a2) = 0 Find the value(s) of p for which the points (p + 1, 2p – 2), (p – 1 ,p) and (p -6, 2p – 6) are collinear. (a) 4 units Then, the coordinates of A and O and (3, 5) and (6, 6) respectively. Find the equation of the locus of all points equidistant from (3, 5) and the x-axis. which is the required equation of the locus. The coordinates of the mid-point are : Hence the vertices of the triangle are A(3, 2), B(-1, 2) and C(1, -4) respectively. Find the value of k, if the points A(7, -2), B(5,1) and C(3,2k) are collinear If A(l, 2), B(4, 3) and €(6,6) are three vertices of parallelogram ABCD, find co-ordinates of D. (b)(0, 7) AP = PQ = BQ = k (Say) Solution: Question 79. Solution: Question 62. Co-ordinates of this point may be : ⇒ x2 + 6x + 9 + y2 – 8y + 16 Solution. Use analytical geometry to prove that the mid-point of the hypotenuse of a right-angled triangle is equidistant from its vertices. ∴ \(\frac{1}{2}\) (- x – 5y + 17) = 0 Hence P divides AB in ratio 1 : 5. ∴ PB = PQ + QB = 2k The co-ordinates of mid-points D, E and F are Using distance formula, find which of them is correct. (d) (-6, -6). (c) 25 units Find the value of y. Let points P(a, a), Q(-a, -a) and R(-a√3, a√3) are given, then If the coordinates of points A and B are (-2, -2) and (2, – 4) respectively, find the coordinates of P such that AP = 3/7 AB, where P lies on the line segment AB. For what value of a, the points (1, 4) (a, -2) and (-3, 16) are collinear? (iii) The given condition is x = y. Let A = (1, -2), B = (3, 0) and C = (1, 2). Solution. The position of the point with abscissa = -5 and ordinate = +4 will be : (a) 0 ⇒ 6 = \(\frac{1}{2}\)[x(2 – 1) + 1(1 – y) + 2 (y – 2)] Solution: Question 80. Find the value of x for which the points (x – 1), (2,1) and (4,5) are collinear Also find the value of x Find the coordinates of a point P, which lies on the line segment joining the points A(-2, -2) and B(2, -4) such that AP =3/7 AB. x = \(\frac{56}{-8}\) = -7 (a) In the first quadrant Adding (i), (ii) and (iii) we get (b) 5 units Solution. In a classroom, 4 friends are seated at the points, A, B, C and Das shown in figure. The coordinates of two points are (6, 0) and (0, 8). (c) (-4, -4), Question 20. Solution. Find a point P on they-axis which is equidistant from the points A(4,8) and B(- 6, 6). Find the values of k so that the area of the triangle with vertices (1, -1), (-4,2k) and (-k, – 5) is 24 sq. Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry. (a) First The negative sign represents external division. (a) 5, Question 16. Question 19. Solution. Solution: Question 114. Question 2. (a) (3, 4), Question 18. Clearly point a is the midpoint of AB. ⇒ -8x = 56 Solution: Question 88. . Prove that the points A(1, -2), B(3, 0) and C(1, 2) are the vertices of an isosceles right-angled triangle. Question 16. . Find the area of a triangle, the coordinates of the mid-points of whose sides are (-2, -1), (1, 6) and (5, 3) respectively. Solution: Question 40. ⇒ 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y ⇒ y2 + 9y – 3y – 27 = 0 Also find the value of y. Question 7. ⇒ \(\frac{1}{2}\)[6(x + 7) + 0(-7 – 9) + (-6)(9 – x)) = 0 AB = AC For the AABC 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. (c) 3rd quadrant. Area of ∆ABC = 4 (area of ∆PQR). ⇒ 2m1 = 4m2 Answer: (5/3 , 1/3) Solution: Question 105. Question 4. Solution: Question 24. 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Answer: ⇒ OA2 = OB2 and m1 = 2, m2 = 3 Solution. The co-ordinates of two points are (-8, 0) and (0, -8). Question 17. (b) In the second quadrant. (a) 5 units Solution: Question 3. Let A(-2, 2) and B(-4, -2) be the given points and let P(x, y) be the variable point, Then. x2 = 0, y2 = x So, Champa is correct. Answer: Hence, from result (1), (2) and (4) ∆ABC is an isosceles right-angled triangle. Answer: Find the value(s) of k for which the points (3k – 1, k – 2), (k, k-1) and (k – 1, – k – 2) are collinear