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y3 + y1 = -2 ……(vi) ⇒ 1 + x = 7 and 8 = 5 + y (a) (√3, – √3) Point P(x, 4) lies on the line segment joining the points A(-5,8) and B(4, -10). Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and Bis (1, 4). Hence P divides AB in ratio 1 : 5. Let the coordinates of a point are (x, y). = i.e., m1 = m2 = 2 : 1. (d) None of these. x3 = -1, y3 = 1 P(x, y) = (11, 15) lie on the circle, Then, we have Any point on X-axis is P(x, 0) Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram. (a) 7 units (b) Second Let the co-ordinates of the third vertex be C(x, y). Find the value of k, if the points P(5,4), Q(7, k) and R(9, -2) are collinear Solution. Solution. Solution. Prove that the three points (1, 2), (3, 3) and (-1, 1) are collinear. Question 5. and \(\frac{y+4}{2}\) = -3 = y + 4 = -6 (a) 5 units, Question 9. ⇒ – 6x – 6y – 12 = 0 ⇒ AP : PB = k : 2k = 1 : 2 ⇒ (x + g)2 + (y + f)2 = a2 According to the question, AD is the median of ∆ABC, therefore D is the midpoint of BC. If P(2, p) is the mid-point of the line segment joining the points A(6, -5) and B(-2,11), find the value of p. Solution: Question 60. (a) (-7, 0) (i) (2, 3), (-1, 0), (2, -4) Solution: Question 38. ⇒ x2 + 6x + 9 + y2 – 8y + 16 ⇒ (x + 2)2 + (y – 2)2 = (x + 4)2 + (y + 2)2 Solution: Question 92. Solution. Find the value of k, if the point P(2,4) is equidistant from the points A(5, k) and B(k, 7). Question 2. x2 + y2 + 2gx + 2fy + c = 0, If the students practice NCERT solutions they can solve the most complex problems easily. The base BC of an equilateral triangle ABC lies on y-axis. ∴ Choice (c) is correct. Question 7. Let (a, b) be the coordinates of B. If x takes the values – 2, -1, 0, 1, 2, then y will also take the values -2, -1, 0, 1, 2 respectively. Also find the value ofy Answer: The co-ordinates of a point Mare (3, 4). Solution. Practice Questions. Solution: Question 111. y = – 10 Find the equation of the locus of all points equidistant from (3, 5) and the x-axis. Is ABCD a square? If A(-3,5), B(-2, -7), C(l, -8) and D(6,3) are the vertices of a quadrilateral ABCD, find its area. Let O(x, y) be the centre of the circle and the points 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. (d) 10 units. If the points A(x, y), B(2, 3) and C(3, 4) are collinear, then prove that x + 5y = 17. 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 Question 3. Solution: Question 56. According to the question, Solution. ⇒ a(-1 -4) + 1(4 – 1) + 11(1 + 1) = 0 Find the equation of the locus of points equidistant from (-2, 2) and (-4, -2). Show that the points (a,a), (-a,-a) and (-√3a, √3 a) are the vertices of an equilateral triangle. Geometry Problems With Solutions and Explanations for Grade 9; High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers (c) (-5, -5) Solution: Question 39. ⇒ 6k – 3 = – k – 1 and – 8k + 10 = 6k + 6 Let A(2, 1), B(5, 2), C(6, 4) and D(3, 3). Solution: Question 101. Answer: x1 + x2 + x3 = 3. The co-ordinates of two points are (9, 4) and (3, 8). The middle point of BC is ∴ The given three points are collinear. ∴ AC and BD will bisect each other. 100 flower pots have been placed at a distance of 1 m from each other along with AD, as shown in figure. Solution: Question 82. Since, the four sides and diagonals are equal. ⇒ x2 + 2x + 4 + y2 – 2y + 4 = x2 + 8x + 16 + y2 + 4y + 4 (c) 10 x = \(\frac{56}{-8}\) = -7 The endpoints of a line segment AB are A(2, 5) and B(1, 9). Hence, coordinates of the point are (1, 3). x1 = -2, x2 = 1, x3 = 5 >. ⇒ x2 = 16 ⇒ 15(m1 + m2) = 20m1 + 5m2 Solution: Question 80. Question 15. The distance between the points (-6, 5) and (-1, 7) is : (c) 16 units Answer: Solution: Question 31. Coordinates of a point on X-axis which is equal distance from the points A(2, -5) and B(-2, 9) will be : Solution. (b) In the second quadrant. The co-ordinates of the midpoint of the line segment joining them will be : Find the value of y for which the distance between the points A(3, -1) and B(11,y) is 10 units. The co-ordinates of two points are (-8, 0) and (0, -8). Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1,2). Area of ∆ = \(\frac{1}{2}\) [x1 (y2 – y3) + x2(y3 – y1)+ x3 y1 – y2)] Solution: Question 41. The line segment AB joining the points A(3, – 4), and B(l, 2) is trisected at the points P(p, – 2) and Q(5/3, q). ∴ \(\frac{1}{2}\) (- x – 5y + 17) = 0 OA2 = (x + 3)2 + (y + 2)2 Solution: Question 54. ⇒ OA2 = OB2 Again, OB = OC Answer: Clearly point a is the midpoint of AB. Find the value of p. ⇒ y2 + 9y – 3y – 27 = 0 If P is equidistant from Q(2, -5) and R(-3, 6), find the coordinates of P. Find the coordinates of the points which divide the line segment joining A (2, – 3) and B(-4, – 6) into three equal parts. Prove that the diagonals of a rectangle ABCD, with vertices A(2, – 1), B(5, – 1), C(5,6) and D(2,6), are equal and bisect each other. If the point P(x, y) is equidistant from the points A (a + b, b – a) and B(a -b,a+ b). Solution: Question 44. 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. Students are suggested to practice these questions by themselves and then verify the answers. The three points are collinear if the area of the triangle formed by them is zero. ∵ Area of ∆ADC = Area of ∆ABD ⇒ a = 5 Solution. Let the coordinates of A be (x1, y1), then co-ordinates of the centroid G of ∆ABC are : Question 8. Fmd the ratio in which point P divides the line segment AB. The mid-point P of the line segment joining the points A(- 10, 4) and B(- 2, 0) lies on the line segment joining the points C(- 9, – 4) and D(- 4, y). Expert teachers at CBSETuts.com collected and solved 2 Marks and 4 mark important questions for Class 10 Maths Chapter 7 Coordinate Geometry. The distance between (-1, -3) and (3, 0) is : ⇒ x + 5y = 17. Solution: Question 3. Solution: Question 95. (d) 10 units. Find the area of a quadrilateral ABCD, the coordinates of whose vertices are A(—3,2), B(5,4), C(7, -6) and D(-5, -4). A point (2, 3) divides the distance between the points (x, y) and (-5, 4) internally in the ratio 1 : 2. Answer: ⇒ 5m1 = 10m2 Solution: Question 112. Solution: Question 17. AP = PQ = BQ = k (Say) ⇒ y2 + 6y + 73 = 100 Find the value of k, if the points A(7, -2), B(5,1) and C(3,2k) are collinear Let the required ratio be m1 : m2 Answer: Find the value of x for which the points (x – 1), (2,1) and (4,5) are collinear A right triangle, B, c and Das shown in figure 3 and x2 -1!.. Maths Important Questions Class 10 Maths Chapter 7 Extra Questions for Class 10 you now the! ) First ( B ), ( v ) and the perpendicular OB taken along the are. Answers and explanations are given below fifth line at a distance of point c are ( 0 -8! Right angled isosceles triangle ) 17 units, Question 19 of two points are collinear 3 – 1 =...., y1 ), Question 18 i ) the condition given in figure. Taken along the x-axis line AB is the mid-point of BC c and Das shown in figure ABC a... 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