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Q1. Using section formula, show that the points A(-3,-1), B(1,3) and C(-1,1) are collinear.

Solution

Let C(-1,1) divides AB in the ratio of k : 1 Using section formula, we have: (k - 3)/(k + 1) = -1 … (1) (3k - 1)/(k + 1) = 1 … (2) From (1), k - 3 = -k - 1 2k = 2 k = 1 Thus, C divides AB in the ratio 1: 1, that is, C is the mid-point of AB. A, B and C are collinear

Q2. Find the ratio in which the line joining points (1, 1) and (2, 4) is divided by the line 4x + y = 6. 

• 1) 1:2
• 2) 4:5
• 3) 1:4
• 4) 1:6

Solution

Let the ratio be k:1. The coordinates of the point dividing the line joining points (1, 1) and (2, 4) are 4x + y = 6 ... (i) Put x =  and y =  in (i) The ratio is 1:6.

Q3. Find the value of 's' if the point P(0,2) is equidistant from Q(3,s) and R(s,5).

Solution

PQ = PR PQ² = PR² (0 - 3)² + (2 - s)² =(0 - s)² + (2 - 5)² 9 + 4 - 4s + s² = s² + 9 4 = 4s s = 1

Q4. Find the value of k if the area of a triangle is 16 sq. units and its vertices are (1, −2), (5, −6) and (k, 4). 

• 1) 3
• 2) 1
• 3) 2
• 4) 4

Solution

Consider x₁ = 1, y₁ = −2, x₂ = 5, y₂ = −6, x₃ = k, y₃ = 4 Area of the triangle = 16 x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂) = 0 ∴ 4k + 20 = 32 ∴ 4k=12   ∴ k = 3

Q5. If A (−2, 4), B (0, 0) and C (4, 2) are the vertices of ∆ABC, then find the length of the median through the vertex A.

• 1) 8
• 2) 2.5
• 3) 5
• 4) 7

Solution

Let AD be the median of ∆ABC. D is the mid-point of BC. The coordinates of D are = (2, 1) 

Q6. The three vertices of a parallelogram ABCD are A (−2, 3), B (6, 7), C (8, 3) and the fourth vertex D is

• 1) (1, 0)
• 2) (−1, 0)
• 3) (0, −1)
• 4) (0, 1)

Solution

Let the fourth vertex be d(x, y). Then Mid-point of AC = Mid-point of BD 6 + x = 6 and 7 + y = 6 x = 0 and y = −1

Q7. Two vertices of ∆ABC are given by A (6, 4) and B (−2, 2), and its centroid is G (3, 4). Find the coordinates of the third vertex C of ∆ABC. 

• 1) (−5, 6)
• 2) (5, 6)
• 3) (5, −6)
• 4) (−5, −6)

Solution

Let the coordinates of C be (x, y). According to the question, 4 + a = 9 a = 5 Also, 6 + b = 12 b = 6 The coordinates of C (5, 6).

Q8. The line segment joining points (−3, −4) and (1, −2) is divided by the y-axis in the ratio 

• 1) 3:2
• 2) 3:1
• 3) 2:3
• 4) 1:3

Solution

Let the y-axis divide the points in the ratio k: 1.Hence, The coordinates of the point are The y-axis divides the line. So, the x coordinate of a point is zero. Hence, the ratio is 3:1.

Q9. Find the value of p so that the points with coordinates (3,5), (p,6) and are collinear.

Solution

Let the given points be A (3,5), B (p,6), C The points will be collinear if ar (ABC) = 0 ar (ABC) =

Q10. Find the coordinates of points which trisect the line segment joining (1, −2) and (−3, 4). 

• 1)
• 2)
• 3)
• 4)

Solution

Let A (1, −2) and B (−3, 4) be the given points.Let the points of trisection be P and Q. Then AP = PQ = QB AP:QB = 1:2 and AQ:QB = 2:1   

Q11. Show that the point P(-4,2) lies on the line segment joining the points A(-4,6) and B (-4,-6).

Solution

We know, if points are collinear the area of the triangle formed by three points as vertices is zero. Area = [-4 (6 + 6) - 4 (-6 - 2) - 4(2 - 6)] = 0 So, the points are collinear.

Q12.

Solution

Q13. Find the ratio in which the line segment joining (8, 9) and (2, 6) is divided by the line 2x + y = 4. 

• 1) 7:2
• 2) ⎼7:2
• 3) 1:7
• 4) 4:5

Solution

Let the ratio be k:1. The coordinates of the point dividing the line segment joining (8, 9) and (2, 6) are 2x + y = 4 … (i) Put x =  and y =  in (i) The ratio is 7:2

Q14. The area of a triangle whose vertices are (-2, -2), (-1,-3) and (x, 0) is 3 square units. Find the value of x.

Solution

_{Area of triangle:} (x₁ (y₂ - y₃) + x₂(y₃ - y₁) + x₃ (y₁ - y₂)] = 3 -2 (-3-0) -1 (0 + 2) + x(-2 + 3) = 6 6 - 2 + x = 6 x = 2

Q15. Points A (a, a), B (−a, −a) and C (0, 0) form the vertices of

• 1) A right-angled triangle
• 2) An equilateral triangle
• 3) An isosceles triangle
• 4) A scalene triangle

Solution

As BC = AC, ∆ABC is an isosceles triangle.

Q16. Find the perimeter of the triangle formed by the points (0,0), (1,0), (0,1).

Solution

  By distance formula, Distance Formula =   Let the points be A, B, C AB = _{= 1 units} CA = Similarly, BC = Perimeter = (2 + ) units

Q17. Find the ratio in which the line segment joining the points (3, 5) and (1, 4) is divided by the line 5x - y = 4. 

• 1) 1:1
• 2) 1:2
• 3) 2:1
• 4) 3:1

Solution

Let the ratio be k:1. The coordinates of the point dividing the line segment joining the points (3, 5) and (1, 4) are 5x - y = 4 … (i) Put x =  and y =  in (i) The ratio is 2:1.

Q18.

Solution

Q19. Determine the ratio in which the line 3x + y - 9 = 0 divides the line segment joining the points (1,3) and (2,7)

Solution

Let the given line divides the line segment in ratio k:1 and point on line 3x + y - 9 = 0 dividing line segment A(1,3) and B(2,7) be (a,b) a = and b = _{And as point P} (a,b) lies on straight line: 3x + y - 9 = 0 3a + b - 9 = 0 k = Required ratio is 3:4.

Q20. Prove that the points (a,0), (0,b) and (1,1) are collinear if,

Solution

Points A(a,0), B(0,b) and C(1,1) are collinear, Area of ABC = 0 Here, x₁ = a, y₁ = 0; x₂ = 0, y₂ = b; x₃ = 1, y₃ = 1  ab - a - b = 0 ab = a + b

Q21. Find the area of the triangle whose vertices are (a, b), (a + b, a + b) and (a - b, a - b). 

• 1) −2
• 2) None of these
• 3) 1
• 4) 2

Solution

Consider x₁ = a, y₁ = b, x₂ = a + b, y₂ = a + b, x₃ = a - b, y₃ = a - b According to the question, Area of the triangle = x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂) = a [a + b - (a - b)] + (a + b)(a - b - b) + (a - b) (b - a - b) = a [a + b - a + b] + (a + b)(a - 2b) + (a - b) (-a) = 0

Q22. Find the coordinates of the point of intersection of the medians of triangle ABC; given A(-2, 3), B(6, 7) and C(4, 1).

Solution

 

Q23. Find the area of the rhombus ABCD whose vertices are A(3,0), B(4,5), C(-1,4) and D(-2,-1).

Solution

Q24. Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right angled isosceles triangle.

Solution

Q25. Find the area of a triangle whose vertices are (4, 12), (3, 9) and (0, 0). 

• 1) 2
• 2) ⎼1
• 3) 1
• 4) 0

Solution

Consider x₁ = 1, y₁ = 2, x₂ = −5, y₂ = 6, x₃ = a, y₃ = −2 According to the question, Area of the triangle = = = 36 - 36 = 0

Q26.

Solution

Q27. Find the area of the triangle whose sides are 17 cm, 25 cm and 26 cm. 

• 1) 204 cm²
• 2) 206 cm²
• 3) 202 cm²
• 4) 208 cm²

Solution

Consider a = 17 cm, b = 25 cm, c = 26 cm Area of the triangle =  = = 204 cm²

Q28. Three consecutive vertices of a parallelogram ABCD are A(1,2), B(1,0) and C(4,0). Find the fourth vertex D.

Solution

Let the co-ordinates of the fourth vertex D be (x, y). We know that diagonals of a parallelogram bisect each other. Mid-point of BD = Mid-point of AC Coordinates of the mid-point of BD are Coordinates of the mid-point of AC are Thus, the co-ordinates of the vertex D are (4, 2).

Q29. If the points A(4,3) and B(x, 5) are on the circle with centre O(2,3); find the value of x.

Solution

OA = OB (Radii of same circle) OA² = OB² (4 - 2)² + (3 - 3)² = (x - 2)² + (5 - 3)² 4 = (x - 2)² + 4 x = 2

Q30. The mid points D, E, F of the sides AB, BC, CA of a triangle ABC are (1, 2), (3, -1) and (5, 0). Find the coordinates of the vertices A, B, C of the triangle.

Solution

Q31. If the point (k, 4) lies on a circle whose centre is at the origin and radius is 5, then k = 

• 1) ±5
• 2) 0
• 3) ±4
• 4) ±3

Solution

Let O (0, 0) be the origin and A (k, 4) be the point on the circle.OQ = 5OQ² = 25(0 - k)² + (0 - 4)² = 25k² + 16 = 25k² = 9k = ±3

Q32.

Solution

Q33. If the point P(x, y) is equidistant from the points A(5, 1) and B(-1, 5), then prove that 3x = 2y.

Solution

Q34. Find the value of x such that PQ = QR, where P, Q and R are the points (2, 5) (x, -3) and (7, 9) respectively.

Solution

PQ = QR PQ² = QR² By distance formula, (x - 2)² + (-3 - 5)² = (7 - x)² + (9 + 3)² x² - 4x + 4 + 64 = 49 + x² - 14x + 144 10x = 193 - 68 = 125 x = 12.5

Q35. Show that the points (a, b + c), (b, c + a) and (c, a + b) are collinear.

Solution

Let the given points be A (a, b + c), B (b, c + a), C (c, a + b). These points will be collinear if the area of triangle ABC is 0. Now, area (ABC) =

Q36. Find the area of the triangle whose sides are 13 cm, 15 cm and 10 cm. 

• 1) 66.36 cm²
• 2) 64.06 cm²
• 3) 68.26 cm²
• 4) 62.56 cm²

Solution

Consider a = 13 cm, b = 15 cm, c = 10 cm Area of the triangle = = = 64.06 cm²

Q37. Find the ratio in which the line segment joining the points (1, -3) and (- 4, 5) is divided by the x-axis.

Solution

Q38. Show that the quadrilateral ABCD with A(3, 1), B(0, -2), C(1, 1) and D(4, 4) is a parallelogram.

Solution

Using the distance formula, we have   AB = CD and BC = DA Since, the opposite sides of a quadrilateral are equal, it is a parallelogram. Hence, ABCD is a parallelogram.

Q39. If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p.

Solution

Q40. The ratio in which the line joining (1, 4) and (6, −4) is divided by the y-axis is 

• 1) 1:6
• 2) 6:1
• 3) 1:3
• 4) 2:3

Solution

Let point C divide the y-axis in the ratio k:1. According to the question, x coordinate of C is zero.So, the ratio is 1:6.

Q41. Find the coordinates of the point B, if the point P(-4,1) divides the line segment joining the points A(2,-2) and B in the ratio 3 : 5.

Solution

Let coordinates of point B be (x,y) By section formula, -4 = x= -14 1 = Co-ordiantes of point B are (-14, 6).

Q42. The base BC of an equilateral ABC lies on y-axis. The co-ordinates of the point C are (0,-3). If origin is the mid-point of BC, find the coordinates of point A and B.

Solution

Let (x, y) be the coordinates of point B. _{Using mid-point formula, we have:} Coordinates of B are (0, 3) BC = 6 units Now, ABC is an equilateral triangle and OB = 3 units, AB = 6 units AO² = 6² - 3² = 27 AO = 3 units Coordinates of A are

Q43. Find the point which divides the line segment joining the two points (6, 3) and (−4, 5) internally in the ratio 3:2.

• 1)
• 2)
• 3)
• 4)

Solution

Section Formula:  

Q44. In the seating arrangement of desks in a class room three students Rohini, Sandhya and Bina are seated at A(3, 1), B(6, 4) and C(8, 6). Do you think they are seated in a line?

Solution

If three students are seated in a line, then AB + BC = AC or AB + AC = BC or AC + BC = AB AB + BC = CA Hence, the three places are in line. Yes, the three students are sitting in a line.

Q45. The points A(2,9), B(a,5) and C(5,5) are the vertices of a triangle ABC, right angled at B. Find the value of a and hence the area of triangle ABC.

Solution

Since, ABC is a right triangle, right angled at B, we have: AB² + BC² = CA² [(2 - a)² + (9 - 5)²] + [(a - 5)² + (5 - 5)²] = (2 - 5)² + (9 - 5)² 4 + a² - 4a + 16 + a² + 25 - 10a = 9 + 16 2a² - 14a + 20 = 0 a² - 7a + 10 = 0 (a - 5)(a - 2) = 0 a = 5 or 2 But a cannot be 5 as then the points B and C will coincide. a = 2 Area = = 6 sq. units

Q46. Show that the points (-4,0), (4,0) and (0,3) are vertices of an isosceles triangle.

Solution

Let A (-4,0), B (4,0), C(0,3) be the given points. AB² = 8² = 64 AB = 8 units BC = units CA = units BC = AC ABC is isosceles

Q47. Find points on the x-axis, which are at a distance of 5 units from the point A(5,-3).

Solution

Let the required point on x-axis be (x, 0). Using distance formula, we have: (x - 5)² + (0 + 3)² = (5)² (x - 5)² = 25 - 9 = x - 5 = x = 9, 1 Required points on x-axis are (9, 0) and (1, 0).

Q48. Find the ratio in which the line 3x + y – 9 = 0 divides the line segment joining the points (1,3) and (2,7).

Solution

Let the line 3x + y – 9 = 0 divide the line segment joining the points A(1,3) and B(2,7) at the point C in the ratio k:1 Then point C by section formula is As C lies on the line3x + y – 9 = 0, we have 6k + 3 + 7k + 3 - 9k - 9 = 0 4k - 3 = 0 The required ratio is 3:4 internally

Q49. Find the values of y, if the distance between the points (2, -3) and (10, y) is 10 units.

Solution

Q50.

Solution

Q51. Find the point on the X-axis which is equidistant from (2, -5) and (-2, 9).

Solution

Q52. If A(3,0), B(4,5), C(-1,4) and D(-2,-1) be four points in a plane, show that ABCD is a rhombus but not a square.

Solution

The given points are A(3,0), B(4,5), C(-1,4) and D(-2,-1). Using distance formula, AB = BC = CD = DA = AC = BD = Since, AB = BC = CD = DA and AC BD, Hence, ABCD is a rhombus but not a square.

Q53.

Solution

Q54. If three points A (0, 0), B (3,) and C (3, k) form an equilateral triangle, then k =

• 1) −4
• 2) None of these
• 3) 2
• 4) −3

Solution

According to the question,AB = BC = ACAB = AC∴ AB² = AC²∴ (0 - 3)² + (0 ⎼ )² = (0 ⎼ 3)² + (0 ⎼ k)² ∴ 9 + 3 = 9 + k²∴ k² = 3∴ k = 

Q55. Find a relation between x and y if the points (x,y), (1,2) and (7,0) are collinear.

Solution

  Points A (x,y), B (1,2) and C(7,0) will be collinear if the area of the triangle formed by these points is zero. Here x₁ = x, y₁ = y, x₂ = 1, y₂ = 2, x₃ = 7, y₃ = 0 Substituting the values, Area of ABC = 2x + 6y -14 Now area of ABC = 0 2x + 6y -14 = 0  x + 3y = 7

Q56. Show that the point P(0,-2), Q(3,1), R(0,4) and S(-3,1) are the vertices of a square PQRS.

Solution

Using distance formula, PQ = QR = RS = PS = So, PQ = QR = RS = SP So, diagonal PR = diagonal QS Hence, PQRS is a square.

Q57. The points A(3, 2) and B(2, -3) are equidistant from a point P(x, y). Find the relation between x and y.

Solution

Given: A(3, 2) and B(2, -3) are equidistant from the point P(x, y) Then, PA = PB PA² = PB² (x - 3)² + (y - 2)² = (x - 2)² + (y + 3)² x² + 9 - 6x + y² + 4 - 4y = x² + 4 - 4x + y² + 9 + 6y -6x - 4y = -4x + 6y -2x - 10y = 0 x + 5y = 0

Q58. The midpoint of the line segment joining points A(x,y + 1) and B(x + 1, y + 2) is C. Find the value of x and y if the coordinates of C are (3/2, 5/2).

Solution

The midpoint of two points (x₁,y₁) and (x₂,t₂) is ( ,) Hence midpoint of A and B is C[(x₁ + x₂)/2, (y₁ + y₂)/2] (x₁ + x₂)/2 = 3/2 2x = 3 - 1 x = 1 Similarly, 2y + 3 = 5 2y = 2 y = 1 Coordinates of midpoint (1,1)

Q59. Find the points on the x-axis which are at a distance of 2 units from the point (7, -4). How many such points are there?

Solution

Let the point on x-axis be P(x, 0). Let the given point be Q = (7, -4). PQ = 2 Using distance formula, PQ² = (x - 7)² + (0 + 4)² = (2)² x² - 14x + 49 + 16 = 20 x² - 14x + 45 = 0 (x - 9) (x - 5) = 0 x = 9, 5 Hence, there are two such points, (9, 0) and (5, 0).

Q60. Find the co-ordinates of a point A, where AB is the diameter of a circle whose centre is (2,-3) and B is (1, 4).

Solution

Q61. In the figure, in ABC, D and E are the mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = AB.

Solution

Coordinates of E = = Coordinates of D = = DE = AB = Hence, DE = AB

Q62. Find the point on x-axis which is equidistant from the points (-2,5) and (2,-3).

Solution

Let the point on x-axis be (a,0) Since PA = PB PA² = PB² (a + 2)² + (0 - 5)² = (a - 2)² + (0 + 3)² (a + 2)² - (a - 2)² = 9 - 25 = -16 8a = -16 a = -2. The required point is (-2,0).

Q63. If A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram ABCD taken in order, find the values of x and y.

Solution

Q64. Show that the points A (a, b+c), B (b, c + a) and C (c, a + b) are collinear.

Solution

If the given points are collinear then the area of the triangle formed by joining these points is zero. Here, x₁ = a, y₁ = b + c; x₂ = b, y₂ = c + a; x₃ = c, y₃ = a + b. Substituting the values in the formula for area of a triangle, you get ac - ab + ab - bc + bc - ac = 0 Hence the given points are collinear.

Q65. Find a point on y-axis which is equidistant from (2,2) and (9,9).

Solution

Let the required point on y-axis be (0, y). According to question, PA = PB PA² = PB² - 4y + 8 = - 18y + 162 14y = 154 Thus, the required point is (0, 11).

Q66. Find the co-ordinates of a point A, where AB is the diameter of circle whose centre is O (2, -3) and B is (1, 4).

Solution

Let the co-ordinates of A be (x, y). The mid-point of AB is O. Using mid-point formula, we have: (x + 1)/ 2 = 2 x = 3 (y + 4)/2 = -3 y = -10 Thus, the coordinates of point A are (3, -10).

Q67. Find the area of a quadrilateral whose vertices, taken in order, are (-4,-2), (-3,-5), (3,-2) and (2,3).

Solution

The vertices of the quadrilateral ABCD are A (-4, -2), B (-3, -5), C (3, -2) and D (2, 3).          Area of quad. ABCD = ar ABC + ar ACD Now area of ABC: Here x₁ = -4, y₁ = -2, x₂ = -3, y₂ = -5, x₃ = 3, y₃ = -2 Substituting the values, Area of ABC = = 10.5sq.units Area of ACD: Here x₁ = -4, y₁ = -2, x₂ = 3, y₂ = -2, x₃ = 2, y₃ = 3 Substituting the values, Area of ACD = = 17.5 sq.units Area of ABCD = ar ABC + ar ACD                          =         10.5 + 17.5 Area of quadrilateral ABCD = 27.5 sq.units

Q68. In what ratio is the line segment joining the points P(-2,-3) and Q(3,7) divided by y-axis?

Solution

The coordinates of point on y-axis are (0, y). Let it divide the line segment joining (-2,-3) and (3,7) in ratio k : 1. Using section formula, we have: Thus, the required ratio is 2: 3.

Q69. Find the co-ordinates of the circum-centre of the triangle ABC, whose vertices A, B and C are (4, 6), (0, 4) and (6, 2) respectively.

Solution

Q70.

Solution

Q71. If A and B are the points (-2,-2) and (2,-4) respectively, find the coordinates of P on the line segment AB such that AP = AB.

Solution

_{Using section formula, we have:} The coordinates of the point P are .