TOPICS
Exercise - 7.1

Question-1 :- Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (– 5, 7), (– 1, 3)
(iii) (a, b), (– a, – b)

Solution :-
i) By using distance formula, 
= √(x₂-x₁)² + (y₂-y₁)²
= √(4-2)² + (1-3)² 
= √(2)² + (-2)² 
= √8 
= 2√2
    
ii) By using distance formula, 
= √(x₂-x₁)² + (y₂-y₁)²
= √(-1+5)² + (3-7)² 
= √(4)² + (-4)² 
= √32
= 4√2
    
iii) By using distance formula, 
= √(x₂-x₁)² + (y₂-y₁)²
= √(-a-a)² + (-b-b)² 
= √(-2a)² + (-2b)² 
= √4a² + 4b² 
= 2√a² + b² 
    

Question-2 :-  Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Solution :-
  Given the points A(0, 0) and B(36, 15),
  By using distance formula,
  AB = √(x₂-x₁)² + (y₂-y₁)²
  AB = √(36-0)² + (15-0)² 
  AB = √(36)² + (15)² 
  AB = √1296 + 225 
  AB = √1521
  AB = 39

  Yes, we can find the distance between the given towns A and B.
  Assume town A at origin point (0, 0).
  Therefore, town B will be at point (36, 15) with respect to town A.
  And hence, as calculated above, the distance between town A and B will be 39km.
    

Question-3 :-  Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Solution :-
  Using the distance formula, we have
  AB =  √(2-1)² + (3-5)² = √(1)² + (-2)² = √5
  BC =  √(-2-2)² + (-11-3)² = √(-4)² + (-14)² = √212
  AC =  √(-2-1)² + (-11-5)² = √(-3)² + (-16)² = √265
  Since, AB + BC ≠ AC 
  So, we can say that the points A, B and C are not collinear. 
  Therefore, they are seated in a line.
    

Question-4 :-  Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Solution :-
  Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the a triangle respectively.
  Using the distance formula, we have
  AB =  √(6-5)² + (4+2)² = √(1)² + (6)² = √37
  BC =  √(7-6)² + (-2-4)² = √(1)² + (-6)² = √37
  AC =  √(7-5)² + (-2+2)² = √(2)² + (0)² = √4
  Since, AB = BC
  As two sides are equal in length, therefore, ABC is an isosceles triangle.
    

Question-5 :-  In a classroom, 4 friends are seated at the points A, B, C and D as shown in Figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct. coordinate geometry

Solution :-
  It can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends.
  Using the distance formula, we have
  coordinate geometry
  AB =  √(6-3)² + (7-4)² = √(3)² + (3)² = √18
  BC =  √(9-6)² + (4-7)² = √(3)² + (-3)² = √18
  CD =  √(6-9)² + (1-4)² = √(-3)² + (-3)² = √18
  AD =  √(6-3)² + (1-4)² = √(3)² + (-3)² = √18
  AC =  √(9-3)² + (4-4)² = √(6)² + (0)² = √36
  BD =  √(6-6)² + (1-7)² = √(0)² + (-6)² = √36

  It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
  Therefore, ABCD is a square and hence, Champa was correct.
    

Question-6 :-  Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution :-
(i) Let the points (–1, –2), (1, 0), (–1, 2) and (–3, 0) are representing the vertices A, B, C and D of the a quadrilateral respectively.
  AB =  √(1+1)² + (0+2)² = √(2)² + (2)² = √8
  BC =  √(-1-1)² + (2-0)² = √(-2)² + (2)² = √8
  CD =  √(-3+1)² + (0-2)² = √(-2)² + (-2)² = √8
  AD =  √(-3+1)² + (0+2)² = √(-2)² + (2)² = √8
  AC =  √(-1+1)² + (2+2)² = √(0)² + (4)² = √16 = 4
  BD =  √(-3-1)² + (0-0)² = √(-4)² + (0)² = √16 = 4

  It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. 
  Therefore, the given points are the vertices of a square.
    
(ii) Let the points (–3, 5), (3, 1), (0, 3) and (–1, –4) are representing the vertices A, B, C and D of the a quadrilateral respectively.
  AB =  √(3+3)² + (1-5)² = √(6)² + (-4)² = √52
  BC =  √(0-3)² + (3-1)² = √(-3)² + (2)² = √13
  CD =  √(-1-0)² + (-4-3)² = √(-1)² + (-7)² = √50
  AD =  √(-1+3)² + (-4-5)² = √(2)² + (-9)² = √85

  It can be observed that all sides of this quadrilateral are of different lengths. 
  Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.
    
(iii) Let the points (4, 5), (7, 6), (4, 3) and (1, 2) are representing the vertices A, B, C and D of the a quadrilateral respectively.
  AB =  √(7-4)² + (6-5)² = √(3)² + (1)² = √10
  BC =  √(4-7)² + (3-6)² = √(-3)² + (-3)² = √18
  CD =  √(1-4)² + (2-3)² = √(-3)² + (-1)² = √10
  AD =  √(1-4)² + (2-5)² = √(-3)² + (-3)² = √18
  AC =  √(4-4)² + (3-5)² = √(0)² + (-2)² = √4 = 2
  BD =  √(1-7)² + (2-6)² = √(-6)² + (-4)² = √52

  It can be observed that opposite sides of this quadrilateral are of the same length.
  However, the diagonals are of different lengths. 
  Therefore, the given points are the vertices of a parallelogram.
    

Question-7 :-  Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Solution :-
  We know that a point on the y-axis is of the form (x, 0). 
  So, let the point P(x, 0) be equidistant from A and B. 
  We are given that AP = BP. 
  So, AP² = BP² 
  Then (x – 2)² + (0 + 5)² = (x + 2)² + (0 – 9)² 
  i.e., x² + 4 - 4x + 25 = x² + 4 + 4x + 81 
  i.e., 8x = -56 
  i.e., x = -7 
  So, the required point is (-7, 0).
    

Question-8 :-  Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Solution :-
  It is given that the distance between (2, −3) and (10, y) is 10.
  (10 - 2)² + (y + 3)² = (10)²
  64 + y² + 9 + 6y = 100
  y² + 6y + 64 + 9 - 100 = 0
  y² + 6y - 27 = 0
  y² + 9y - 3y - 27 = 0
  y(y + 9) - 3(y + 9) = 0
  (y - 3)(y + 9) = 0
  y - 3 = 0, y + 9 = 0
  y = 3, y = -9
    

Question-9 :-  If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution :-
  Given that Q(0, 1) is equidistant from P(5, –3) and R(x, 6). 
  Therefore, PQ = QR
  (0 - 5)² + (1 + 3)² = (0 - x)² + (1 - 6)²
              25 + 16 = x² + 25
                   41 = x² + 25
                   x² = 16
                   x = ±4
  Therefore, point R is (4, 6) or (−4, 6).

  When point R is (4, 6)
  PR = √(4-5)² + (6+3)² = √(-1)² + (9)² = √82
  QR = √(4-0)² + (6-1)² = √(4)² + (5)² = √41

  When point R is (-4, 6)
  PR = √(-4-5)² + (6+3)² = √(-9)² + (9)² = √162 = 9√2
  QR = √(-4-0)² + (6-1)² = √(-4)² + (5)² = √41
    

Question-10 :-  Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Solution :-
  Let P(x, y) be equidistant from the points A(3, 6) and B(– 3, 4).
  We are given that AP = BP. 
  So, AP² = BP² 
  i.e., (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)² 
  i.e., x² + 9 - 6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y
  i.e.,-6x - 12y + 45 = 6x - 8y + 25
  i.e.,12x + 4y = 20
  i.e., 3x + y = 5
  i.e., 3x + y - 5 = 0
  which is the required relation.
    
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