TOPICS
Exercise - 7.2

Question-1 :- Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Solution :-
  Let P(x, y) be the required point. 
  Given that : x₁ = -1, y₁ = 7 ; x₂ = 4, y₂ = -3 ; m₁ = 2, m₂ = 3
  Using the section formula, we get
section formula
  x = [(2x4 + 3x(-1))/(2 + 3)] = (8 - 3)/5 = 5/5 = 1
  y = [(2x(-3) + 3x7)/(2 + 3)] = (-6 + 21)/5 = 15/5 = 3
  Therefore, (1, 3) is the required point.
    

Question-2 :-  Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

Solution :-
  Let P and Q be the points of trisection of AB i.e., AP = PQ = QB 
  line
  Therefore, P divides AB internally in the ratio 1 : 2.
  Therefore, the coordinates of P, by applying the section formula, are
  [(1x(-2) + 2x4)/(1 + 2), (1x(-3) + 2x(-1))/(1 + 2)]
  i.e., [(-2+8)/3, (-3-2)/3] 
  i.e., (3, -5/3)

  Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
  [(2x(-2) + 1x4)/(2 + 1), (2x(-3) + 1x(-1))/(2 + 1)] 
  i.e., [(-4+4)/3, (-6-1)/3] 
  i.e., (0, -7/3)

  Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).
    

Question-3 :-  To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? coordinate geometry

Solution :-
  It can be observed that Niharika posted the green flag at 1/4	of the distance AD 
  i.e., (1/4 x 100)m = 25m from the starting point of 2nd line. 
  Therefore, the coordinates of this point G is (2, 25).

  Similarly, Preet posted red flag at 1/5 of the distance AD 
  i.e., (1/5 x 100)m = 20m from the starting point of 8th line. 
  Therefore, the coordinates of this point R are (8, 20). 
  Distance between these flags by using distance formula = GR

= √(8-2)² + (25-20)² 
= √(6)² + (5)² 
= √61 m

  The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. 
  Let this point be A (x, y).
  x = (2+8)/2, y = (25+20)/2
  x = 5, y = 22.5
  Hence, A(x, y) = (5, 22.5)
  Therefore, Rashmi should post her blue flag at 22.5m on 5th line
    

Question-4 :-  Find the ratio in which the line segment joining the points (–3, 10) and (6, – 8) is divided by (– 1, 6).

Solution :-
  Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be k : 1.
  Therefore, 
  (-1, 6) = [(6k - 3)/(k + 1), (-8k + 10)/(k + 1)]
  
  -1 = (6k - 3)/(k + 1)
  -k - 1 = 6k - 3
  -k - 6k = -3 + 1
  -7k = -2
    k = 2/7
  Therefore, Required ratio is 2:7.
    

Question-5 :-  Find the ratio in which the line segment joining A(1, – 5) and B(–4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution :-
  Let the ratio be k : 1. 
  Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are
  [(-4k + 1)/(k + 1), (5k - 5)/(k + 1)]
  This point lies on the x-axis, and we know that on the x-axis the ordinate is 0.
  Therefore,
  (5k - 5)/(k + 1) = 0
  5k - 5 = 0
  5k = 5
   k = 1
  i.e., the ratio is 1 : 1. 
  Putting the value of k = 1, we get the point of intersection as 
= [(-4 + 1)/2, (5 - 5)/2]     
= (-3/2, 0)
    

Question-6 :-  If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution :-
  Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. 
  Intersection point O of diagonal AC and BD also divides these diagonals.
  parallelogram
  Therefore, O is the mid-point of AC and BD.
  If O is the mid-point of AC, then the coordinates of O are [(1+x)/2, (2+6)/2] = [(x+1)/2, 4]
  If O is the mid-point of BD, then the coordinates of O are [(4+3)/2, (5+y)/2] = [7/2, (y+5)/2]
  Since both the coordinates are of the same point O,
  (x+1)/2 = 7/2 ; 4 = (y+5)/2
  x + 1 = 7 ; 8 = y + 5
  x = 7 - 1 ; y = 8 - 5
  x = 6; y = 3
    

Question-7 :-  Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).

Solution :-
  Let the coordinates of point A be (x, y).
  Mid-point of AB is (2, −3), which is the center of the circle.
  Therefore,
  (2, -3) = [(x+1)/2, (y+4)/2]
  2 = (x+1)/2 ; -3 = (y+4)/2
  4 = x + 1 ; -6 = y + 4
  x = 4 - 1 ; y = -6 - 4
  x = 3; y = -10
  Hence, the coordinates of A are (3, -10).
    

Question-8 :-  If A and B are (–2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3AB/7 and P lies on the line segment AB.

Solution :-
  The coordinates of point A and B are (−2, −2) and (2, −4) respectively.
  AP = 3AB/7
  AP/AB = 3/7
  PB = AB - AP = 7 - 3 = 4
  So, AP/PB = 3/4
  Therefore, AP: PB = 3:4
  Point P divides the line segment AB in the ratio 3:4.
  Coordinates of P 
= [(3x2 + 4x(-2))/(3 + 4), (3x(-4) + 4x(-2))/(3 + 4)] 
= [(6 - 8)/7, (-12 - 8)/7] 
= (-2/7, -20/7)  
    

Question-9 :-  Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Solution :-
  From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.
  coordinate geometry
  
  Now, Coordinates of P = [(1x2 + 3x(-2))/(1 + 3), (1x8 + 3x2)/(1 + 3)] = (-1, 7/2)
  Also, Coordinates of Q = [(1x2 + 1x(-2))/(1 + 1), (1x8 + 1x2)/(1 + 1)] = (0, 5)
  Also, Coordinates of R = [(3x2 + 1x(-2))/(3 + 1), (3x8 + 1x2)/(3 + 1)] = (1, 13/2)
    

Question-10 :-  Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Solution :-
  Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.
rhombus
  Now, Length of diagonal of AC 
= √(3+1)² + (0-4)² 
= √(4)² + (4)² 
= √32 
= 4√2

  Also, Length of diagonal of BD 
= √(4+2)² + (5+1)² 
= √(6)² + (6)²
= √72 
= 6√2

  Therefore, Area of Rhombus ABCD = 1/2 x 4√2 x 6√2 = 24 square units.
    
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