TOPICS
Exercise - 7.4 (Optional)

Question-1 :- Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

Solution :-
  Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k:1.
  Coordinates of the point of division = [(3k+2)/(k+1), (7k-2)/(k+1)]
  This point also lies on 2x + y − 4 = 0
  Therefore,
  2(3k+2)/(k+1) + (7k-2)/(k+1) - 4 = 0
  (6k + 4 + 7k - 2 - 4k - 4)/(k+1) = 0
  9k - 2 = 0
  k = 2/9
  Therefore, the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, −2) and B(3, 7) is 2:9.
    

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

Solution :-
  If the given points are collinear, then the area of triangle formed by these points will be 0.
  By using the area of triangle formula, is given by 
  0 = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
  0 = 1/2[x(2 - 0) + 1(0 - y) + 7(y - 2)]
  0 = 1/2[2x - y + 7y - 14]
  0 = 1/2[2x + 6y - 14]
  0 = 1/2 x 2[x + 3y - 7] 
  0 = x + 3y - 7
  x + 3y - 7 = 0 
  This is the required relation between x and y.
    

Question-3 :-  Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

Solution :-
  Let O (x, y) be the centre of the circle. 
  And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.
  Therefore,
  OA = √(x-6)² + (y+6)²
  OB = √(x-3)² + (y+7)²
  OC = √(x-3)² + (y-3)²

  However, OA = OB (Radii of the same circle)
  √(x-6)² + (y+6)² = √(x-3)² + (y+7)²
  (x-6)² + (y+6)² = (x-3)² + (y+7)²
  x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 49 + 14y
  72 - 12x + 12y = 58 - 6x + 14y
  -12x + 6x + 12y - 14y = 58 - 72
  -6x - 2y = -14
    3x + y = 7  .....(i)

  Similarly, OA = OC (Radii of the same circle)
  √(x-6)² + (y+6)² = √(x-3)² + (y-3)²
  (x-6)² + (y+6)² = (x-3)² + (y-3)²
  x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 9 - 6y
  72 - 12x + 12y = 18 - 6x - 6y
  -12x + 6x + 12y + 6y = 18 - 72
  -6x + 18y = -54
  -3x + 9y = -27 .....(ii)

  By adding eq. (i) and eq. (ii),
  3x + y - 3x + 9y = 7 - 27
  10y = -20
    y = -2
  Put in eq. (i),
  3x - 2 = 7
  3x = 7 + 2
   x = 9/3
   x = 3

  Therefore, the centre of the circle is (3, −2).
    

Question-4 :-  The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

Solution :-
  Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively.
  Let (x, y), (x₁, y₁) be the coordinate of vertex B and D respectively. 
coordinate geometry
  We know that the sides of a square are equal to each other. 
  Therefore, AB = BC
  √(x+1)² + (y-2)² = √(x-3)² + (y-2)²
  x² + 2x + 1 + y² - 4y + 4 = x² + 9 - 6x + y² + 4 - 4y
  5 + 2x = 13 - 6x
  2x + 6x = 13 - 5
  8x = 8
   x = 1

  We know that in a square, all interior angles are of 90°.
  Now, In ΔABC,
  AB² + BC² = AC²
  [√(1+1)² + (y-2)²]² + [√(1-3)² + (y-2)²]² = [√(3+1)² + (2-2)²]²
  4 + y² - 4y + 4 + 4 + y² - 4y + 4 = 16
  16 + 2y² - 8y = 16
  2y² - 8y = 0
  2y(y - 4) = 0
  2y = 0; y - 4 = 0
   y = 0; y = 4

  We know that in a square, the diagonals are of equal length and bisect each other at 90°. 
  Let O be the mid-point of AC. 
  Therefore, it will also be the mid-point of BD.
  Coordinate of point O = [(-1+3)/2, (2+2)/2] = (1, 2)
  Now, [(1+x₁)/2, (y+y₁)/2] = (1, 2)
  (1+x₁)/2 = 1 
  1 + x₁ = 2 
  x₁ = 2 - 1
  x₁ = 1

  (y+y₁)/2 = 2
  y + y₁ = 4
  if, y = 0, then y₁ = 4
  if, y = 4, then y₁ = 0
  Therefore, the required coordinates are (1, 0) and (1, 4). 
    

Question-5 :-  The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the Figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of Δ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe? coordinate geometry

Solution :-
(i) Taking A as origin, we will take AD as x-axis and AB as y-axis. 
  It can be observed that the coordinates of point P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.

  Now, Area of triangle PQR 
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[4(2 - 5) + 3(5 - 6) + 6(6 - 2)]
= 1/2[-12 - 3 + 24]
= 1/2 x 9
= 9/2 
  Therefore, area of triangle PQR = 9/2 square units.
    
(ii) Taking C as origin, CB as x-axis, and CD as y-axis.
  The coordinates of vertices P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.

  Now, Area of triangle PQR 
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[12(6 - 3) + 13(3 - 2) + 10(2 - 6)]
= 1/2[36 + 13 - 40]
= 1/2 x 9
= 9/2 
  Therefore, area of triangle PQR = 9/2 square units.

  It can be observed that the area of the triangle is same in both the cases.
    

Question-6 :-  The vertices of a Δ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the Δ ADE and compare it with the area of Δ ABC.

Solution :-
  Given that : AD/AB = AE/AC = 1/4
  AD/(AD+DB) = AE/(AE+EC) = 1/4
  AD/DB = AE/EC = 1/3
  coordinate geometry
  Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3.

  Coordinates of point D = [(1x1 + 3x4)/(1+3), (1x5 + 3x6)/(1+3)] = [(1+12)/4, (5+18)/4] = (13/4, 23/4)
  Coordinates of point E = [(1x7 + 3x4)/(1+3), (1x2 + 3x6)/(1+3)] = [(7+12)/4, (2+18)/4] = (19/4, 20/4)

  Now, Area of triangle ADE 
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[4(23/4 - 20/4) + 13/4(20/4 - 6) + 19/4(6 - 23/4)]
= 1/2[3 - 13/4 + 19/16]
= 1/2[(48 - 52 + 19)/16]
= 15/32 square units

  Now, Area of triangle ABC 
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[4(5 - 2) + 1(2 - 6) + 7(6 - 5)]
= 1/2[12 - 4 + 7]
= 1/2 x 15 
= 15/2 square units

  Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16.
    

Question-7 :-  Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of Δ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do yo observe? [Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of Δ ABC, find the coordinates of the centroid of the triangle.

Solution :-
coordinate geometry
(i) Median AD of the triangle will divide the side BC in two equal parts.
  Therefore, D is the mid-point of side BC.
  Coordinates of D = [(6+1)/2, (5+4)/2] = (7/2, 9/2)
    
(ii) Point P divides the side AD in a ratio 2:1.
  Coordinates of P = [(2x7/2 + 1x4)/(2+1), (2x9/2 + 1x2)/(2+1)] = (11/3, 11/3)
    
(iii) Median BE of the triangle will divide the side AC in two equal parts.
  Therefore, E is the mid-point of side AC.
  Coordinates of E = [(4+1)/2, (2+4)/2] = (5/2, 3)

  Point Q divides the side BE in a ratio 2:1.
  Coordinates of Q = [(2x5/2 + 1x6)/(2+1), (2x3 + 1x5)/(2+1)] = (11/3, 11/3)

  Median CF of the triangle will divide the side AB in two equal parts. 
  Therefore, F is the mid-point of side AB.
  Coordinates of F = [(4+6)/2, (2+5)/2] = (5, 7/2)

  Point R divides the side CF in a ratio 2:1.
  Coordinates of R = [(2x5 + 1x1)/(2+1), (2x7/2 + 1x4)/(2+1)] = (11/3, 11/3)
    
(iv) It can be observed that the coordinates of point P, Q, R are the same.
  Therefore, all these are representing the same point on the plane. 
  i.e., the centroid of the triangle.
    
(v) Consider a triangle, ΔABC, having its vertices as A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).
  Median AD of the triangle will divide the side BC in two equal parts. 
  Therefore, D is the mid-point of side BC.

  Coordinates of D = [(x₂+x₃)/2, (y₂+y₃)/2]
  Let the centroid of this triangle be O.
  Point O divides the side AD in a ratio 2:1.
  Coordinates of O = [(x₁+x₂+x₃)/2, (y₁+y₂+y₃)/2]
    

Question-8 :- ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Solution :-
  P is the midpoint of side AB.
  coordinate geometry
  Therefore, the coordinates of P are [(-1-1)/2, (-1+4)/2] = (-1, 3/2)
  Similarly, the coordinates of Q, R and S are (2, 4), (5, 3/2) and (2, -1) respectively.
  Length of PQ = √(-1-2)² + (3/2-4)² = √61/4
  Length of QR = √(2-5)² + (4-3/2)² = √61/4
  Length of RS = √(5-2)² + (3/2+1)² = √61/4
  Length of SP = √(2+1)² + (-1-3/2)² = √61/4
  Length of PR = √(-1-5)² + (3/2-3/2)² = 6
  Length of QS = √(2-2)² + (4+1)² = 5
  It can be observed that all sides of the given quadrilateral are of the same measure.
  However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.
    
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