TOPICS
Unit-7(Examples)

Example-1 :-  Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.

Solution :-
  Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. 
  We have
  PQ =  √(3+2)² + (2+3)² = √(5)² + (5)² = √50
  QR =  √(-2-2)² + (-3-3)² = √(-4)² + (-6)² = √52
  PR =  √(3-2)² + (2-3)² = √(1)² + (-1)² = √2
  Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle.
  Also, PQ² + PR² = QR², 
  (√50)² + (√2)² = (√52)²
  50 + 2 = 52
  52 = 52
  L.H.S = R.H.S
  by the converse of Pythagoras theorem, we have ∠ P = 90°. Therefore, PQR is a right triangle.
    

Example-2 :-  Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.

Solution :-
  Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. 
  One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. 
  Now, 
  AB =  √(1-4)² + (7-2)² = √(3)² + (5)² = √34
  BC =  √(4+1)² + (2+1)² = √(5)² + (3)² = √34
  CD =  √(-1+4)² + (-1-4)² = √(3)² + (-5)² = √34
  DA =  √(1+4)² + (7-4)² = √(5)² + (3)² = √34
  AC =  √(1+1)² + (7+1)² = √(2)² + (8)² = √68
  BD =  √(4+4)² + (2-4)² = √(8)² + (-2)² = √68
  Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. 
  Thereore, ABCD is a square. 
    

Example-3 :-  Figure shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer. coordinate geometry

Solution :-
  Using the distance formula, we have
  AB =  √(6-3)² + (4-1)² = √(3)² + (3)² = √18 = 3√2
  BC =  √(8-6)² + (6-4)² = √(2)² + (2)² = √8 = 2√2
  AC =  √(8-3)² + (6-1)² = √(5)² + (5)² = √50 = 5√2
  Since, AB + BC = AC, 
  3√2 + 2√2 = 5√2
  5√2 = 5√2
  L.H.S = R.H.S
  So, we can say that the points A, B and C are collinear. 
  Therefore, they are seated in a line.
    

Example-4 :-  : Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).

Solution :-
  Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
  We are given that AP = BP. 
  So, AP² = BP² 
  i.e., (x – 7)² + (y – 1)² =( x – 3)² + (y – 5)² 
  i.e., x² – 14x + 49 + y² – 2y + 1 = x² – 6x + 9 + y² – 10y + 25 
  i.e., x – y = 2
  which is the required relation.
    

Example-5 :-  Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).

Solution :-
  We know that a point on the y-axis is of the form (0, y). 
  So, let the point P(0, y) be equidistant from A and B. 
  We are given that AP = BP. 
  So, AP² = BP² 
  Then (6 – 0)² + (5 – y)² = (– 4 – 0)² + (3 – y)²
  i.e., 36 + 25 + y² – 10y = 16 + 9 + y² – 6y 
  i.e., 4y = 36 
  i.e., y = 9 
  So, the required point is (0, 9).
    

Example-6 :-  Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.

Solution :-
  Let P(x, y) be the required point. 
  Given that : x₁ = 4, y₁ = -3 ; x₂ = 8, y₂ = 5 ; m₁ = 3, m₂ = 1
  Using the section formula, we get
section formula
  x = [(3x8 + 1x4)/(3 + 1)] = (24 + 4)/4 = 28/4 = 7
  y = [(3x5 + 1x(-3))/(3 + 1)] = (15 - 3)/4 = 12/4 = 3
  Therefore, (7, 3) is the required point.
    

Example-7 :-  In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?

Solution :-
  Let (– 4, 6) divide AB internally in the ratio m1 : m2. 
  Using the section formula, we get
  (– 4, 6) = [(3m₁ - 6m₂)/(m₁ + m₂), (-8m₁ + 10m₂)/(m₁ + m₂)]
  -4 = (3m₁ - 6m₂)/(m₁ + m₂)
  -4m₁ - 4m₂ = 3m₁ + 6m₂
  -4m₁ - 3m₁ = -6m₂ + 4m₂
        -7m₁ = -2m₂
       m₁/m₂ = 2/7
       m₁:m₂ = 2:7
  Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.
    

Example-8 :-  Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).

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(-7) + 2x2)/(1 + 2), (1x4 + 2x(-2))/(1 + 2)] i.e., (-1, 0)

  Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
  [(2x(-7) + 1x2)/(2 + 1), (2x4 + 1x(-2))/(2 + 1)] i.e., (-4, 2)

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

Example-9 :-  Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.

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
  [(-k + 5)/(k + 1), (-4k - 6)/(k + 1)]
  This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
  Therefore,
  (-k + 5)/(k + 1) = 0
  -k + 5 = 0
  -k = -5
   k = 5
  i.e., the ratio is 5 : 1. 
  Putting the value of k = 5, we get the point of intersection as (0, -13/3).
    

Example-10 :-  If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Solution :-
  We know that diagonals of a parallelogram bisect each other. 
  So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
  [(6 + 9)/2, (1 + 4)/2] = [(8 + p)/2, (2 + 3)/2]
  (15/2, 5/2) = [(8 + p)/2, 5/2]
  15/2 = (8 + p)/2
  15 = 8 + p
   p = 15 - 8
   p = 7
    

Example-11 :-  Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5).

Solution :-
  The area of the triangle formed by the vertices A(1, –1), B(– 4, 6) and C (–3, –5), 
  By using the area of triangle formula, is given by 
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[1(6 + 5) + (-4)(-5 + 1) + (-3)(-1 - 6)]
= 1/2[11 + 16 + 21]
= 1/2 x 48 
= 24
  So, the area of the triangle is 24 square units.
    

Example-12 :-  Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, – 4).

Solution :-
  The area of the triangle formed by the vertices A(5, 2), B(4, 7) and C (7, – 4), 
  By using the area of triangle formula, is given by 
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[5(7 + 4) + 4(-4 - 2) + 7(2 - 7)]
= 1/2[55 - 24 - 35]
= 1/2 x (-4)
= -2
  Since area is a measure, which cannot be negative, we will take the numerical value of – 2, i.e., 2. 
  So, the area of the triangle is 2 square units.
    

Example-13 :-  Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2) and R(–3, 4).

Solution :-
  The area of the triangle formed by the vertices P(–1.5, 3), Q(6, –2) and R(–3, 4), 
  By using the area of triangle formula, is given by 
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[(-1.5)(-2 - 4) + 6(4 - 3) + (-3)(3 + 2)]
= 1/2[9.0 + 6 - 15]
= 1/2 x 0
= 0 
  If the area of a triangle is 0 square units, then its vertices will be collinear.
    

Example-14 :- Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

Solution :-
  Since the given points are collinear, the area of the triangle formed by them must be 0, i.e.,
  1/2[2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
  1/2[2k + 6 - 24 + 18 - 6k] = 0
  1/2[-4k + 0] = 0
  -4k = 0
    k = 0
    

Example-15 :-  If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Solution :-
  By joining B to D, you will get two triangles ABD and BCD.
  Now, the area of Δ ABD 
= 1/2[(-5)(-5 - 5) + (-4)(5 - 7) + 4(7 + 5)]
= 1/2[50 + 8 + 48]
= 1/2 x 106
= 53 square units

  Also, the area of Δ BCD 
= 1/2[(-4)(-6 - 5) + (-1)(5 + 5) + 4(-5 + 6)]
= 1/2[44 - 10 + 4]
= 1/2 x 38
= 19 square units

  So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.
    
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