FINDING AREA OF TRIANGLES WITH THREE VERTICES

Problem 1 :

If the three points (h, 0) (a, b) and (0, k) lie on a straight line, then using the area of the triangle formula show that (a/h) + (b/k) = 1, where h, k ≠ 0

Solution :

Let A(h, 0) B (a, b) and C (0, k) are the three points

Since the three points A (h, 0) B (a, b) and C (0, k) lie on a straight line we can say that the three points are collinear

So , area of triangle ABC = 0

(1/2) [ (h b + a k + 0) – (0 + 0 + kh) ] = 0

[ h b + a k  – k h ] = 0 x 2

h b + a k - k h = 0

h b + a k  = k h

Divided (k h) on both sides,

(h b)/(k h) + (a k)/(k h)  = (k h)/(k h)

(b/k) + (a/h) = 1

(a/h) + (b/k) = 1

Problem 2 :

Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are (0, -1) (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Solution :

Let A (0,-1) B (2,1) and C (0,3) are the vertices of the triangle. D,E and F are the midpoints of the side AB,BC and CA respectively.

Midpoint of AB  =  (x₁ + x₂)/2 , (y₁ + y₂)/2

  =  (0 + 2)/2 , (-1 + 1)/2

  =  2/2 , 0/2

  =  D (1,0)

Midpoint of BC  =  (x₁ + x₂)/2 , (y₁ + y₂)/2

  =  (2 + 0)/2 , (1 + 3)/2

  =  2/2 , 4/2

  =  E (1,2)    

Midpoint of CA  =  (x1+x2)/2 , (y1+y2)/2

  =  (0 + 0)/2 , (3 + (-1))/2

  =  0/2 , 2/2

  =  F (0,1)    

  =  (1/2) [(0 + 6 + 0) – (-2 + 0 + 0)]

  =  (1/2) [6+2]

  =  (1/2) [8]

  =  4 square units

  =  (1/2) [(2 + 1 + 0) – (0 + 0 + 1)]

  =  (1/2) [3-1]

  =  (1/2) [2]

  =  1 square units

Area of triangle ABC: Area of triangle DEF

4 : 1

Problem 3 :

If the points A(-3, 9), B (a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.

Solution :

Since these point are collinear,

(-3b - 5a + 36) - (9a + 4b + 15) = 0

-3b - 5a + 36 - 9a - 4b - 15 = 0

-14a - 7b + 21 = 0

Dividing by 7, we get 

-2a - b + 3 = 0

-2a - b = -3

2a + b = 3 --------(1)

Given that  a + b = 1 --------(2)

(1) - (2)

2a + b - a - b = 3 - 1

a = 2

Applying the value of a, we get

2(2) + b = 3

4 + b = 3

b = 3 - 4

b = -1

So, the values of a and b are 2 and -1 respectively.

Problem 4 :

The area of a triangle is 5 sq.units. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) where

y = x + 3

Find the coordinates of the third vertex.

Solution :

Area of the triangle formed by by the above vertices

= 5 square units

(1/2)  [(-4 + 3y + x) - (3 - 2x + 2y)] = 5

[-4 + 3y + x - 3 + 2x - 2y] = 10

3x + y - 7 = 10

3x + y = 17 ----(1)

y = x + 3 ----(2)

Applying the value of y in (1), we get

3x + x + 3 = 17

4x = 17 - 3

4x = 14

x = 14/4

x = 3.5

Problem 5 :

Find the area of a triangle formed by the lines 3x + y - 2 = 0, 5x + 2y - 3 = 0 and 2x - y - 3 = 0

Solution :

3x + y - 2 = 0 ----(1)

5x + 2y - 3 = 0 ------(2)

2x - y - 3 = 0 ------(3)

(1) ⋅ 2 - (2)

6x + 2y - 4 - (5x + 2y - 3) = 0

6x + 2y - 4 - 5x - 2y + 3 = 0

1x - 1 = 0

x = 1

Applying the value of x in (1), we get

3(1) + y - 2 = 0

3 + y - 2 = 0

1 + y = 0

y = -1

So, the point of intersection of (1) and (2) is (1, -1).

(2) + 2 ⋅ (3)

5x + 2y - 3 + 2(2x - y - 3) = 0

5x + 2y - 3 + 4x - 2y - 6 = 0

9x - 9 = 0

9x = 9

x = 1

Applying the value of x in (2), we get

5(1) + 2y - 3 = 0

5 + 2y - 3 = 0

2y + 2 = 0

2y = -2

y = -1

So, the point of intersection of (2) and (3) is (1, -1).

(1) + (3)

3x + y - 2 + 2x - y - 3 = 0

5x - 5 = 0

5x = 5

x = 5/5

x = 1

Applying the value of x n (1)

3(1) + y - 2 = 0

3 - 2 + y = 0

1 + y = 0

y = -1

So, the point of intersection of (1) and (3) is (1, -1).

Since all points are the same, these are collinear it may not create a triangle. Then its area will become 0.

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