Two friends Seema and Aditya work in the same office at Delhi. In the Christmas vacations, both decided to go to their hometowns represented by Town A and Town B respectively in the figure given below. Town A and Town B are connected by trains from the same station C (in the given figure)in Delhi. Based on the given situation, answer the following questions:

(i) Who will travel more distance, Seema or Aditya, to reach to their hometown?

(ii) Seema and Aditya planned to meet at a location D situated at a point D. represented by the mid-point of the line joining the points represented by Town A and Town B. Find the coordinates of the point represented by the point D.

(iii) Find the area of the triangle formed by joining the points represented by A, B and C.

(i) Aditya has to travel more.

(ii) D ≡ (2.5, 4.5)

(iii) A (∆ABC) = 17 cm²

Formulae –

1. Distance between two points –

If A ≡ (x₁, y₁) & B ≡ (x₂, y₂), then distance between points AB is

2. Mid Point Formula –

If A ≡ (x₁, y₁) & B ≡ (x₂, y₂), then co-ordinates of the mid point joining points A & B are

3. Area of triangle by Hero’s formula –

If s is semi perimeter of a triangle and a,b,c are three sides of that triangle then area of the triangle is given by

Answer –

(i) From figure, co-ordinates of points A, B, C are

A ≡ (1, 7) , B ≡ (4, 2) , C ≡ (-4, 4)

Distance between points A and B is

………..(1)

Hence, length of segment AB = √34 cm

Distance between points A and C is

………..(2)

Hence, length of segment AC = √34 cm

Distance between points B and C is

………..(3)

Hence, length of segment BC = √68 cm cm

As town B is far from Station C as compared to town A.

Hence, has to travel more as compared to Seema.

(ii) By midpoint formula, co-ordinates of midpoint D of segment AB are

∴ D (x,y) = (2.5, 4.5)

Hence, co-ordinates of D ≡ (2.5, 4.5).

(iii) Now, perimeter of ∆ABC is

P = √34 + √34 + 2√17

∴ P = 2√34 + 2√17

Therefore, semi perimeter of ∆ABC is

By Hero’s formula, area of ∆ABC is

Here, a = 2√17, b = √34, c = √34

∴ s – a = √34 + √17 – 2√17 = √34 – √17

∴ s – b = √34 + √17 – √34 = √17

∴ s – c = √34 + √17 – √34 = √17

∴ A (∆ABC) = 17 cm²