Question

Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC. Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.

Answer 1

Let the vertices of the ∆ABC be A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)

Mid point of AB = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

(11, 7) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

`(x_1 + x_2)/2` = 11 x1 + x2 = 22   ...(1)

`(y_1 + y_2)/2` = 7 y1 + y2 = 14  ...(2)

Mid point of BC = `((x_2 + x_3)/2, (y_2 + y_3)/2)` 

⇒ (13.5, 4) = `((x_2 + x_3)/2, (y_2 + y_3)/2)`

`(x_2 + x_3)/2` = 13.5 x2 + x3 = 27   ...(3)

`(y_2 + y_3)/2` = 4 y2 + y3 = 8   ...(4)

Mid point of AC = `((x_1 + x_3)/2, (y_1 + y_3)/2)`

(9.5, 4) = `((x_1 + x_2)/2, (y_1 + y_3)/2)`

`(x_1 + x_3)/2` = 9.5 x1 + x3 = 19   ...(5)

`(y_1 + y_3)/2` = 4 y1 + y3 = 8   ...(6)

Add (1), (3) and (5)

2x₁ + 2x₂ + 2x₃ = 22 + 27 + 19

2(x₁ + x₂ + x₃) = 68

x₁ + x₂ + x₃ = 34

From (1) ⇒ x₁ + x₂ = 22

x₃ = 34 – 22 = 12

From (3) ⇒ x₂ + x₃ = 27

x₁ = 34 – 27 = 7

From (5) ⇒ x₁ + x₃ = 19

 x₂ = 34 – 19 = 15

Add (2), (4) and (6)

2y₁ + 2y₂ + 2y₃ = 14 + 8 + 8

2(y₁ + y₂ + y₃) = 30

y₁ + y₂ + y₃ = 15

From (2) ⇒ y₁ + y₂ = 14

y₃ = 15 – 14 = 1

From (4) ⇒ y₂ + y₃ = 18

y₁ = 15 – 8 = 7

From (6) ⇒ y₁ + y₃ = 8

y₂ = 15 – 8 = 7

The vertices of a ΔABC are A(7, 7), B(15, 7) and C(12, 1)

Area of ΔABC = `1/2[(x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3)]`

= `1/2[(7 + 84 + 105) - (84 + 15 + 49)]`

= `1/2[196 - 148]`

= `1/2 xx 48`

= 24 sq. units

Area of ΔPRQ = `1/2[(44 + 8 + 94.5) - (66.5 + 54 + 44)]`

= `1/2[176.5 - 164.5]`

= `1/2 xx 12`

= 6 sq. units