Question

The point A divides the join of P (–5, 1)  and Q (3, 5) in the ratio k : 1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C (7, –2) is equal to 2 units.

Answer 1

GIVEN: point A divides the line segment joining P (–5, 1) and Q (3, –5) in the ratio k : 1

Coordinates of point B (1, 5) and C (7, –2)

TO FIND: The value of k

PROOF: point A divides the line segment joining P (–5, 1) and Q (3, –5) in the ratio k : 1

So the coordinates of A are `[(3k+1)/(k+1),(5k+1)/(k+1)]`  

We know area of triangle formed by three points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by   

Δ `= 1/2|(x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)|` 

Now Area of ΔABC = 2 sq units. 

Taking three points A `[(3k+1)/(k+1),(5k+1)/(k+1)]`, B(1, 5) and C(7, −2) 

`2=1/2 {((3k-5)/(k+1))5-2+7((5k+1)/(k+1))}-((5k+1)/(k+1)) +35-2((3k-5)/(k+1))`

`2=1/2 {((15k-25)/(k+1))-2+((35k+7)/(k+1))}-((5k+1)/(k+1))+35-((6k-10)/(k+1))` 

`2=1/2 ((15k-25-2k+35k+7)/(k+1))-((5k+1+35k + 35-6k+10)/(k+1))` 

`2=1/2 (((48k-20)-(34k+46))/(k+1))`

`2=1/2 ((14k-66)/(k+1))` 

`2=((7k-33)/(k+1))` 

`+2=((7k-33)/(k+1))` 

⇒ 7k – 33 = ±2(k + 1) 

⇒ 7k – 33 = 2(k + 1), 7k – 33 = –2(k + 1) 

⇒ 7k – 33 = 2k + 2, 7k – 33 = –2k – 2 

⇒ 7k – 2k = 33 + 2, 7k + 2k = +33 – 2 

⇒ 5k = 35, 9k = 31 

⇒ `k = 35/5, k = 31/9 `

⇒ `k = 7, k = 31/9` 

Hence k = 7 or `31/9`