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 (x_1,y₁)(x₂,y₂) and (x₃,y₃)s given by   

Δ `= 1/2([x1y2+x2y3+x3y1)-(x2y1+x3y2+x1y3)]` 

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-6k+10)/(k+1))]` 

`2=1/2[((48-20-(34k+46))/(k+1))]` 

`2=1/2[((48k-20)-(34k+46))/(k+1))]` 

`2=1/2[((14k-66)/(k+1))]` 

`2=[((7k-33)/(k-1))]` 

`+-=((7k-33)/(k+1))]` 

⇒ 7k -33= ±2(k+1) 

⇒ 7k -33= 2(k+1), 7k-33= -2(k+1) 

⇒7k -33=2k +2, 7k+2k= +33- 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`