#### 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 Δ*ABC* where *B* is (1, 5) and *C*(7, −2) is equal to 2 units.

#### Solution

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,}y_{1})(x_{2},y_{2}) and (x_{3},y_{3})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`