Question

the point P divides the line segment joining A ( - 5,1 )and B (3, 5 )in ratio K:1 internal. Q and R (1, 5 )and( 7,-2) respectively if the area of triangle PQR is 2 then find the value of K.

Answer 1

A (-5, 1)....(k).......P ......(1)......B (3,5)

using section formula,

point A =[(3K - 5)/(1 + K) , (5K + 1)/(K + 1) ]

now, P , Q and R points form a triangle PQR.

using formula,

area of triangle = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

so, area of ∆PQR =1/2|(3K - 5)/(K + 1)(5+2) + 1{-2 - (5K+1)/(K+1)} +7{(5K+1)/(K+1)) -5} |

⇒ ± 2 × 2 = [ (3K-5)(7)/(1+K) +(-2K-2-5K-1)/(1+K) +7(5K+1-5K-5)/(1+K)]

⇒ ±4(1+K) ={21K -35 -7K -3 -28 }

⇒ ±4(1 +K) =14K -66

now,

4( 1 + K) = 14K -66

⇒4 + 4K = 14K -66

⇒ 70 =10 K

K =7

again,

⇒ -4( 1+ K ) = 14K -66

⇒-4 -4K = 14 K -66

⇒ 62 = 18K

⇒ K = 31/9

hence , K = 7 , 31/9.

Answer 2

Answer:

P (-5, 1) ==========A =======Q(3,5)

Use section formula,

Point A =$\dfrac{(3K-5)}{(1+K)} ,\dfrac {(5K+1)}{(K+1)}$

now, A , B, and C points form ∆

use area of ∆ formula ,

ar∆ =${\dfrac{1/2|(3K-5)}{(K+1)(5+2)}} +1 {\dfrac{-2-(5K+1)}{(K+1)}} +7{\dfrac{(5K+1)}{(K+1)) -5}}$

$±4 ={\dfrac{(3K-5)(7)}{(1+K)}} +{\dfrac{(-2K-2-5K-1)}{(1+K)}} +{\dfrac{7(5K+1-5K-5)}{(1+K)}}$

±4(1+K) ={21K -35 -7K -3 -28 }

±4(1 +K) =14K -66

Now,

4( 1 + K) = 14K -66

4 + 4K = 14K -66

70 =10 K

K =7

Now again,

-4( 1+ K ) = 14K -66

-4 -4K = 14 K -66

62 = 18K

K = 31/9

Hence ,

$\bf{K = 7}, {\dfrac{31}{9}}$