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 a ABC where B is (1, 5) and C (7, -2) is equal to 2 units.
coordinates A can be given by using section formula for internal division,
A = ( 
, 
)
and B (1,5), C (7,−2)
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Area of ∆ABC
= 
| 
(7) + 1(−2 − 
) + 7( 
- 5 ) |
But Area of ∆ABC = 2
∴ 
| 
(7) + 1(−2 − 
) + 7( 
- 5 ) | = 2
Solving above we get,
| 
| = 4
Taking positive sign, 14k−66=4k+4
10k = 70
k=7
Taking negative sign we get,
14k−66=−4k−4
18k = 62
k = 
= 
7