The point A divid

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 = =