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 a ABC where B is (1, 5) and C (7, -2) is equal to 2 units.

Answer 1

coordinates A can be given by using section formula for internal division,

A = \((\frac{-5+3k}{k+1},\frac{1+5k}{k+1})\)

and B (1,5), C (7,−2) 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) 

= \(\frac{1}2\) |x₁(y₂ - y₃)+x₂(y₃ - y₁)+x₃(y₁ - y₂)| 

Area of ∆ABC

Solving above we get,

\(|\frac{14k - 66}{k+1}|\) = 4

Taking positive sign, 14k−66=4k+4 

10k = 70 

k = 7 

Taking negative sign we get, 

14k−66=−4k−4 

18k = 62

k = \(\frac{62}{18}\) = \(\frac{31}9\)