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

• A. undefined
• B. undefined
• C. undefined
• D. undefined

Answer 1

Using section formula, the coordinates of point A which divides the join of P(-5, 1) and Q(3, 5) in the ratio k:1 is
$(\frac{3k-5}{k+1},\frac{5k+1}{k+1})$

Fig. C - 9
Also B is (1, 5) and C is (7, -2).
Area of $\Delta ABC=\frac{1}{2}\left\{x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right\}$
$=\pm 2\left(given\right)$
$or\frac{1}{2}\left\{\frac{3x-5}{k+1}\right(5+2)+(-2-\frac{5x+1}{k+1})+7(\frac{5k+1}{k+1}-5)\}$
$=\pm 2$
$or21k-35-7k-3-28=\pm 4(k+1)$
$\therefore 14k-66=4k+4$
$or14k-66=-4k-4$
$\therefore 10k=70or18k=62$
$\therefore k=7or\frac{31}{9}$