Question

The coordinates of the points A,B and C are (6,3), (-3,5) (4,-2) respectively. P (x,y) is any point in the plane. Show that $ar\left(PBC\right)÷ar\left(triangleABC\right)=|(x+y-2)÷\left(7\right)|$

Answer 1

Area of $\Delta PBC==\frac{1}{2}|x(5-(-2\left)\right)+(-3)(-2-y)+4(y-5)|$
$=\frac{1}{2}|7x+6+3y+4y-20|$
$=\frac{1}{2}|7x+7y-14|=\frac{7}{2}|x+y-2|$
$||{|}^{ly}$, Area of $\Delta ABC=\frac{1}{2}|6(5-(-2\left)\right)+(-3)(-2-3)+4(3-5)|$
$=\frac{1}{2}|6\times 7+(-3)(-5)+4(-2)|$
$=\frac{1}{2}|42+15-8|=\frac{1}{2}|49|=\frac{49}{2}$
$\therefore \frac{Areaof\Delta PBC}{Areaof\Delta ABC}=\frac{\frac{7}{2}|x+y-2|}{\frac{49}{2}}=\frac{|x+y-2|}{7}$
$=\left|\frac{x+y-2}{7}\right|$