Question

In $\Delta ABC,A\equiv (6,3),B\equiv (-3,5),C\equiv (4,-2)$. $P$ is any point $(x,y)$. Show how that $\frac{\Delta \left(PBC\right)}{\Delta \left(ABC\right)}=\left|\frac{x+y-2}{7}\right|$

Answer 1

In $\Delta$ABC,

$A=(6,3)$;

$B(-3,5)$;

$C=(4,-2)$;

$P=(x,y)$

$Ar\left(\Delta ABC\right)=\frac{1}{2}\times \left(BC\right)\times h_1$

$Ar\left(\Delta PBC\right)=\frac{1}{2}\times \left(BC\right)\times h_2$

BC, equation

$\Rightarrow m=\frac{7}{-7}=-1$

$x+y=2$

$h_1=\frac{|6+3-2|}{\sqrt{1^2+1^2}}=\frac{7}{\sqrt{2}}$

$h_2=\frac{|x+y-z|}{\sqrt{2}}$

$\frac{\Delta \left(PBC\right)}{\Delta \left(ABC\right)}=\frac{\frac{1}{2}\times BC\times h_2}{\frac{1}{2}\times BC\times h_1}$

$=\frac{h_2}{h_1}=\frac{|x+y-2|}{7}$.