If I have a right angled triangle, with the following vertices
$$v1 = (0,0)$$ $$v2 = (8,0)$$ $$v3 = (0,6)$$ $$v4 = (2,2) \text{ - this point lies within the triangle}$$
I know the area of the whole triangle is $24$. However, I am struggling to calculate the area of the $3$ inner triangles. ($v1v2v4$, $v1v3v4$ and $v2v3v4$).
How can I calculate this?
$V_1V_3V_4$: $$S=\frac12 2 \times 6=6$$
$V_1V_2V_4$: $$S=\frac12 2 \times 8=8$$
$V_2V_3V_4$: $$S=24-6-8=10$$
In another way:
$$S_234=\frac12 (V_4H)\times (V_3V_2)$$
$$V_3V_2=\sqrt{6^2+8^2}=10$$
$$V_3V_2: y=6-\frac34 x$$
$$V_4H: y=-\frac23+\frac43 x$$
H is intersect of $V_3V_2$ and $V_4H$:
$$6-\frac34 x=-\frac23+\frac43 x$$
$$x_H=\frac{16}5$$ $$y_H=\frac{18}5$$
$$V_4 H=\sqrt{(\frac{16}5-2)^2+(\frac{18}5-2)^2}=2$$
$$S_234=\frac12 (V_4H)\times (V_3V_2)=\frac12 10 \times 2 =10$$