Find the area of a triangle $\triangle FGH$

Question

In triangle $\triangle FGH$, $GM$ is a median that lies on $4x-y=27$; height $HA$ lies on $x-y+3=0$; $F$ is $(4,5)$. Find the area of the triangle $\triangle FGH$. My attempt: F is not on any of the lines. Intersection of the lines (10,13). Then, i struggle.

Answer 1

With GF$\perp$HA, the slope of GF is $-\frac14$ and GF lies on $4y+x-24 =0$. G is the intersection of GF and GM, which is $G(\frac{12}5,\frac{27}5)$.

Let $H(a,4a-27)$. Then, the midpoint is $M(\frac {a+4}2, \frac{4a-22}2)$ and it satisfies $x-y+3=0$, or

$$\frac {a+4}2-\frac{4a-22}2+3=0 $$

which yields $a=\frac{32}3$ and $H(\frac{32}3, \frac{47}3)$. With the coordinates of $F,G,H$ known, the area is given by the formula below

$$Area = \frac12| x_F(y_G-y_H)+ x_G(y_H-y_F)+ x_H(y_F-y_G)| $$