I have a problem related to triangles. Please give me some hint to progress.
Suppose we have the following coordinates $A(-2,3)$,$B(1,-1)$,$C(-1,-1)$.
From point $A$, draw a line which divides the area of triangle $ABC$ in the ratio $1:2$.
I have calculated the length of the sides:
$[AB]=5$,$[AC]=\sqrt{17}$ and $[BC]=2$.
I have searched online for what this line should be, like medians, altitudes or etc and found this,
where is written:
If one median of a triangle is drawn, the second median to be drawn will divide the areas of the two triangles formed by the first median in the ratio 1:2.
But is it what I need? I can write the equation of lines which go through $AC$,$BC$,$AB$, but how do I find the equation of the required line?
Best Answer
Let $D$ be a point on $\overline{BC}$ between $B$ and $C$. Observe that $\triangle BDA$ and $\triangle DCA$ have the same height, $3-(-1)=4$. Let $x=|BD|$ and $y=|DC|$. The area of $\triangle BDA$ is $2x$, the area of $\triangle DCA$ is $2y$, and $x+y=|BC|=2$, so $y=2-x$. Now all you have to do is find $x$ so that $2(2-x)=2(2x)$.