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?


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 $2(2-x)=2(2x)$.

  • $\begingroup$ some question please why BDA and DCA have same height? $\endgroup$ – dato datuashvili Jul 4 '12 at 19:58
  • $\begingroup$ @dato: Because the height is measured perpendicular to the baseline, which is the line $y=-1$ for both triangles (and also for $\triangle ABC$). $\endgroup$ – Brian M. Scott Jul 4 '12 at 20:00
  • $\begingroup$ ok can we generalize this problem?so for example if instead of given coordinates,we know that $A(x_1,y_1)$,$B(x_2,y_2)$,$C(x_3,y_3)$?how could i do in this case? $\endgroup$ – dato datuashvili Jul 4 '12 at 20:04
  • $\begingroup$ @dato: Find the equation of the line $\overline{BC}$ and calculate the perpendicular distance from $A$ to that line to get the common height of $\triangle BDA,\triangle DCA$, and $\triangle ABC$; the rest would then be similar to the easier case above. $\endgroup$ – Brian M. Scott Jul 4 '12 at 20:07
  • 1
    $\begingroup$ @dato: That’s right: you can get any desired ratio using the same basic idea. $\endgroup$ – Brian M. Scott Jul 4 '12 at 20:11

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