quotient between areas of triangles I am confused  for answer from the following link,  problem 67:
http://www.naec.ge/images/doc/EXAMS/exams-2011-gat-5-ivlisi.pdf 

The problem states that the vertices A and C of a quadrilateral  ABCD  are  on $y-$axis, and the coordinates  of B and D  are given.
We are asked to find the ratio of areas of ABC and ADC.
From the given figure it is clear that it will be the ratio between heights  of both triangles or 5.4/3  which is equal to 1.8, but the answer is different (1.5). Why? thanks
 A: You have the right idea, but you’re using the wrong heights. The side that the two triangles share is $\overline{AC}$. Measured from this side, the height of $\triangle ABC$ is 9, and the height of $\triangle ADC$ is 6, so the ratio is $\frac{9}{6} = 1.5$.
A: The area of ABC is $\frac{\overline{AC} \times 9}2$ and the area of $ABD = \frac{\overline{AC} \times 6}2$, which explains the ratio. You were using the $y$ coordinates instead of the $x$'s.
A: Sorry, the answer $1.5$ is right, $1.8$ is wrong.
Look at $\triangle ABC$, viewed as having base $AC$.  Then its height is the perpendicular distance from $B$ to the $y$-axis.  this perpendicular distance is the absolute value of the $x$-coordinate, namely $9$.
Why? Imagine dropping a perpendicular from $B$ to the $y$-axis. The length of the perpendicular is the distance you must travel in the $x$-direction to get to the $y$-axis.  that's $9$.
Similarly, if you view $\triangle ADC$ as having base $AC$, the height is $6$.
The ratio of the areas is therefore $9/6$.  
