We need to answer two questions based on the following paragraph.
At times the methods of coordinates become effective in solving problems of properties of triangles. We may choose one vertex of the triangle as origin and one side passing through this vertex as x-axis. Thus without loss of generality, we can assume that every triangle ABC has a vertex situated at $(0,0)$, another at $(x,0)$ and third vertex at $(h,k)$.
Question $1$: If in triangle ABC, AC=$3$, BC=$4$, medians AD and BE are perpendicular then find the area of triangle ABC.
Question $2$: Suppose the bisector AD of the interior angle A of triangle ABC divides side BC into segments BD=$4$, DC=$2$ then
A) $b\gt c$ and $c\lt4$
B) $2\lt b\lt6$ and $c\lt1$
C) $2\lt b\lt6$ and $4\lt c\lt12$
D) $b\lt c$ and $c\gt4$
For the first question, I drew perpendicular medians, used the fact that the centroid divides median in the ratio $2:1$, used Pythagoras and then Herons formula to calculate the area.
For second question, I used the sum of two sides greater than the third side. Also, angle bisector divides the opposite side in the same ratio as the ratio of the sides containing the angle.
How to do these questions using the method of coordinates?
We have to draw a diagram to help us to visualize the problem and what we should do to solve it. We selected the vertices as suggested by the problem statement. Our first aim is to find the $x\text{-}$ and $y\text{-coordinates}$ of the vertex $A$, i.e., $h$ and $k$. Since there are two unknowns, we need to derive two equations using methods in coordinate geometry.
To derive the first of the two sought equations do the following.
It is given that the medians $AD$ and $BE$ are perpendicular to each other. Hence we have, $$m_{AD}\times m_{BE} = -1,$$ where $m_{AD}$ and $m_{BE}$ are the slopes of the medians $AD$ and $BE$ respectively
To determine the slope of a line segment, we need to know the coordinates of two points lying on that line. Therefore, we have to find the coordinates of the points $D$ and $E$, the mid points of $BC$ and $CA$ respectively.
The second equation can be deduced as shown below.
Since we now know the coordinates of $A$ and $C$, we can write down the Euclidean Distance Formula to express the length of the line segment $AC$. When we equate this expression to the length of $AC$ (which is 3), we get the second equation.
Now we have two equations we needed to determine $h$ and $k$.
Finally, to find the area of the $\triangle ABC$, we can use the Determinant Method given blow. $$\text{Area of Triangle}=\dfrac{1}{2} \begin{vmatrix} x_A & y_A & 1\\ x_B & y_B & 1\\ x_C & y_C & 1\\ \end{vmatrix} $$
If you do not like this method, you can find the length $d$ of the perpendicular drop from the vertex $A$ to side $BC$, the equation of which is $y=0$. Then we have, $$\text{Area of Triangle}=\dfrac{d\times BC}{2}.$$
The values of $h$, $k$, and the area are $\dfrac{3}{2}$, $\dfrac{\sqrt{11}}{2}$, and $\sqrt{11}$ respectively.
Since you now know how to solve this type of problems in coordinate geometry, we hope you would be able to find the solution to the second question.