Question
The coordinates of the four vertices of a quadrilateral are $$\left( { - 2,\,4} \right),\,\left( { - 1,\,2} \right),\,\left( {1,\,2} \right)$$ and $$\left( {2,\,4} \right)$$ taken in order. The equation of the line passing through the vertex $$\left( { - 1,\,2} \right)$$ and dividing the quadrilateral in two equal areas is :
A.
$$x+1=0$$
B.
$$x+y=1$$
C.
$$x-y+3=0$$
D.
none of these
Answer :
$$x-y+3=0$$
Solution :
The points $$A,\,B,\,C,\,D$$ form a trapezium whose line of symmetry is the $$y$$-axis.
Let $$BE$$ be the required line where $$ED = \lambda .$$
$$\eqalign{ & {\text{ar}}\left( {BCDE} \right) = \frac{1}{2}{\text{ar}}\left( {ABCD} \right) \cr & \Rightarrow \frac{1}{2}.\left( {2 + \lambda } \right).2 = \frac{1}{2}.\frac{1}{2}\left( {4 + 2} \right).2 \cr & \Rightarrow \lambda = 1 \cr & \therefore {\text{ the point }}E = \left( {1,\,4} \right) \cr} $$
So, the required line $$BE$$ is $$y - 2 = \frac{{4 - 2}}{{1 + 1}}\left( {x + 1} \right).$$
The points $$A,\,B,\,C,\,D$$ form a trapezium whose line of symmetry is the $$y$$-axis.
Let $$BE$$ be the required line where $$ED = \lambda .$$
$$\eqalign{ & {\text{ar}}\left( {BCDE} \right) = \frac{1}{2}{\text{ar}}\left( {ABCD} \right) \cr & \Rightarrow \frac{1}{2}.\left( {2 + \lambda } \right).2 = \frac{1}{2}.\frac{1}{2}\left( {4 + 2} \right).2 \cr & \Rightarrow \lambda = 1 \cr & \therefore {\text{ the point }}E = \left( {1,\,4} \right) \cr} $$
So, the required line $$BE$$ is $$y - 2 = \frac{{4 - 2}}{{1 + 1}}\left( {x + 1} \right).$$