Answer: The correct option is (D) (4, -6).
Step-by-step explanation: Given that the areas of the triangles ADC and DCB are in the ratio 3 : 4.
We are to find the co-ordinates of point C.
From the diagram, we note that
the co-ordinates of point A and B are A(1, -9) and B(8,-2).
So, the length of the line segment AB, calculated by distance formula, is
[tex]AB=\sqrt{(1-8)^2+(-9+2)^2}=\sqrt{49+49}=7\sqrt2.[/tex]
Now, area of ΔADC is
[tex]A_{ABC}=\dfrac{1}{2}\times AC\times CD,[/tex]
and area of ΔDCB is
[tex]A_{DCE}=\dfrac{1}{2}\times BC\times CD.[/tex]
According to the given information, we have
[tex]A_{ADC}:A_{DCB}=3:4\\\\\\\Rightarrow \dfrac{A_{ADC}}{A_{DCE}}=\dfrac{3}{4}\\\\\\\Rightarrow \dfrac{\frac{1}{2}\times AC\times CD}{\frac{1}{2}\times BC\times CD}=\dfrac{3}{4}\\\\\\\Rightarrow \dfrac{AC}{BC}=3:4.[/tex]
So, the point C divides the line segment AB internally in the ratio 3 : 4.
We know that
if a point divides a line segment with end-points (a, b) and (c, d) internally in the ration m : n, then its co-ordinates are
[tex]\left(\dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n}\right).[/tex]
Since point C divides the line segment AB with end-points A(1, -9) and B(8, -2) internally, so the co-ordinates of point C will be
[tex]\left(\dfrac{3\times 8+4\times 1}{3+4},\dfrac{3\times (-2)+4\times (-9)}{3+4}\right)\\\\\\=\left(\dfrac{28}{7},\dfrac{-42}{7}\right)\\\\\\=(4, -6).[/tex]
Thus, the co-ordinates of point C are (4, -6).
Option (D) is correct.
