Question

Four points A (6, 3), B (−3, 5), C (4, −2) and D (x, 3x) are given in such a way that \[\frac{\Delta DBC}{\Delta ABC} = \frac{1}{2}\]. Find x.

Answer 1

We know that the area of a triangle with vertices \[\left( x_1 , y_1 \right), \left( x_2 , y_2 \right)\text{ and }\left( x_3 , y_3 \right)\] is given by:

\[\text{ Area }= \frac{1}{2}\left\{ x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right\}\]

\[\therefore\text{ Area of ∆ DBC }= \frac{1}{2}\left| - 3\left( - 2 - 3x \right) + 4\left( 3x - 5 \right) + x\left( 5 + 2 \right) \right|\]

\[\Rightarrow\text{ Area of ∆ DBC }= 7\left( 2x - 1 \right)\]

\[\therefore\text{ Area of ∆ ABC }= \frac{1}{2}\left| 6\left( 5 + 2 \right) - 3\left( - 2 - 3 \right) + 4\left( 3 - 5 \right) \right|\]

\[= \frac{49}{2}\]
It is given that \[\frac{\Delta DBC}{\Delta ABC} = \frac{1}{2}\].

\[\therefore \frac{7\left( 2x - 1 \right) \times 2}{49} = \frac{1}{2}\]

\[\therefore x = \frac{11}{8}\]