Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given

Solution:

GIVEN: four points $A(6,3), B(-3,5) C(4,-2)$ and $D(x, 3 x)$ such that $\frac{\Delta D B C}{\Delta A B C}=\frac{1}{2}$

TO FIND: the value of x

PROOF:

We know area of the triangles formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by Area of triangle $=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

Now

Area of triangle DBC taking $D(x, 3 x), B(-3,5), C(4,-2)$

$\Delta D B C \Rightarrow \frac{1}{2}|x(5-(-2))+(-3)((-2)-3 x)+(4)(3 x-5)|$

$\triangle D B C \Rightarrow \frac{1}{2} \mid 7 x+6+9 x+12 x-20$

$\triangle D B C \Rightarrow \frac{1}{2}|28 x-14|$

$\triangle D B C \Rightarrow \frac{1}{2}|14(2 x-1)|$

$\triangle D B C \Rightarrow|7(2 x-1)|$......(1)

Area of triangle ABC taking, A (6, 3), B (−3, 5), C (4, −2)

$\Rightarrow \frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

$\Rightarrow \frac{1}{2}|6(5-(-2))+(-3)((-2)-3)+(4)(3-5)|$

$\Rightarrow \frac{1}{2} 6(7)+(-3)(-5)+(4)(-2)$

$\Rightarrow \frac{1}{2}|42+15-8|$

$\Rightarrow \frac{49}{2}$.....(2)

Also it is given that

$\frac{\Delta D B C}{\Delta A B C}=\frac{1}{2}$

Substituting the values from (1) and (2) we get

$\frac{\Delta D B C}{\Delta A B C}=\frac{1}{2}$

$\frac{\pm 7(2 x-1)}{\frac{49}{2}}=\frac{1}{2}$

$\frac{2 \times 7(2 x-1)}{49}=\frac{1}{2}$ or $\frac{-2 \times 7(2 x-1)}{49}=\frac{1}{2}$

$2 x=\frac{7}{4}+1 \quad$ or $2 x=\frac{7}{4}-1$

$2 x=\frac{11}{4} \quad$ or $2 x=\frac{-3}{4}$

$x=\frac{11}{8}$ or $x=\frac{-3}{8}$