Four points $A (6, 3), B (-3, 5), C (4, -2)$ and $D (x, 3x)$ are given in such a way that $\frac{\triangle DBC}{\triangle ABC}=\frac{1}{2}$ , find x.
Given:
Four points $A (6, 3), B (-3, 5), C (4, -2)$ and $D (x, 3x)$ are given in such a way that $\frac{\triangle DBC}{\triangle ABC}=\frac{1}{2}$.
To do:
We have to find the value of $x$.
Solution:
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC =\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\( =\frac{1}{2}[6(5+2)+(-3)(-2-3)+4(3-5)] \)
\( =\frac{1}{2}[6 \times 7+(-3)(-5)+4(-2)] \)
\( =\frac{1}{2}[42+15-8]=\frac{49}{2} \)
Area of \( \Delta \mathrm{DBC}=\frac{1}{2}[x(5+2)+(-3)(-2-3 x)+4(3 x-5)] \)
\( =\frac{1}{2}[7 x+6+9 x+12 x-20] \)
\( =\frac{1}{2}[28 x-14]=14 x-7 \)
Therefore,
\( \frac{\Delta \mathrm{DBC}}{\Delta \mathrm{ABC}}=\frac{14 x-7}{\frac{49}{2}} \)
\( =\frac{2(14 x-7)}{49} \)
\( \left|\frac{2(14 x-7)}{49}\right|=\frac{1}{2} \)
\( \Rightarrow 4|14 x-7|=\pm 49 \)
If \( 56 x-28=49 \)
\( \Rightarrow 56 x=49+28 \)
\( \Rightarrow 56x=77 \)
\( \Rightarrow x=\frac{77}{56} \)
\( \Rightarrow x=\frac{11}{8} \)
If \( 4(14 x-7)=-49 \)
\( \Rightarrow 56 x-28=-49 \)
\( \Rightarrow 56 x=-49+28 \)
\( \Rightarrow 56x=-21 \)
\( \Rightarrow x=\frac{-21}{56} \)
\( \Rightarrow x=\frac{-3}{8} \)
The value of \( x \) is \( \frac{11}{8} \) or \( \frac{-3}{8} \).
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