MCQ
$A(6, 3), B(-3, 5), C(4, -2)$ and $D(x, 3x)$ are four points. If $\triangle\text{DBC} : \triangle\text{ABC}= 1 : 2,$ then $x$ is equal to:
- ✓$\frac{11}{8}$
- B$\frac{8}{11}$
- C$3$
- DNone of these
✓
Answer
Correct option: A.
$\frac{11}{8}$
The area of a triangle with vertices $D(x, 3x), B(-3, 5)$ and $C(4, -2)$ is given below:
Area of $\triangle\text{DBC}=\frac{1}{2}\{\text{x}(5+2)-3(-2-3\text{x})+4(3\text{x}-5)\}$
$\Rightarrow$ Area of $\triangle\text{DBC}=(14\text{x}-7)\text{sq units}$
Similarly, the area of a triangle with vertices $A(6, 3), B(-3, 5)$ and $C(4, -2)$ is given below:
$\triangle\text{ABC}=\frac{1}{2}\{6(5+2)-3(-2-3)+4(3-5)\}$
$\triangle\text{ABC}=\frac{49}{2}\text{sq units}$
Given:
$\triangle\text{DBC}:\triangle\text{ABC}=1:2$
$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$
$\Rightarrow8\text{x}-4=7$
$\Rightarrow\text{x}=\frac{11}{8}$
Area of $\triangle\text{DBC}=\frac{1}{2}\{\text{x}(5+2)-3(-2-3\text{x})+4(3\text{x}-5)\}$
$\Rightarrow$ Area of $\triangle\text{DBC}=(14\text{x}-7)\text{sq units}$
Similarly, the area of a triangle with vertices $A(6, 3), B(-3, 5)$ and $C(4, -2)$ is given below:
$\triangle\text{ABC}=\frac{1}{2}\{6(5+2)-3(-2-3)+4(3-5)\}$
$\triangle\text{ABC}=\frac{49}{2}\text{sq units}$
Given:
$\triangle\text{DBC}:\triangle\text{ABC}=1:2$
$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$
$\Rightarrow8\text{x}-4=7$
$\Rightarrow\text{x}=\frac{11}{8}$
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