Suppose the vertices of a triangle are given by $\mathrm{A}(0,3)$, $\mathrm{B}(-2,0)$ and $\mathrm{C}(6,1)$. For $(\alpha, \alpha+1)$ to lie inside the triangle, $\alpha$ should lie in the interval

• A

  $\left(\frac{-6}{7}, 4\right)$
• B

  $\left(\frac{4}{5}, 4\right)$
• C

  $\left(-\infty, \frac{-6}{7}\right) \cup(4, \infty)$
• Correct answer
  D

  $\left(\frac{-6}{7}, \frac{3}{2}\right)$

Solution

Given points $\mathrm{A}(0,3)$ $\mathrm{B}(-2,0)$ and $\mathrm{C}(6,1)$ as vertices of triangle $\mathrm{ABC}$
how Line $A C=x+3 y=9$...(i) and $\mathrm{BC}=\mathrm{x}-8 \mathrm{y}=-2$...(ii) for $(\alpha, \alpha+1)$ lies in side te triangle $\mathrm{ABC}$ Let $\alpha=\mathrm{x} \Rightarrow \mathrm{y}=\mathrm{x}-1$...(iii) On solving (i) and (iii) we get $\left(\frac{3}{2}, \frac{5}{2}\right)$ and solving (ii) and (iii) we get $\left(\frac{-6}{7}, \frac{1}{7}\right)$ $\Rightarrow \alpha \Sigma\left(\frac{-6}{7}, \frac{3}{2}\right)$

Hence, the correct answer is option: 'D'.