Question

The range of value of $\alpha$ such that $(0,\alpha )$ lies on or inside the triangle formed by the lines $y+3x+2=0,3y-2x-5=0,4y+x-14=0$ is

• A. $5<\alpha \le 7$
• B. $\frac{1}{2}\le \alpha \le 1$
• C. $\frac{5}{3}\le \alpha \le \frac{7}{2}$
• D. None of these

Answer 1

Answer: (C)

$\frac{5}{3}\le \alpha \le \frac{7}{2}$

Triangle formed by lines $y+3x+2=0,3y-2x-5=0,4y+x-14=0$ is :-

To calculate the range of $\alpha$ such that $(0,\alpha )$ lies on inside of triangle $△ABC$ is calculated by substituting values of $(0,\alpha )$ in equations.

If point ties inside equation $ax+bx\alpha +c<0$

(i) At $(0,\alpha )$ inside$y+3x+2=0$,

On substituting, $\alpha +3\times 0+2<0\alpha <-2$

(ii) At $(0,\alpha )$ inside$3y-2x-5=0$,

On substituting, $3\times \alpha -0-5\ge 0\alpha \ge \frac{5}{3}$

(ii) At $(0,\alpha )$ inside$4y+x-14=0$,

On substituting, $4\times \alpha +0-14\le 0\alpha \le \frac{7}{2}$

Hence $\frac{5}{3}\le \alpha \le \frac{7}{2}$