Question

The set of values of $\beta$ so that the point $(0,\beta )$ lies inside or on the triangle having the sides $3x+y+2=0,2x-3y+5=0$ and $x+4y-14=0$ is,

• A. $\left[\frac{5}{3},\infty \right)$
• B. $(-\infty ,\frac{7}{2}]$
• C. $\left[\frac{5}{3},\frac{7}{2}\right]$
• D. None of these

Answer 1

The correct option is C $\left[\frac{5}{3},\frac{7}{2}\right]$

Equation of sides of triangle are

$3x+y+2=0,2x-3y+5=0$ and $x+4y-14=0$

Point of intersection of sides of triangle are $A(2,3),B(-2,4),C(-1,1)$.

Now, the point $P(0,\beta )$ will lie inside or on the triangle $ABC,$ if the following three conditions hold simultaneously.

(i) $A$ and $P$ lie on the same side of $BC,$

(ii) $B$ and $P$ lie on the same side of $AC,$

(iii) $C$ and $P$ lie on the same side of $AB.$

Now, $A$ and $P$ will lie on the same side of $BC,$ if

$(3\times 2+3+2)(3\times 0+\beta +2)\ge 0$

$\Rightarrow 11(\beta +2)\ge 0\Rightarrow \beta +2\ge 0\Rightarrow \beta \ge -2\cdots \left(i\right)$

$B$ and $P$ will lie on the same side of $AC,$ if

$(-2\times 2-3\times 4+5)(2\times 0-3\beta +5)\ge 0$

$\Rightarrow -11(-3\beta +5)\ge 0\Rightarrow 3\beta -5\ge 0\Rightarrow \beta \ge \frac{5}{3}\cdots \left(ii\right)$

$C$ and $P$ will lie on the same side of $AB,$ if

$(-1+1\times 4-14)(0+4\beta -14)\ge 0$

$\Rightarrow -22(2\beta -7)\ge 0\Rightarrow 2\beta -7\le 0\Rightarrow \beta \le \frac{7}{2}\cdots \left(iii\right)$

From $\left(i\right),\left(ii\right)and\left(iii\right),$ we obtain that

$\frac{5}{3}\le \beta \le \frac{7}{2}i.e.\beta \in \left[\frac{5}{3},\frac{7}{2}\right]$