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Derive the conditions to be imposed on $\beta$ so that $\mathrm{(0, \beta)}$ should lie inside the triangle having sides $\mathrm{y+3 x+2=0,3 y-2 x-5=0 \; and \; 4 y+x-14=0}$
Option: 1
$\mathrm{\frac{-5}{2}<\beta <\frac{-3}{2}}$

Option: 2
$\mathrm{\frac{5}{3}<\beta <\frac{7}{2}}$

Option: 3
$\mathrm{\frac{5}{2}<\beta <\frac{7}{2}}$

Option: 4
$\mathrm{5<\beta <7}$

Answers (1)

best_answer

Let the equation of BC be

$\mathrm{4 y+x-14=0}\: \: \: \: \: \: \: \: \: \: \: \: \: ....(1)$

Equation of CA be

$\mathrm{y + 3x+ 2 = 0}\: \: \: \: \: \: \: \: \: \: \: \: \: ....(2)$

And equation of AB be

$\mathrm{3y - 2x - 5 = 0}\: \: \: \: \: \: \: \: \: \: \: \: \: ....(3)$

Let $\mathrm{P \equiv(0, \beta)}$

Solving these equations two by two, we get

$\mathrm{A \equiv(-1,1), B \equiv(2,3) \text { and } C \equiv(-2,4) \text {. }}$

For point $\mathrm{\mathrm{A}(-1,1), 4 \mathrm{y}+\mathrm{x}-14=-11<0}$

For point $\mathrm{P(0, \beta), 4 y+x-14=4 \beta-14}$

For point $\mathrm{{B}(2,3), y+3 x+2=3+6+2=11>0}$

For point $\mathrm{P(0, \beta), y+3 x+2=\beta+2}$

For point $\mathrm{C(-2,4), 3 y-2 x-5=11>0}$

For point $\mathrm{P(0, \beta), 3 y-2 x-5=3 \beta-5}$

Point $\mathrm{P(0, \beta)}$ will lie inside $\mathrm{\triangle \mathrm{ABC}}$ only when P, A lie on the same side of BC, P, B lie on the same side of AC and P, C lie on the same side of AB i.e. only when

$\mathrm{4\beta -14 <0\Rightarrow \beta \frac{7}{2}}\; \; \; \; \; \; \; \; \; \; \; \; ...(4)$

$\mathrm{\beta+2>0 \Rightarrow \beta>-2}\; \; \; \; \; \; \; \; \; \; \; \; ...(5)$

and $\mathrm{3 \beta-5>0 \Rightarrow \beta>\frac{5}{3}}\; \; \; \; \; \; \; \; \; \; \; \; ...(6)$

From (4), (5) and (6), we have $\mathrm{\frac{5}{3}<\beta<\frac{7}{2}}$

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