Question 1: Find the values of $\alpha$ so that the point $P (\alpha^2, \alpha)$ lies inside or on the triangle formed by the lines $x -5y + 6 =0, \ \ x - 3y + 2=0$ and $x -2y - 3 =0$

Let the triangle be $\triangle ABC$. Let,

$AB : x-5y+6=0$   … … … i)

$BC : x - 3y + 2 = 0$   … … … ii)

$CA : x - 2y - 3 = 0$   … … … iii)

Solving i), ii) and iii) we get the vertices:    $A ( 9, 3), \ \ B( 4, 2), \ \ C( 13, 5)$

Given point $P ( \alpha^2, \alpha)$

The point $P ( \alpha^2, \alpha)$ will lie inside or on $\triangle ABC$ is the following three conditions hold hold simultaneously.

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

ii) $B$ and $P$ lie on the same side of $CA$

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

If $A$ and $P$ lie on the same side of $BC$ then

$[ 9 - (3)(3) + 2]( \alpha^2 - 3 \alpha + 2 ) \geq 0$

$\Rightarrow ( \alpha-2) ( \alpha - 1) \geq 0$

$\Rightarrow \alpha \in ( -\infty, 1] \cup [ 2, \infty)$     … … … … … iv)

If $B$ and $P$ lie on the same side of $CA$ then

$[ 4 - (2)(2) -3]( \alpha^2 - 2 \alpha -3 ) \geq 0$

$\Rightarrow ( \alpha-3) ( \alpha + 1) \leq 0$

$\Rightarrow \alpha \in [-1, 3]$    … … … … … v)

If $C$ and $P$ lie on the same side of $AB$ then

$[ 13 - (5)(5) +6]( \alpha^2 - 5 \alpha +6 ) \geq 0$

$\Rightarrow ( \alpha-3) ( \alpha -2) \leq 0$

$\Rightarrow \alpha \in [2, 3]$     … … … … … vi)

From iv), v) and vi) we get $\alpha \in [2, 3]$

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Question 2: Find the values of the parameter a so that the point $(a, 2)$ is an-interior point of the triangle formed by the lines $x + y -4 = 0 , \ \ 3x - 7y -8 =0$ and $4x-y-31 = 0$.

Let the triangle be $\triangle ABC$. Let,

$AB : x + y -4 = 0$   … … … i)

$BC : 3x - 7y -8 =0$   … … … ii)

$CA : 4x-y-31 = 0$   … … … iii)

Solving i), ii) and iii) we get the vertices:    $A ( 7, -3), \ \ B($ $\frac{18}{5}$ $,$ $\frac{2}{5}$ $), \ \ C($ $\frac{209}{25}$ $,$ $\frac{61}{25}$ $)$

Given point $P (a, 2)$

The point $P ( a, 2)$ will lie inside or on $\triangle ABC$ is the following three conditions hold hold simultaneously.

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

ii) $B$ and $P$ lie on the same side of $CA$

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

If $A$ and $P$ lie on the same side of $BC$ then

$[ (3)(7) - (7)(-3) -8]( 3a-14-8 ) > 0$

$\Rightarrow a >$ $\frac{22}{3}$     … … … … … iv)

If $B$ and $P$ lie on the same side of $CA$ then

$\Big[ 4 \times$ $\frac{18}{5}$ $-$ $\frac{2}{5}$ $-4 \Big]( 4a-2-31 ) > 0$

$\Rightarrow a <$ $\frac{33}{4}$ … … … … … v)

If $C$ and $P$ lie on the same side of $AB$ then

$\Big[$ $\frac{209}{25}$ $+$ $\frac{61}{25}$ $- 4 \Big]( a+2-4 ) \geq 0$

$\Rightarrow a> 2$     … … … … … vi)

From iv), v) and vi) we get $\alpha \in ($ $\frac{22}{3}$ $,$ $\frac{33}{4}$ $)$

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Question 3: Determine whether the point $(-3 , 2)$ lies inside or outside the triangle whose sides are given by  the equations  $x+y - 4 = 0, \ \ 3x-7y+8 = 0 , \ \ 4x-y - 31 = 0$.

Let the triangle be $\triangle ABC$. Let,

$AB : x + y - 4 = 0$   … … … i)

$BC : 3x-7y+8 = 0$   … … … ii)

$CA : 4x-y - 31 = 0$   … … … iii)

Solving i), ii) and iii) we get the vertices:    $A ( 7, -3), \ \ B( 2, 2), \ \ C( 9, 5)$

Given point $P (2,2)$

The point $P ( 2,2)$ will lie inside or on $\triangle ABC$ is the following three conditions hold hold simultaneously.

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

ii) $B$ and $P$ lie on the same side of $CA$

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

If $A$ and $P$ lie on the same side of $BC$ then

$[ 3(7) - (7)(-3) + 8]( -9-14+8 ) >0$

$\Rightarrow ( 50) ( -15) \geq 0$ : This is FALSE

If $B$ and $P$ lie on the same side of $CA$ then

$[ 4(2) - 2 -31]( -12-2-31 ) > 0$

$\Rightarrow ( -25) ( -45) > 0$

$\Rightarrow 1125 > 0$  : This is TRUE

If $C$ and $P$ lie on the same side of $AB$ then

$[ 9+5-4]( -3+2-4 ) >0 0$

$\Rightarrow -50 > 0$ : This is FALSE

Therefore we can say that the point $P(2,2)$ lies outside the $\triangle ABC$.