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 .

Answer:

2021-01-30_19-53-05Let 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 .

Answer:

2021-01-30_19-53-20Let 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 .

Answer:

2021-01-30_19-53-34Let 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 .