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Question

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(a) (1, 2)

(b) (2, 1)

(c) (2, 3)

(d) (-3, 2)

Answer

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According to the given information we have equation $2x + 3y - 12 \leqslant 0$

We know that the graphical representation of the given equation is given by

Fir let’s put (0, 0) in the given equation we get

$2\left( 0 \right) + 3\left( 0 \right) - 12 \leqslant 0$

$ \Rightarrow $$0 + 0 - 12 \leqslant 0$

$ \Rightarrow $$ - 12 \leqslant 0$

Since – 12 is less than 0 therefore the equation satisfies the statement

Hence, the region of the equation of line will be towards the origin

Now substituting the given points to check which of the points lie on the region of the equation.

For: (1, 2)

We get,

2 (1) + 3 (2) – 12 = 0

$ \Rightarrow $2 + 6 – 12 = 0

$ \Rightarrow $8 – 12 = 0

$ \Rightarrow $– 4 = 0

Since the result is negative it means the point lie on the region of the given line

For: (2, 1)

We get,

2 (2) + 3 (1) – 12 = 0

$ \Rightarrow $4 + 3 – 12 = 0

$ \Rightarrow $4 + 3 – 12 = 0

$ \Rightarrow $7 – 12 = 0

$ \Rightarrow $ – 5 = 0

Since the result is negative it means the point lie on the region of the given line

For: (2, 3)

We get,

2 (2) + 3 (3) – 12 = 0

$ \Rightarrow $4 + 9 – 12 = 0

$ \Rightarrow $1 = 0

Since the result is positive it means the point doesn’t lie on the region of the given line

For: (-3, 3)

We get,

2 (-3) + 3 (3) – 12 = 0

$ \Rightarrow $-6 + 9 – 12 = 0

$ \Rightarrow $- 9 = 0

Since the result is negative it means the point lie on the region of the given line

So, we know that accept (2, 3) every point lies on the region of the given line

Hence, (2, 3) is the point which doesn’t lie in the half plane.