Question

Find the number which does not lie in the half plane $2x + 3y - 12 \leqslant 0$.
A) (1, 2)
B) (2, 1)
C) (2, 3)
D) (-3, 2)

Answer 1

Hint: We know that if a point lies in the plane, it will definitely satisfy the equation. So, we will just put in each of the given points and see if they result in a correct result or not.

Complete step-by-step answer:
We know if we say that this point lies on the line or in the plane, it must satisfy their equation without giving rise to any problem.
So, we will just start with the first point and put in the values of each coordinate in the given equation of the plane to see if it does satisfy that. If the point satisfies the equation, it means it lies in the given plane but if it gives rise to a false statement or any contradicting statement, then it does not lie in the given plane.
We have the first point in option (A), which is: (1, 2).
So, we will just put in x = 1 and y = 2 in the given plane $2x + 3y - 12 \leqslant 0$.
Putting the values, we have on the LHS:
$2x + 3y - 12 = 2(1) + 3(2) - 12 = 2 + 6 - 12 = - 4$
And we know that $ - 4 \leqslant 0$ is definitely true.
Hence, (1, 2) lies in the given plane $2x + 3y - 12 \leqslant 0$.
We have the second point in option (B), which is: (2, 1).
So, we will just put in x = 2 and y = 1 in the given plane $2x + 3y - 12 \leqslant 0$.
Putting the values, we have on the LHS:
$2x + 3y - 12 = 2(2) + 3(1) - 12 = 4 + 3 - 12 = - 5$
And we know that $ - 5 \leqslant 0$ is definitely true.
Hence, (2, 1) lies in the given plane $2x + 3y - 12 \leqslant 0$.
We have the third point in option (C), which is: (2, 3).
So, we will just put in x = 2 and y = 3 in the given plane $2x + 3y - 12 \leqslant 0$.
Putting the values, we have on the LHS:
$2x + 3y - 12 = 2(2) + 3(3) - 12 = 4 + 9 - 12 = 1$
And we know that $1 \leqslant 0$ is definitely absurd.
Hence, (2, 3) does not lie in the given plane $2x + 3y - 12 \leqslant 0$.
We have the fourth point in option (C), which is: (-3, 2).
So, we will just put in x = -3 and y = 2 in the given plane $2x + 3y - 12 \leqslant 0$.
Putting the values, we have on the LHS:
$2x + 3y - 12 = 2( - 3) + 3(2) - 12 = - 6 + 6 - 12 = - 12$
And we know that $ - 12 \leqslant 0$ is definitely true.
Hence, (-3, 2) lies in the given plane $2x + 3y - 12 \leqslant 0$.
Hence, only the point (2, 3) does not lie in the given plane $2x + 3y - 12 \leqslant 0$.
Hence, the answer is option (C).

Note: The students might misread the question and answer it as (A), (B) and (D). So, take care of that.
You also may draw the equation of line and then select the portion and see it visually. That is an alternate way of solving this question.
Let us solve it in this way now briefly by drawing its diagram. It will be as follows:-

We see this is the graph. The region included will be the region below the line.
Now, you may check the point’s inclusion for all points.