Answer:
Step-by-step explanation:
To get the solution lying in the shaded region,
Justify the equation with the given options,
Option (1)
[tex]\frac{x^2}{16}+\frac{y^2}{4}\leq 1[/tex]
For (1, 1),
[tex]\frac{1^2}{16}+\frac{1^2}{4}\leq 1[/tex]
[tex]\frac{1}{16}+\frac{1}{4}\leq 1[/tex]
[tex]\frac{1+4}{16}\leq 1[/tex]
[tex]\frac{5}{16}\leq 1[/tex]
True.
Therefore, point (1, 1) justifies the equation.
Option (2)
For (4, 1),
[tex]\frac{x^2}{16}+\frac{y^2}{4}\leq 1[/tex]
[tex]\frac{4^2}{16}+\frac{1^2}{4}\leq 1[/tex]
[tex]1+\frac{1}{4}\leq 1[/tex]
[tex]1\frac{1}{4}\leq 1[/tex]
False.
Therefore, point (4, 1) doesn't justify the equation.
Option (3)
For (-1, 2),
[tex]\frac{x^2}{16}+\frac{y^2}{4}\leq 1[/tex]
[tex]\frac{(-1)^2}{16}+\frac{(2)^2}{4}\leq 1[/tex]
[tex]\frac{1}{16}+1\leq 1[/tex]
[tex]1\frac{1}{16}\leq 1[/tex]
False.
Therefore, point (-1, 2) doesn't justify the equation.
Option (4)
For (2, -2),
[tex]\frac{x^2}{16}+\frac{y^2}{4}\leq 1[/tex]
[tex]\frac{2^2}{16}+\frac{(-2)^2}{4}\leq 1[/tex]
[tex]\frac{1}{4}+1\leq 1[/tex]
[tex]1\frac{1}{4}\leq 1[/tex]
False.
Therefore, point (2, -2) doesn't justify the equation.
Option (1) will be the answer.
