The region represented by |x - y| leq 2 and | | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given inequalities are $|x - y| \leq 2$ and $|x + y| \leq 2$.These can be written as $-2 \leq x - y \leq 2$ and $-2 \leq x + y \leq 2$.This re

Vedclass · Accepted Answer

square of side length $2\sqrt{2}$ units

Vedclass · Answer

(D) The given inequalities are $|x - y| \leq 2$ and $|x + y| \leq 2$.These can be written as $-2 \leq x - y \leq 2$ and $-2 \leq x + y \leq 2$.This represents the region bounded by the lines $x - y = 2$,$x - y = -2$,$x + y = 2$,and $x + y = -2$.The vertices of this region are the intersection points of these lines:$1$) $x - y = 2$ and $x + y = 2$ gives $(2, 0)$.$2$) $x + y = 2$ and $x - y = -2$ gives $(0, 2)$.$3$) $x - y = -2$ and $x + y = -2$ gives $(-2, 0)$.$4$) $x + y = -2$ and $x - y = 2$ gives $(0, -2)$.The distance between consecutive vertices (e.g.,$(2, 0)$ and $(0, 2)$) is $\sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.Since all sides are equal to $2\sqrt{2}$ and the diagonals are perpendicular (slopes are $1$ and $-1$),the region is a square.The area of the square is $(	ext{side})^2 = (2\sqrt{2})^2 = 8$ sq. units.Thus,the region is a square of side length $2\sqrt{2}$ units.

The region represented by $|x - y| \leq 2$ and $|x + y| \leq 2$ is bounded by a

Similar Questions

Solve the following system of inequalities: $\frac{2x+1}{7x-1} > 5$ and $\frac{x+7}{x-8} > 2$.

$A$ solution is to be kept between $30^{\circ} C$ and $35^{\circ} C$. What is the range of temperature in degree Fahrenheit,if the conversion formula is $F = \frac{9}{5} C + 32$?

The region formed by the inequalities $2x + 3y - 5 \leq 0$,$4x - 3y + 2 \leq 0$,and $x \geq 0$ is:

The set of all real values of $x$ for which $\frac{x^2-1}{(x-4)(x-3)} \geq 1$ is

The solution set of the inequation $\sqrt{x^2+x-2} > (1-x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module