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If equation \(|x| + |y| = 5\) encloses a certain region on the graph, what is the area of that region?

A. 5 B. 10 C. 25 D. 50 E. 100

X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to \(\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50\).

If equation \(|x| + |y| = 5\) encloses a certain region on the graph, what is the area of that region?

A. 5 B. 10 C. 25 D. 50 E. 100

X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to \(\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50\).

Answer: D

Another way to find the area of the square, if you don't know how to find it from the diagonal, is to take the area of the four triangles that are made from the x and y axis. You end up with \((\frac{5*5}{2})*4\). = Additionally, you can find the difference between any two points with the formula \(\sqrt{(X2-X1)^2+(Y2-Y1)^2}\). Plug in (5,0) and (0,5) into that equation, then square the result to find the area.

pate13 But you would be getting a different area wouldn't you? For example +/- 3 and +/- 2 solves the equation but would not yield an area of 50. Let me know if I am still missing something.

pate13 But you would be getting a different area wouldn't you? For example +/- 3 and +/- 2 solves the equation but would not yield an area of 50. Let me know if I am still missing something.

Hi,

The graph includes all points as mentioned.. the solution of |x|+|y|=5 will be as it has been shown above and x=4 when y=1 and so on are the points which we have joined to get the sketch..

We take values when x= WHAT when y=0 and y=WHAT when x=0 because the line being formed by values of x and y change direction at these points otherwise it continues as straight line..

I think this is a High quality question and I agree with the explanation.

However, I have a small question. Can someone please help me with an answer?

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say

I think this is a High quality question and I agree with the explanation.

However, I have a small question. Can someone please help me with an answer?

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say

TIA!

|x|+|y|=5 has infinitely may solutions. For any value of x from -5 to 5 there will exist some value of y which will satisfy |x|+|y|=5. Since there are infinitely many real numbers from -5 to 5, then there are infinitely many pairs of (x, y) which satisfy |x|+|y|=5.

(0, 5), (0, -5), (5, 0), and (-5, 0) are x and y-intercepts of |x|+|y|=5. x-intercept is the value of x when y = 0 and y-intercept is the value of y when x = 0. If you plug x = 0, you'll get two value of y and if you plug y = 0, you'll get two value of x. Thus 4 points.

A million thanks to you Bunuel for the quick answer. I fully understand this now. Awesome solution to an equally awesome question!

Bunuel wrote:

susheelh wrote:

I think this is a High quality question and I agree with the explanation.

However, I have a small question. Can someone please help me with an answer?

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say

TIA!

|x|+|y|=5 has infinitely may solutions. For any value of x from -5 to 5 there will exist some value of y which will satisfy |x|+|y|=5. Since there are infinitely many real numbers from -5 to 5, then there are infinitely many pairs of (x, y) which satisfy |x|+|y|=5.

(0, 5), (0, -5), (5, 0), and (-5, 0) are x and y-intercepts of |x|+|y|=5. x-intercept is the value of x when y = 0 and y-intercept is the value of y when x = 0. If you plug x = 0, you'll get two value of y and if you plug y = 0, you'll get two value of x. Thus 4 points.