Finding the set of points (a,b) in an xy-plane for which |a|+|b| = 5

Question

I came across this question in an SAT Math Level 2 Subject Test book and the answer confuses me:

Question: Which of the following describes the set of points (a,b) for which |a|+|b| = 5 in the xy-plane?

Answer Choices:

A) A circle with radius 5

B) A circle with radius 5√2

C) A square with sides of length 5√2

D) A square with sides of length 10

E) A regular hexagon with sides of length 5

Correct Answer: (C)

I thought that the answer would be (A) because all points on the circle would be equidistant from the center. Can someone tell me how I'm mistaken?

Answer 1

That depends on which norm are you using. If you are using $2$-norm, then the corresponding equation is $$a^2+b^2=5^2$$ which is a circle.

$|a|+|b|=5$ uses $1$-norm, in particular, if you consider the first quadrant, that is $a>0, b>0$, then you can see that in the first quadrant, it corresponds to $a+b=5$ which is a line segment. You can consider other quadrant by cases and trace out a square of size $5\sqrt{2}$ .

Answer 2

The distance between two points $P=(p_1,p_2)$ and $Q=(q_1,q_2)$ in the plane is $\sqrt{(p_1-q_1)^2+(p_2-q_2)^2}$, by Pythagoras on the triangle with sides parallel to the axes and $P$ and $Q$ at opposite ends of the hypotenuse.

Here, you instead have a set that is defined by the sum distance to the $x$-axis and to the $y$-axis being constant. You can also think of this as the distance along the $x$-axis to the point $(a,0)$, added then the distance $b$ is the distance from $(a,0)$ to $(a,b)$. I.e. if $a=x$ and $0<x<5$, then we could have $b=5-x$, which gives (part of) the line $a+b=5$. We could also have $b=x-5$, since $\lvert b \rvert$ remains the same, which gives part of the line $a-b=5$. Similarly, if $-5<a<0$, we obtain the parts of the lines $-a+b=5$ and $-a-b=5$. Putting all these parts together gives a square, since all the lines have gradient $\pm 1$.

Answer 3

Nice problem.

$|x| + |y| =5.$

1) $1$st quadrant: $x,y \ge 0.$

$x+y = 5;$ or

$y= -x + 5$, a straight line with $y-$intercept $5$ , slope $m_1=-1$.

2) $3$rd quadrant: $x, y \le 0.$

$-x - y = 5$; or

$ y = -x -5$, a straight line with $y-$intercept $-5$ slope $m_1=-1.$

3) $2$nd quadrant: $x \le 0$, $y\ge 0.$

$-x +y =5$; or

$y = x +5,$ a straight line with $y-$intercept $5$, slope $m_2 = 1$.

4) $4$th quadrant: $x \ge 0$, $y \le 0.$

$x- y =5;$ or

$y = x -5$, a straight line with $y-$intercept $-5$, slope $m_2=1.$

Summary:

2 sets of parallel lines.

They are intersecting at $(0,5), (0,-5), (5,0)$ and $(-5,0)$.

A square (why?) with a diagonal of length $10$ units.

Left to do: Find side length.