Sketch the graph of |x| + |y| = 1

Question

Sketch the graph of $|x| + |y| = 1$ .

Solution — Step by Step

The absolute value function changes sign based on whether the input is positive or negative. For $|x|$ : when $x \geq 0$ , $|x| = x$ ; when $x < 0$ , $|x| = -x$ . Same logic applies to $|y|$ .

So the expression $|x| + |y| = 1$ behaves differently in each of the four quadrants of the coordinate plane. We handle each quadrant separately.

Quadrant I ( $x \geq 0$ , $y \geq 0$ ): $|x| = x$ , $|y| = y$ , so the equation becomes $x + y = 1$ .

Quadrant II ( $x < 0$ , $y \geq 0$ ): $|x| = -x$ , $|y| = y$ , so the equation becomes $-x + y = 1$ , i.e., $y = x + 1$ .

Quadrant III ( $x < 0$ , $y < 0$ ): $|x| = -x$ , $|y| = -y$ , so $-x - y = 1$ , i.e., $x + y = -1$ .

Quadrant IV ( $x \geq 0$ , $y < 0$ ): $|x| = x$ , $|y| = -y$ , so $x - y = 1$ , i.e., $y = x - 1$ .

Each quadrant gives a line segment. Let’s find where these lines intersect the axes and each other:

$x = 1$ , $y = 0$ → $(1, 0)$ — lies on Q1/Q4 boundary (x-axis, positive)
$x = 0$ , $y = 1$ → $(0, 1)$ — lies on Q1/Q2 boundary (y-axis, positive)
$x = -1$ , $y = 0$ → $(-1, 0)$ — lies on Q2/Q3 boundary (x-axis, negative)
$x = 0$ , $y = -1$ → $(0, -1)$ — lies on Q3/Q4 boundary (y-axis, negative)

These four points are the vertices of the shape.

Connecting $(1,0) \to (0,1) \to (-1,0) \to (0,-1) \to (1,0)$ with straight lines gives a diamond shape. Specifically, it is a square rotated 45° (a rhombus with equal diagonals), centred at the origin.

The diagonals of this square lie along the coordinate axes, each of length 2 (from $-1$ to $1$ on each axis).

The figure is symmetric about both axes and about the origin — which makes sense because replacing $x$ with $-x$ or $y$ with $-y$ in $|x| + |y| = 1$ doesn’t change the equation.

Why This Works

The graph of $|x| + |y| = 1$ is a specific case of the $L^1$ norm in two dimensions. More generally, $|x| + |y| = r$ gives a square with diagonals of length $2r$ along the axes, for any positive constant $r$ .

The key insight is that absolute value equations always produce shapes with corners (because the derivative is discontinuous at the origin of the absolute value). The “corner” of $|x|$ is at $x = 0$ . Our graph has corners at $(1,0)$ , $(0,1)$ , $(-1,0)$ , and $(0,-1)$ — precisely at the axis intercepts.

Compare with $x^2 + y^2 = 1$ (the unit circle, smooth and round). The $|x| + |y| = 1$ diamond is the “unit ball” in the taxicab metric — the set of points whose taxicab distance from the origin is exactly 1.

Alternative Method

By symmetry: Note that $|x| + |y| = 1$ is symmetric under reflections across both axes and under 90° rotations. So we only need to sketch the first-quadrant portion ( $x + y = 1$ from $(1,0)$ to $(0,1)$ ) and then reflect it three times to complete the diamond.

This is faster in an exam: draw the segment from $(1,0)$ to $(0,1)$ , then mirror it into the other three quadrants.

Common Mistake

Students often try to solve $|x| + |y| = 1$ by squaring both sides or by algebraic manipulation, and end up with complicated expressions. The right approach is always to consider cases based on the signs of $x$ and $y$ — i.e., consider each quadrant separately. The moment you see an absolute value equation in graphing, split into cases first.

Compare two important graphs: $|x| + |y| = 1$ gives a diamond (square tilted 45°), while $\max(|x|, |y|) = 1$ gives a square aligned with the axes (side length 2, vertices at $(\pm 1, \pm 1)$ ). These are the $L^1$ and $L^\infty$ unit balls — a beautiful duality that sometimes appears in JEE Advanced.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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