Equations of diagonals of the square formed by the lines x=0, y=0, x=1 and y=1 are (A) y=x,y+x=1 (B) y=x,x+y=2

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x=0

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Concept understanding — Coordinate Axes Equations

Coordinate Axes Equations

Imagine a blank sheet of paper. You want to describe exactly where a point is. The most natural way is to draw two perpendicular lines — one horizontal, one vertical — and measure distances from them. Those two lines are the coordinate axes.

The horizontal line is the x-axis. The vertical line is the y-axis. Where they cross is the origin, the point $(0,0)$ .

Now, here's the key idea: every point on the x-axis has a y-coordinate of 0. Why? Because to be on that horizontal line, you haven't moved up or down at all — your vertical distance from the axis is zero. Similarly, every point on the y-axis has an x-coordinate of 0 — you haven't moved left or right from the vertical line.

That's the entire intuition. Let's make it precise.

The Equations

\text{x-axis: } y = 0

That's it. These are the equations of the coordinate axes.

$y = 0$ means: "all points where the y-coordinate is zero, regardless of x." That's the entire horizontal line through the origin.
$x = 0$ means: "all points where the x-coordinate is zero, regardless of y." That's the entire vertical line through the origin.

Watch out

A common mistake is to think $y=0$ means "the point $(0,0)$ ". No — it means every point with $y=0$ , like $(5,0)$ , $(-3,0)$ , $(100,0)$ , etc. It's a whole line, not a single point.

Why This Matters

These two equations are the foundation for everything in coordinate geometry. Every other line, curve, or shape is described relative to these axes. For example:

The line $y = 2$ is a horizontal line parallel to the x-axis, shifted up by 2 units.
The line $x = -3$ is a vertical line parallel to the y-axis, shifted left by 3 units.

So when you see $y = 0$ or $x = 0$ , think: "the original, unshifted axes themselves." …

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The square is bounded by $x=0$ , $y=0$ , $x=1$ , and $y=1$ . Its diagonals join opposite vertices: $(0,0)$ to $(1,1)$ gives $y=x$ , and $(0,1)$ to $(1,0)$ gives $y+x=1$ . The correct option is (A).

The first step is always to see the shape. The lines $x=0$ , $y=0$ , $x=1$ , and $y=1$ form a square in the first quadrant, with sides parallel to the axes. Its vertices are at $(0,0)$ , $(1,0)$ , $(1,1)$ , and $(0,1)$ . A diagonal of a square connects two opposite vertices — vertices that are not adjacent.

Identify the opposite vertices.

In this square, the pairs of opposite vertices are:
- $(0,0)$ and $(1,1)$
- $(0,1)$ and $(1,0)$
These are the endpoints of the two diagonals.
Find the equation of the first diagonal.

The line through $(0,0)$ and $(1,1)$ has slope $m = \frac{1-0}{1-0} = 1$ .

Using $y = mx + c$ , with $c=0$ (since it passes through the origin), we get:

y = x

Find the equation of the second diagonal. The line through $(0,1)$ and $(1,0)$ has slope $m = \frac{0-1}{1-0} = -1$ . Using point-slope form with $(0,1)$ :

y - 1 = -1(x - 0) \quad \Rightarrow \quad y = -x + 1

Rearranging:

$y + x = 1$ …

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Q.Equations of diagonals of the square formed by the lines x=0x=0x=0, y=0y=0y=0, x=1x=1x=1 and y=1y=1y=1 are (A) y=x, y+x=1y=x,\ y+x=1y=x, y+x=1 (B) y=x, x+y=2y=x,\ x+y=2y=x, x+y=2 (C) 2y=x, y+x=132y=x,\ y+x=\dfrac{1}{3}2y=x, y+x=31​ (D) y=2x, y+2x=1y=2x,\ y+2x=1y=2x, y+2x=1

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Q.Equations of diagonals of the square formed by the lines $x=0$ , $y=0$ , $x=1$ and $y=1$ are
(A) $y=x,\ y+x=1$
(B) $y=x,\ x+y=2$
(C) $2y=x,\ y+x=\dfrac{1}{3}$
(D) $y=2x,\ y+2x=1$

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