The line which cuts off equal intercept from the axes and pass through the point (1,-2) is __.

NCERT Exemplar · Q43

(1,-2)

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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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A line cutting equal intercepts from both axes has the form $x + y = a$ ; substituting the given point yields the equation $x + y + 1 = 0$ .

When a line cuts off equal intercepts from the coordinate axes, it means the $x$ -intercept and $y$ -intercept have the same absolute value. If we denote both intercepts as $a$ (they could be equal or opposite in sign), the intercept form of a line is:

\frac{x}{a} + \frac{y}{b} = 1

For equal intercepts, we have two cases to consider: either both intercepts are $a$ (same sign), or they are $a$ and $-a$ (opposite signs).

Case 1: Both intercepts equal $a$

The intercept form becomes:

\frac{x}{a} + \frac{y}{a} = 1

Simplifying:

x + y = a

This represents a family of lines with slope $-1$ .

Case 2: Intercepts are $a$ and $-a$

The intercept form becomes:

\frac{x}{a} + \frac{y}{-a} = 1

Simplifying:

x - y = a

This represents a family of lines with slope $1$ .

Now we determine which case applies by using the condition that the line passes through $(1, -2)$ .

Testing Case 1: Substitute $(1, -2)$ into $x + y = a$ :

1 + (-2) = a

a = -1

So the line is $x + y = -1$ , or equivalently $x + y + 1 = 0$ .

Testing Case 2: Substitute $(1, -2)$ into $x - y = a$ :

1 - (-2) = a

a = 3

So the line would be $x - y = 3$ , or $x - y - 3 = 0$ .

Verifying the intercepts:

For $x + y + 1 = 0$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04214 marksMCQ
$(a,b,c)$
›Reveal solutionSolution

A line through $(a,b,c)$ parallel to the $x$ -axis has $y$ - and $z$ -direction ratios zero.

A line parallel to the $x$ -axis has direction ratios proportional to $(1,0,0)$ , i.e. the $y$ - and $z$ -denominators are $0$ . Among the options, only (A) has both $y$ - and $z$ -denominators equal to $0$ :

$\frac{x-a}{a}=\frac{y-b}{0}=\frac{z-c}{0}.$ …
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KEAM 2025Set eng-2025-04264 marksMCQ
$\dfrac{x+3}{11} = \dfrac{y-2}{-1} = \dfrac{z+1}{3}$
›Reveal solutionSolution
[!TLDR]

Parametrise the line, set $y=0$ (the $zx$ -plane), find $t=2$ , and read off $(19,0,5)$ .

Concept

A line in symmetric form $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}=t$ can be written parametrically. The $zx$ -plane (also called the $xz$ -plane) is the set of points with $y=0$ — an NCERT/CBSE three-dimensional-geometry standard.

Solution

Set each ratio equal to $t$ :

$x=-3+11t,\qquad y=2-t,\qquad z=-1+3t.$ …
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KEAM 2025Set eng-2025-04274 marksMCQ
$x=0$
›Reveal solutionSolution

The vertex lies on the axis $x=0$ and is $3$ above the $x$ -axis, so it is $(0,3)$ .

The axis of the parabola is the line $x=0$ (the $y$ -axis), so the vertex has $x$ -coordinate $0$ . Since it is at distance $3$ from the origin above the $x$ - …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$(2,1,0),(5,0,1)$
›Reveal solutionSolution
The plane meets the x-axis at $(3,0,0)$ .

Concept and Intuition

Find the plane through the three points, then set $y=z=0$ for the x-intercept.

Step-by-Step Solution

Vectors in the plane: $(3,-1,1)$ and $(2,0,1)$ ; normal $=(-1,-1,2)$ .

Plane through $(2,1,0)$ : $-(x-2)-(y-1)+2z=0\Rightarrow x+y-2z=3$ . …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$(-1, -2, -3)$
›Reveal solutionSolution
The plane is $x=-1$ .

Concept and Intuition

A plane perpendicular to the $x$ -axis has the $x$ -axis as normal direction, so it is of the form $x=c$ , fixed by the point it passes through.

Step-by-Step Solution

Normal vector $=\hat i$ , giving equation $x=c$ .

The plane passes through $(-1,-2,-3)$ , so $c=-1$ .

Equation: $x=-1$ .

Common Mistakes …
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Q.The line which cuts off equal intercept from the axes and pass through the point (1,−2)(1,-2)(1,−2) is ____.

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Q.The line which cuts off equal intercept from the axes and pass through the point $(1,-2)$ is ____.

Unlock everything free for 14 days