The distance of point $P(a, b, c)$ from the $y$-axis is: (A) $b$ (B) $b^2$ (C) $\sqrt{a^2 + c^2}$ (D) $a^2 + c^2$

The distance of point P(a, b, c) from the y-axis is: (A) b (B) b² (C) √(a² + c²) (D) a² + c²

Question

P(a, b, c)

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

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Concept understanding — Distance From Y Axis

Distance from the Y‑Axis: The Intuition

Imagine you are standing in a large, empty hall. The floor is marked with two perpendicular lines that cross at the centre: one running north‑south (the Y‑axis) and one running east‑west (the X‑axis). Now, I ask you: how far are you from the north‑south line?

You would look at your feet, measure the shortest straight‑line distance to that line, and give me a number. That number — the perpendicular distance from you to the Y‑axis — is exactly what we mean by "distance from the Y‑axis" in coordinate geometry.

Note

The Y‑axis is the vertical line $x = 0$ . Distance is always measured perpendicularly (at a right angle) to the axis, never along a slant.

The Precise Statement

In the Cartesian plane, any point is written as $(x, y)$ . The distance of a point from the Y‑axis is simply the absolute value of its $x$ ‑coordinate.

\text{Distance from Y‑axis} = |x|

Why? Because the Y‑axis is the line $x = 0$ . The perpendicular distance from any point $(x, y)$ to the line $x = 0$ is the horizontal gap between $x$ and $0$ , which is $|x - 0| = |x|$ .

\text{Distance from Y‑axis} = |x|

Examples to Lock It In

Point	$x$ ‑coordinate	Distance from Y‑axis
$(3, 5)$	$3$	$3$ units
$(-4, 2)$	$-4$	$4$ units (distance is always positive)
$(0, 7)$	$0$	$0$ units (point lies on the Y‑axis)
$(-2.5, -1)$	$-2.5$	$2.5$ units

Watch out

A common mistake: thinking the $y$ ‑coordinate matters. It does not. The $y$ ‑coordinate tells you how far the point is from the X‑axis, not the Y‑axis. The two distances are independent.

Why This Matters

This concept is the foundation for:

Finding the abscissa (the $x$ ‑coordinate) of a point. …

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The distance from a point to the $y$ -axis is the length of the perpendicular from the point to the axis; since the $y$ -axis is the set of all points $(0, y, 0)$ , we measure how far the point is in the $xz$ -plane, giving $\sqrt{a^2 + c^2}$ .

Understanding Distance from an Axis

When we talk about the distance of a point from an axis in three-dimensional space, we mean the perpendicular distance—the shortest straight-line path from the point to that axis.

The $y$ -axis consists of all points of the form $(0, y, 0)$ where $y$ can be any real number. Think of it as a vertical line running through the origin, where both the $x$ -coordinate and $z$ -coordinate are always zero.

For a point $P(a, b, c)$ , finding the distance to the $y$ -axis means finding the closest point on the $y$ -axis and measuring that distance. The closest point on the $y$ -axis to $P$ will have the same $y$ -coordinate as $P$ (namely $b$ ), but will have $x = 0$ and $z = 0$ . So the closest point is $Q(0, b, 0)$ .

Step-by-Step Solution

Identify the closest point on the $y$ -axis.

The point on the $y$ -axis nearest to $P(a, b, c)$ is $Q(0, b, 0)$ . This is because the perpendicular from $P$ to the $y$ -axis drops straight down in the $xz$ -plane while maintaining the same $y$ -coordinate.
Apply the distance formula in 3D.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Calculate the distance from $P$ to $Q$ .

Substituting $P(a, b, c)$ and $Q(0, b, 0)$ :

d = \sqrt{(0 - a)^2 + (b - b)^2 + (0 - c)^2}

d = \sqrt{a^2 + 0 + c^2}

d = \sqrt{a^2 + c^2}

Interpret geometrically. …

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

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The distance from x-axis to point (x, y, z) will be:$
(a) \sqrt{x^2+y^2}
(b) \sqrt{x^2+z^2}
(c) \sqrt{y^2+z^2}
(d) \sqrt{x^2+y^2+z^2}
›Reveal solutionSolution

The distance from a point to the $x$ -axis is the distance to its foot of perpendicular on the axis, which only involves the $y$ and $z$ coordinates.

Concept: The $x$ -axis consists of all points of the form $(t, 0, 0)$ . The perpendicular distance from a point $P(x,y,z)$ to the $x$ -axis is the distance from $P$ to its foot of perpendicular $M(x,0,0)$ on the axis (since $M$ has the same $x$ -coordinate).

Working: …

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CBSE 2024Set 65/1/11 markMCQ
$P(a, b, c)$
›Reveal solutionSolution
The distance from a point to the $y$ -axis is the length of the perpendicular from the point to the axis; since the $y$ -axis is the set of all points $(0, y, 0)$ , we measure how far the point is in the $xz$ -plane, giving $\sqrt{a^2 + c^2}$ .

Understanding Distance from an Axis

When we talk about the distance of a point from an axis in three-dimensional space, we mean the perpendicular distance—the shortest straight-line path from the point to that axis.

The $y$ -axis consists of all points of the form $(0, y, 0)$ where $y$ can be any real number. Think of it as a vertical line running through the origin, where both the $x$ -coordinate and $z$ -coordinate are always zero.

For a point $P(a, b, c)$ , finding the distance to the $y$ -axis means finding the closest point on the $y$ -axis and measuring that distance. The closest point on the $y$ -axis to $P$ will have the same $y$ -coordinate as $P$ (namely $b$ ), but will have $x = 0$ and $z = 0$ . So the closest point is $Q(0, b, 0)$ .

Step-by-Step Solution

Identify the closest point on the $y$ -axis.

The point on the $y$ -axis nearest to $P(a, b, c)$ is $Q(0, b, 0)$ . This is because the perpendicular from $P$ to the $y$ -axis drops straight down in the $xz$ -plane while maintaining the same $y$ -coordinate.

Apply the distance formula in 3D.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Calculate the distance from $P$ to $Q$ .

Substituting $P(a, b, c)$ and $Q(0, b, 0)$ :

$d = \sqrt{(0 - a)^2 + (b - b)^2 + (0 - c)^2}$ $d = \sqrt{a^2 + 0 + c^2}$ $d = \sqrt{a^2 + c^2}$

Interpret geometrically. …
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CBSE 2024Set 65/2/11 markMCQ
$(0, 1, 2)$
›Reveal solutionSolution
The foot of the perpendicular from a point to an axis is its projection onto that axis. For the x-axis, this means setting the y and z coordinates to zero while keeping the x-coordinate. Thus, for $(0, 1, 2)$ , the foot of the perpendicular on the x-axis is $(0, 0, 0)$ .

When we talk about the "foot of the perpendicular" from a point to a line (or an axis in 3D space), we are essentially looking for the point on that line which is closest to the given point. Geometrically, if you imagine dropping a plumb line from the given point straight down to the axis, where it touches the axis is the foot of the perpendicular. This concept is also known as the orthogonal projection of the point onto the line.

Consider a point $P(x_0, y_0, z_0)$ in 3D space.

The x-axis is a line where all points have their y-coordinate and z-coordinate equal to zero. Any point on the x-axis can be represented as $(x, 0, 0)$ .

When we drop a perpendicular from $P(x_0, y_0, z_0)$ to the x-axis, the resulting point on the x-axis, let's call it $F$ , will have its y and z coordinates equal to $0$ . The key insight is that the x-coordinate of $F$ will be exactly the same as the x-coordinate of $P$ . This is because moving along the y or z directions does not change the x-position relative to the x-axis.

Think of it this way: to find the point on the x-axis closest to $P(x_0, y_0, z_0)$ , we need to eliminate the "deviation" in the y and z directions. We do this by setting $y=0$ and $z=0$ . The x-coordinate remains untouched because the x-axis itself extends along the x-direction.
$P(x_0, y_0, z_0)$
Let's apply this understanding to the given problem.

Identify the given point: We are given the point $P(0, 1, 2)$ . Here, $x_0 = 0$ , $y_0 = 1$ , and $z_0 = 2$ .

Identify the target axis: We need to find the foot of the perpendicular on the x-axis. …
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CBSE 2024Set ANNUAL1 markMCQ
$P(a,b,c)$
(a) $\sqrt{b^2+c^2}$
(b) $\sqrt{a^2+c^2}$
(c) $\sqrt{a^2+b^2}$
(d) None of these
›Reveal solutionSolution

The foot of perpendicular from P(a,b,c) to the x-axis is (a,0,0); the distance formula then leaves only b and c.

…

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CBSE 2020Set 65/1/11 markMCQ
$(4, -7, 3)$
›Reveal solutionSolution
The distance from a point to the $y$ -axis is the perpendicular distance in the $xz$ -plane, ignoring the $y$ -coordinate. For $(4, -7, 3)$ , this distance is $\sqrt{4^2 + 3^2} = 5$ units. The correct option is (C).

The key idea: the $y$ -axis is the set of all points where $x = 0$ and $z = 0$ , with $y$ free. So the perpendicular from any point to the $y$ -axis lands at the point that keeps the same $y$ -coordinate but sets $x$ and $z$ to zero. The distance is then just the straight-line distance in the $xz$ -plane from $(x, z)$ to $(0, 0)$ .

Think of it this way: if you stand at $(4, -7, 3)$ and walk straight toward the $y$ -axis, you move only in the $x$ and $z$ directions — your $y$ doesn't change because the axis runs parallel to your $y$ direction. The shortest path is the hypotenuse of a right triangle with legs $4$ (the $x$ -distance) and $3$ (the $z$ -distance).

Identify the foot of the perpendicular.

The $y$ -axis consists of points $(0, y, 0)$ . The perpendicular from $(4, -7, 3)$ to this axis will have the same $y$ -coordinate as the point, because the axis is vertical in $y$ . So the foot is $(0, -7, 0)$ .

Compute the distance between the point and the foot.

Use the 3D distance formula:

$\text{Distance} = \sqrt{(4 - 0)^2 + (-7 - (-7))^2 + (3 - 0)^2}$
The $y$ -difference is $0$ , so it simplifies to:
$\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Interpret geometrically. …
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CBSE 2020Set 65/2/11 markMCQ
$(2, -3, 4)$
›Reveal solutionSolution
The foot of the perpendicular from any point to the y-axis has the form $(0, y, 0)$ ; we preserve only the y-coordinate of the original point. The answer is $(0, -3, 0)$ .

The y-axis is the set of all points of the form $(0, y, 0)$ where $y$ ranges over all real numbers. When we drop a perpendicular from a point in space to the y-axis, we're finding the closest point on that axis.

The key insight is geometric: a perpendicular from a point to a coordinate axis lands at the point on that axis which shares the same coordinate as the original point along that axis direction. Think of it as projecting the point straight onto the axis, collapsing the other two dimensions.

For the y-axis specifically, we keep the y-coordinate and set both x and z to zero.

Identify the structure of points on the y-axis

Any point on the y-axis has coordinates $(0, y, 0)$ for some value of $y$ . The x and z coordinates are always zero.

Understand what "foot of the perpendicular" means

The foot of the perpendicular from point $P = (2, -3, 4)$ to the y-axis is the point $Q$ on the y-axis such that the line segment $PQ$ is perpendicular to the y-axis.

Apply the perpendicularity condition …
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CBSE 2020Set 65/3/11 markMCQ
$(-2, 8, 7)$
›Reveal solutionSolution
The foot of the perpendicular from any point to the XZ-plane is found by setting the $y$ -coordinate to zero while keeping $x$ and $z$ unchanged; the answer is $(-2, 0, 7)$ .

The XZ-plane is the set of all points where $y = 0$ . Think of it as the "floor" that contains the $x$ -axis and $z$ -axis but has no height in the $y$ -direction. When we drop a perpendicular from a point in space to this plane, we're asking: what point on the XZ-plane is closest to our given point?

The key insight is that a perpendicular to the XZ-plane must be parallel to the $y$ -axis (since the XZ-plane is perpendicular to the $y$ -axis). So dropping a perpendicular means moving straight up or down in the $y$ -direction until we hit the plane where $y = 0$ . The $x$ and $z$ coordinates don't change during this vertical drop.

Let's work through this systematically.

Identify the given point and the target plane.

We have the point $P = (-2, 8, 7)$ and we need to find where the perpendicular from $P$ meets the XZ-plane.

Recall the equation of the XZ-plane.

The XZ-plane consists of all points $(x, y, z)$ satisfying $y = 0$ . This is a plane perpendicular to the $y$ -axis.

Understand the geometry of the perpendicular.

A line perpendicular to the XZ-plane must be parallel to the normal vector of that plane. The normal to the XZ-plane is $\hat{j} = (0, 1, 0)$ , pointing in the $y$ -direction. Therefore, the perpendicular from $P$ is a vertical line (in the $y$ -sense) passing through $P$ . …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. Shortest distance between Point (a, b, c) and y-axis is:$
(a) √(a²+b²)
(b) √(b²+c²)
(c) √(c²+a²)
(d) √(a²+b²+c²)
›Reveal solutionSolution

Every point on the $y$ -axis has the form $(0,t,0)$ ; the perpendicular distance from $(a,b,c)$ to the axis ignores the $y$ -coordinate.

The foot of the perpendicular from $(a,b,c)$ to the $y$ -axis is $(0,b,0)$ , so the shortest distance is …

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CBSE 2019Set ANNUAL1 markMCQ
$(3, 4, 5)$
(a) $3$
(b) $5$
(c) $\sqrt{41}$
(d) None of these
›Reveal solutionSolution

Distance $=\sqrt{y^2+z^2}=\sqrt{41}$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$(3, 4, 5)$
(a) $4$
(b) $3$
(c) $5$
(d) $0$
›Reveal solutionSolution

Distance from the XZ-plane is $|y|=4$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$(a, b, c)$
(a) $\sqrt{b^2+c^2}$
(b) $\sqrt{a^2+b^2}$
(c) $\sqrt{c^2+a^2}$
(d) $\sqrt{a^2+b^2+c^2}$
›Reveal solutionSolution

Distance of $(a,b,c)$ from the X-axis is $\sqrt{b^2+c^2}$ .

A general point on the X-axis is $(t,0,0)$ . The nearest such point to $(a,b,c)$ is $(a,0,0)$ , so the shortest distance …

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CBSE 2018Set ANNUAL1 mark
$(a-b)x^2 + (a+b)y^2 = a^2 - b^2$
›Reveal solutionSolution

Divide the whole equation by a²−b² = (a−b)(a+b) to get x²/(a+b) + y²/(a−b) = 1.

We start with (a−b)x² + (a+b)y² = a²−b².

Step 1: Factor the right side. a²−b² = (a−b)(a+b).

Step 2: Divide both sides by (a−b)(a+b):

(a−b)x² / [(a−b)(a+b)] + (a+b)y² / [(a−b)(a+b)] = 1

Step 3: Cancel the common factors: …

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Q.The distance of point P(a,b,c)P(a, b, c)P(a,b,c) from the yyy-axis is: (A) bbb (B) b2b^2b2 (C) a2+c2\sqrt{a^2 + c^2}a2+c2​ (D) a2+c2a^2 + c^2a2+c2

Understanding Distance from an Axis

Step-by-Step Solution

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Q.The distance of point $P(a, b, c)$ from the $y$ -axis is: (A) $b$ (B) $b^2$ (C) $\sqrt{a^2 + c^2}$ (D) $a^2 + c^2$

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