If the origin is the centriod of a triangle ABC having vertices A(a,1,3), B(-2,b,-5) and C(4,7,c), find the va

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A(a,1,3)

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Concept understanding — Coordinate Geometry

Coordinate Geometry: Where Algebra Meets Geometry

Imagine you're telling a friend where you left your book in a library. You don't say "near the window" — you say "third shelf, second row, fourth book from the left." You're using numbers to pin down an exact location.

Coordinate geometry does the same thing, but for points on a flat surface. It gives every point a precise address — a pair of numbers — so we can describe shapes, distances, and positions using algebra.

The Big Idea

Before coordinate geometry, geometry was about drawing shapes and proving things with logic alone. Algebra was about numbers and equations. These two worlds seemed separate.

Then René Descartes (a French mathematician) had a simple but revolutionary idea: draw two perpendicular number lines that cross at zero. Now every point on the plane has a unique pair of numbers — its coordinates.

That's it. That's the entire foundation.

The Coordinate System

Take a horizontal line — call it the x-axis. Take a vertical line — call it the y-axis. They cross at a point called the origin, labelled $O$ .

Any point $P$ is located by two numbers:

Its x-coordinate: how far right (positive) or left (negative) from the origin
Its y-coordinate: how far up (positive) or down (negative) from the origin

We write this as an ordered pair: $(x, y)$ .

Note

The order matters. $(3, 5)$ is not the same point as $(5, 3)$ . The first number is always the horizontal position; the second is always the vertical.

A Concrete Example

Plot the point $A(2, 3)$ :

Start at the origin $(0, 0)$ .
Move 2 units to the right along the x-axis.
From there, move 3 units up (parallel to the y-axis).
Mark the point.

Now plot $B(-1, 4)$ :

Start at the origin.
Move 1 unit left (negative x-direction).
Move 4 units up.
Mark the point.

Every point on the plane has exactly one such address. And every pair of numbers corresponds to exactly one point. This one-to-one matching is what makes coordinate geometry powerful.

The Four Quadrants

The axes divide the plane into four regions, called quadrants:

Quadrant	x-sign	y-sign	Example
I	$+$	$+$	$(2, 3)$
II	$-$	$+$	$(-1, 4)$
III	$-$	$-$	$(-3, -2)$
IV	$+$	$-$	$(5, -1)$

Points on the axes themselves (where either coordinate is zero) don't belong to any quadrant.

Why This Matters

Once every point has a number address, we can:

Calculate distances between points using the Pythagorean theorem
Find midpoints by averaging coordinates
Describe lines with equations like $y = mx + c$
Solve geometric problems using algebra instead of drawing

Important

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

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

This is just the Pythagorean theorem in disguise.

The Precise Statement

Coordinate geometry (also called analytic geometry) is the study of geometry using a coordinate system. It establishes a correspondence between:

Points on a plane and ordered pairs of real numbers
Geometric figures (lines, circles, curves) and algebraic equations …

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The centroid of a triangle is the average of its vertices’ coordinates. Setting the origin $(0,0,0)$ equal to the centroid gives three simple equations, yielding $a = -2$ , $b = -8$ , $c = 2$ .

We are told that the origin $(0,0,0)$ is the centroid of triangle $ABC$ . The centroid of a triangle in 3D (or any dimension) is simply the arithmetic mean of the coordinates of its three vertices. That is the core idea — no fancy geometry, just averaging.

If you ever forget the centroid formula, think of it as the “balance point”: if equal masses are placed at the vertices, the centre of mass is the average position. That’s exactly what we use here.

Let’s write the vertices:

A(a,1,3),\quad B(-2,b,-5),\quad C(4,7,c)

The centroid $G$ has coordinates:

G = \left( \frac{x_A + x_B + x_C}{3},\; \frac{y_A + y_B + y_C}{3},\; \frac{z_A + z_B + z_C}{3} \right)

We are given that $G = (0,0,0)$ . So each coordinate average must equal zero.

For the $x$ -coordinate:

\frac{a + (-2) + 4}{3} = 0

Multiply by 3:

a - 2 + 4 = 0 \quad\Rightarrow\quad a + 2 = 0

Hence:

a = -2

For the $y$ -coordinate:

\frac{1 + b + 7}{3} = 0

Multiply by 3:

1 + b + 7 = 0 \quad\Rightarrow\quad b + 8 = 0

Hence:

b = -8

For the $z$ -coordinate:

\frac{3 + (-5) + c}{3} = 0

Multiply by 3:

3 - 5 + c = 0 \quad\Rightarrow\quad c - 2 = 0 …

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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 22 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$O$
›Reveal solutionSolution

The line is the perpendicular to $OP$ at its foot; use the normal form $x\cos\alpha+y\sin\alpha=p$ with $p=OP=5$ .

Since $OP$ is perpendicular to the line, $OP$ is the normal from the origin and $p=OP=5$ . The angle is obtuse with $\sin\alpha=\tfrac35$ , so $\cos\alpha=-\sqrt{1-\tfrac{9}{25}}=-\tfrac45$ . …
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KEAM 2026Set eng-2026-04174 marksMCQ
$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$
›Reveal solutionSolution

The shared focal distance $c$ satisfies $c^2=13$ ; for the hyperbola $a^2+b^2=c^2$ .

For the ellipse $\dfrac{x^2}{49}+\dfrac{y^2}{36}=1$ , $c^2=49-36=13$ . …
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KEAM 2026Set eng-2026-04174 marksMCQ
$\dfrac{1}{2}$
›Reveal solutionSolution

From $2c=4$ and $e=\tfrac12$ get $a,b^2$ , then $\dfrac{2b^2}{a}$ .

Distance between foci $2c=4\Rightarrow c=2$ . Eccentricity $e=\dfrac{c}{a}=\dfrac12\Rightarrow a=4$ .

$b^2=a^2-c^2=16-4=12$ . …
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KEAM 2026Set eng-2026-04184 marksMCQ
$C$
›Reveal solutionSolution

The perpendicular bisector is $x=5$ ; the point 8 units from $(5,-6)$ in the first quadrant is $(5,2)$ .

$A(-3,-6)$ and $B(13,-6)$ share $y=-6$ , so their midpoint is $M=(5,-6)$ and the perpendicular bisector is the vertical line $x=5$ . …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\dfrac{1}{5}$
›Reveal solutionSolution

Combine LR $=\tfrac{2b^2}{a}$ with $b^2=a^2(1-e^2)$ to get $a=5$ , so major axis $2a=10$ .

With $e=\tfrac15$ , $b^{2}=a^{2}(1-e^{2})=a^{2}\big(1-\tfrac{1}{25}\big)=\tfrac{24}{25}a^{2}$ . …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\dfrac{(3x-7)^{2}}{9}-\dfrac{(4y+3)^{2}}{8}=1$
›Reveal solutionSolution

Normalizing to $(x-\tfrac73)^2/1-(y+\tfrac34)^2/\tfrac12=1$ gives $a^2=1$ , $b^2=\tfrac12$ , so LR $=1$ .

Write $3x-7=3(x-\tfrac73)$ and $4y+3=4(y+\tfrac34)$ :

$\frac{9(x-\tfrac73)^2}{9}-\frac{16(y+\tfrac34)^2}{8}=\frac{(x-\tfrac73)^2}{1}-\frac{(y+\tfrac34)^2}{1/2}=1.$ …
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KEAM 2026Set eng-2026-04214 marksMCQ
$\frac{2\sqrt{2}}{3}$
›Reveal solutionSolution

Major axis $=3\times$ minor axis means $a=3b$ , giving $e=\frac{2\sqrt2}{3}$ .

Length of major axis $=2a$ , minor axis $=2b$ . Given $2a=3(2b)$ , so $a=3b$ . Then …
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KEAM 2026Set eng-2026-04214 marksMCQ
$O$
›Reveal solutionSolution

The midpoint of $A(-2,1)$ and $B(4,-3)$ is $(1,-1)$ , giving $\overrightarrow{OC}=\hat i-\hat j$ .

Midpoint:

$C=\left(\frac{-2+4}{2},\frac{1+(-3)}{2}\right)=(1,-1).$ …
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KEAM 2026Set eng-2026-04224 marksMCQ
$x$
›Reveal solutionSolution

Get the base line's intercepts, scale them, then form the intercept-form equation.

For $3x+2y=6$ : x-intercept $=2$ (set $y=0$ ), y-intercept $=3$ (set $x=0$ ).

New x-intercept $=3\times2=6$ ; new y-intercept $=4\times3=12$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$(\pm 2, 0)$
›Reveal solutionSolution

Read $a$ from the vertices and $c$ from the foci; eccentricity is $c/a$ .

For the ellipse, $a=3$ (vertices $(\pm3,0)$ ) and $c=2$ (foci $(\pm2,0)$ ). …
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KEAM 2026Set eng-2026-04224 marksMCQ
$9x^{2} - 16y^{2} = 144$
›Reveal solutionSolution

Put the hyperbola in standard form, find $a$ and $c$ , then $AS=c-a$ .

$9x^2-16y^2=144\Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1$ , so $a^2=16$ , $b^2=9$ . …
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KEAM 2025Set eng-2025-04234 marksMCQ
$m$
›Reveal solutionSolution
The major axis length is $9$ .

Concept and Intuition

With major axis $m=2a$ and minor axis $n=2b$ , $m^2-n^2=4(a^2-b^2)=4a^2e^2$ . Use the given eccentricity to solve for $a$ .

Step-by-Step Solution

$m^2-n^2=4a^2-4b^2=4(a^2-b^2)=4a^2e^2$ .

$e^2=5/9$ , so $4a^2\cdot\tfrac{5}{9}=45\Rightarrow \tfrac{20a^2}{9}=45\Rightarrow a^2=20.25$ , $a=4.5$ .

Major axis $m=2a=9$ . …
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Q.If the origin is the centriod of a triangle ABC having vertices A(a,1,3)A(a,1,3)A(a,1,3), B(−2,b,−5)B(-2,b,-5)B(−2,b,−5) and C(4,7,c)C(4,7,c)C(4,7,c), find the values of aaa, bbb, ccc.

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Q.If the origin is the centriod of a triangle ABC having vertices $A(a,1,3)$ , $B(-2,b,-5)$ and $C(4,7,c)$ , find the values of $a$ , $b$ , $c$ .

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