Find the coordinate of the points which trisect the line segment joining the points A(2,1,-3) and B(5,-8,3).

NCERT Exemplar · Q15

A(2,1,-3)

Kerala DhseShort· 2mImportance★★★★★est

51% · 37/72 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

To trisect a line segment means to divide it into three equal parts using two points. Using the section formula in 3D with ratios $1:2$ and $2:1$ , the trisecting points are $P_1(3, -2, -1)$ and $P_2(4, -5, 1)$ .

When we trisect a line segment, we're looking for the two points that divide it into three equal pieces. Think of cutting a rope into three equal lengths—you need two cuts, and those cut-points are what we're after.

The section formula is the natural tool here. If a point divides a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ internally, its coordinates are:

\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)

For trisection, the first point $P_1$ (closer to $A$ ) divides $AB$ in the ratio $1:2$ , and the second point $P_2$ divides $AB$ in the ratio $2:1$ .

Finding the trisecting points

1. First trisecting point $P_1$ (ratio $1:2$ from $A$ to $B$ )

Here $m = 1$ , $n = 2$ , with $A(2, 1, -3)$ and $B(5, -8, 3)$ .

For the $x$ -coordinate:

x_1 = \frac{1 \cdot 5 + 2 \cdot 2}{1 + 2} = \frac{5 + 4}{3} = \frac{9}{3} = 3

For the $y$ -coordinate:

y_1 = \frac{1 \cdot (-8) + 2 \cdot 1}{3} = \frac{-8 + 2}{3} = \frac{-6}{3} = -2

For the $z$ -coordinate:

z_1 = \frac{1 \cdot 3 + 2 \cdot (-3)}{3} = \frac{3 - 6}{3} = \frac{-3}{3} = -1

So $P_1 = (3, -2, -1)$ .

2. Second trisecting point $P_2$ (ratio $2:1$ from $A$ to $B$ )

Now $m = 2$ , $n = 1$ . …

Start your 14-day free trial to unlock the full solution →

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 $\cos\alpha=-\sqrt{1-\tfrac{9}{25}}=-\tfrac45$ . …
Start your 14-day free trial to see the full solution →
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 $\dfrac{x^2}{49}+\dfrac{y^2}{36}=1$ , $c^2=49-36=13$ . …
Start your 14-day free trial to see the full solution →
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 $e=\dfrac{c}{a}=\dfrac12\Rightarrow a=4$ .

$b^2=a^2-c^2=16-4=12$ . …
Start your 14-day free trial to see the full solution →
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 $x=5$ . …
Start your 14-day free trial to see the full solution →
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 $e=\tfrac15$ , $b^{2}=a^{2}(1-e^{2})=a^{2}\big(1-\tfrac{1}{25}\big)=\tfrac{24}{25}a^{2}$ . …
Start your 14-day free trial to see the full solution →
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 $4y+3=4(y+\tfrac34)$ :

$\frac{9(x-\tfrac73)^2}{9}-\frac{16(y+\tfrac34)^2}{8}=\frac{(x-\tfrac73)^2}{1}-\frac{(y+\tfrac34)^2}{1/2}=1.$ …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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:

$C=\left(\frac{-2+4}{2},\frac{1+(-3)}{2}\right)=(1,-1).$ …
Start your 14-day free trial to see the full solution →
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 $=4\times3=12$ . …
Start your 14-day free trial to see the full solution →
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 $(\pm2,0)$ ). …
Start your 14-day free trial to see the full solution →
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 $a^2=16$ , $b^2=9$ . …
Start your 14-day free trial to see the full solution →
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 $m=2a=9$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ14 — Three vertices of a Parallelogram ABCD are A(1,2,3), B(-1,-2,-1) and C…→Next questionQ16 — If the origin is the centriod of a triangle ABC having vertices A(a,1,…

Q.Find the coordinate of the points which trisect the line segment joining the points A(2,1,−3)A(2,1,-3)A(2,1,−3) and B(5,−8,3)B(5,-8,3)B(5,−8,3).

Finding the trisecting points

🎓Unlock everything free for 14 days

Q.Find the coordinate of the points which trisect the line segment joining the points $A(2,1,-3)$ and $B(5,-8,3)$ .

Unlock everything free for 14 days