The mid-points of the sides of a triangle are (5,7,11), (0,8,5) and (2,3,-1). Find its vertices.

NCERT Exemplar · Q13

(5,7,11)

Puducherry CbseShort· 3mImportance★★★★★est

49% · 35/72 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 →

Treating each given midpoint as the average of two vertices and solving the resulting equations gives the vertices $(-3,4,-7)$ , $(7,2,5)$ and $(3,12,17)$ .

Why This Works

The midpoint of a segment is just the average of its two endpoints. If we know the midpoints of all three sides of a triangle, each midpoint ties together two unknown vertices. Because every vertex is shared by exactly two sides, three midpoints give us enough equations to recover all three vertices.

Setting Up

Let the vertices be $A(x_1,y_1,z_1)$ , $B(x_2,y_2,z_2)$ , $C(x_3,y_3,z_3)$ , and assign the given midpoints to the sides:

$D(5,7,11)$ = midpoint of $BC$
$E(0,8,5)$ = midpoint of $CA$
$F(2,3,-1)$ = midpoint of $AB$

By the midpoint formula in space,

D=\frac{B+C}{2},\qquad E=\frac{C+A}{2},\qquad F=\frac{A+B}{2}.

Isolating Each Vertex

Add the equations for $E$ and $F$ and subtract $D$ :

E+F-D=\frac{C+A}{2}+\frac{A+B}{2}-\frac{B+C}{2}=\frac{2A}{2}=A.

The same cancellation gives the other two vertices:

A=E+F-D,\qquad B=F+D-E,\qquad C=D+E-F.

Vertex $A=E+F-D$ :

A=(0+2-5,\;8+3-7,\;5+(-1)-11)=(-3,\,4,\,-7).

Vertex $B=F+D-E$ :

B=(2+5-0,\;3+7-8,\;-1+11-5)=(7,\,2,\,5).

Vertex $C=D+E-F$ :

C=(5+0-2,\;7+8-3,\;11+5-(-1))=(3,\,12,\,17).

Verification

Check each midpoint against the recovered vertices: …

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

←Previous questionQ12 — Find the centroid of a triangle, the mid-point of whose sides are D(1,…→Next questionQ14 — Three vertices of a Parallelogram ABCD are A(1,2,3), B(-1,-2,-1) and C…

Q.The mid-points of the sides of a triangle are (5,7,11)(5,7,11)(5,7,11), (0,8,5)(0,8,5)(0,8,5) and (2,3,−1)(2,3,-1)(2,3,−1). Find its vertices.

Why This Works

Setting Up

Isolating Each Vertex

Verification

🎓Unlock everything free for 14 days

Q.The mid-points of the sides of a triangle are $(5,7,11)$ , $(0,8,5)$ and $(2,3,-1)$ . Find its vertices.

Unlock everything free for 14 days