The mid-point of the sides of a triangle are (1,5,-1), (0,4,-2) and (2,3,4). Find its vertices. Also find the

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(1,5,-1)

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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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Each midpoint is the average of two vertices. Summing the three midpoint relations gives the vertex sum, and subtracting recovers each vertex: $A(3,4,5)$ , $B(-1,6,-7)$ , $C(1,2,3)$ . The centroid is $\left(1,4,\tfrac{1}{3}\right)$ .

Let the vertices be $A$ , $B$ , $C$ , with the given midpoints

M_{AB}=(1,5,-1),\quad M_{BC}=(0,4,-2),\quad M_{CA}=(2,3,4).

1. Midpoint (sum) relations. Multiplying each midpoint by $2$ :

A+B=(2,10,-2),\quad B+C=(0,8,-4),\quad C+A=(4,6,8)

2. Vertex sum. Adding all three:

2(A+B+C)=(6,24,2)\ \Rightarrow\ A+B+C=(3,12,1)

3. Recover each vertex by subtracting the opposite-side sum:

A=(A+B+C)-(B+C)=(3,12,1)-(0,8,-4)=(3,4,5)

B=(A+B+C)-(C+A)=(3,12,1)-(4,6,8)=(-1,6,-7)

$C=(A+B+C)-(A+B)=(3,12,1)-(2,10,-2)=(1,2,3)$ …

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Q.The mid-point of the sides of a triangle are (1,5,−1)(1,5,-1)(1,5,−1), (0,4,−2)(0,4,-2)(0,4,−2) and (2,3,4)(2,3,4)(2,3,4). Find its vertices. Also find the centriod of the triangle.

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Q.The mid-point of the sides of a triangle are $(1,5,-1)$ , $(0,4,-2)$ and $(2,3,4)$ . Find its vertices. Also find the centriod of the triangle.

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