SPI Mathematics · Coordinate Geometry

Section Formula, Midpoint and Centroid

The point that divides a join in the ratio m : n is a weighted average of the two ends; the midpoint is the 1 : 1 case and the centroid is the average of a triangle's three corners.

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Why this matters

Twelve questions in the bank, eight EASY and four MODERATE, including the chapter's one question from the 2022 paper, a centroid. The MODERATE ones run the formulas backwards: find an end from the dividing point, find the ratio an axis makes, or find a parallelogram's fourth corner.

Concept 1 of 3: The section formula

A point that splits a join in the ratio m : n lies closer to the end with the smaller share. Its coordinates are a weighted average: each end is weighted by the OTHER end's share of the ratio.

Definition

P divides the join of $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m : n$ (AP : PB).
$P = \left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\right)$ : m multiplies B's coordinates, n multiplies A's.
Ratio made by an axis: call the ratio $k : 1$ . On the x-axis the y-coordinate is 0; on the y-axis the x-coordinate is 0. Solve for k.
If P is known and an end is not, put the end in as $(x, y)$ and solve each coordinate.

Section formula

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

m : nthe ratio AP : PB
$(x_1, y_1)$ the first end, A
$(x_2, y_2)$ the second end, B

Worked example

Find the point that divides the join of A(2, −3) and B(7, 7) in the ratio 2 : 3.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Coordinate Geometry · Section Formula, Midpoint and Centroid

If the point P

(-1, -1)

divides the join of A

(-3, 3)

and B internally in the ratio 2 : 3 then the coordinates of B will be

Putting m with the wrong end

\frac{mx_2 + nx_1}{m + n}

, m multiplies the SECOND point. Swapping them gives the point that divides the join in the ratio n : m, and that point is usually one of the wrong options.

Setting the wrong coordinate to zero

A point on the x-axis has y = 0, so for 'the x-axis divides the join' solve the y-equation. Using the x-equation answers a different question.

Drill 3 more on the section formula

Concept 2 of 3: The midpoint and its uses

The midpoint is the section formula with a 1 : 1 ratio: just average the two x values and the two y values. It finds a circle's centre from a diameter and a missing corner of a parallelogram.

Definition

Midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ : $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ .
The centre of a circle is the midpoint of any diameter.
The diagonals of a parallelogram ABCD bisect each other, so $A + C = B + D$ (add the x values, then the y values).
In a right triangle, the midpoint of the hypotenuse is the same distance from all three corners.

Midpoint formula

M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right) \qquad D = A + C - B

Worked example

Find the midpoint of the join of (−3, 8) and (5, −2).

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Coordinate Geometry · Section Formula, Midpoint and Centroid

If A

(2, -1)

(3, 2)

(-4, 5)

are the vertices of the parallelogram ABCD, then the coordinates of D will be.

Adding the wrong pair of corners

In parallelogram ABCD the diagonals are AC and BD, so D = A + C − B. Writing D = A + B − C pairs corners that are next to each other, and gives a point that makes a different shape.

Drill 3 more on the midpoint and its uses

Concept 3 of 3: The centroid of a triangle

The three medians of a triangle meet at one point, the centroid, its balancing point. Its coordinates are simply the averages of the three corners.

Definition

Centroid of $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ : $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$ .
The medians meet at the centroid, and it divides each median in the ratio 2 : 1 from the vertex.
To find a missing vertex or letter, write the average equal to the given centroid and solve, one coordinate at a time.

Centroid

G = \left(\frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3}\right)

Worked example

Find the centroid of the triangle with vertices (1, 2), (4, 6) and (7, −2).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q73Easy

Example 3 · Coordinate Geometry · Section Formula, Midpoint and Centroid

The third vertex of a triangle, if two of its vertices are (5,2) and (-1,4) and its medians intersect at (0,-3) is ____.

Dividing by 2 instead of 3

The centroid averages THREE corners, so divide the sum by 3. Dividing by 2 is the midpoint habit, and the point it gives is not the centroid.

Forgetting to multiply back

To find a missing vertex, multiply the centroid by 3 first: the three x values must add to 3 times the centroid's x. Subtracting the known vertices straight from the centroid gives the wrong point.

Drill 3 more on the centroid of a triangle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The section formula
Section formula
$P = \left(\frac{mx_2 + nx_1}{m + n},\ \frac{my_2 + ny_1}{m + n}\right)$
The midpoint and its uses
Midpoint formula
$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right) \qquad D = A + C - B$
The centroid of a triangle
Centroid
$G = \left(\frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3}\right)$

Watch out for (5)

Putting m with the wrong end
→ The section formula
Setting the wrong coordinate to zero
→ The section formula
Adding the wrong pair of corners
→ The midpoint and its uses
Dividing by 2 instead of 3
→ The centroid of a triangle
Forgetting to multiply back
→ The centroid of a triangle

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