NDA Maths · Lines

Triangles, Quadrilaterals & Polygons

Coordinate geometry applied to figures: the area of a triangle from its vertices, the triangle centres (centroid, incentre, circumcentre), constructing vertices from medians/altitudes, and quadrilateral relations.

All Lines notes NDA Maths strategy

Why this matters

This is the chapter's largest subtopic and its capstone — it combines slope, distance, section and area into figure problems. Knowing the area determinant and the centre formulas cold makes most of it routine.

Concept 1 of 4

Area of a triangle and collinearity

Intuition

The area of a triangle from its three vertices is half the absolute value of a determinant. When that determinant is zero the 'triangle' has collapsed — the points are collinear.

Definition

Area $=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ $=\dfrac12\left|\begin{smallmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{smallmatrix}\right|$ . Collinear iff this is $0$ . The same determinant gives the condition three points lie on a line.

Area of a triangle from vertices

\text{Area}=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Worked example

Find the area of the triangle with vertices

(0,0),(4,0),(0,3)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1LinesHARD

What is the area of the triangle with vertices

\left(x_1,\frac{1}{x_1}\right), \left(x_2,\frac{1}{x_2}\right), \left(x_3,\frac{1}{x_3}\right)

[Q61 · Sep · 2018]

Don't forget the $\tfrac12$ and the absolute value — and collinearity is area $=0$

Two routine slips on the area formula

\tfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

: dropping the leading **

\tfrac12

(which doubles the answer), and omitting the modulus** (a clockwise vertex order makes the bare determinant negative — area can't be negative). When the expression equals

0

, the three points are collinear, so the 'area' test and the collinearity test are the same computation.

Drill 4 more on area of a triangle and collinearity

Concept 2 of 4

Centroid, incentre, circumcentre

Intuition

Each triangle centre has a formula: the centroid is the plain average of the vertices; the incentre is the side-length-weighted average; the circumcentre is equidistant from all three vertices (intersection of perpendicular bisectors).

Definition

Centroid: $\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$ .
Incentre: $\dfrac{a\,A+b\,B+c\,C}{a+b+c}$ , with $a,b,c$ the side lengths opposite $A,B,C$ .
Circumcentre: equidistant from all vertices — solve two perpendicular-bisector equations (for a right triangle it is the hypotenuse's midpoint).

Centroid and incentre

G=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)\qquad I=\dfrac{a\,A+b\,B+c\,C}{a+b+c}

Worked example

Find the centroid of the triangle with vertices

(1,2),(3,4),(5,0)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2LinesMODERATE

The incentre of the triangle with vertices

A(1, \sqrt{3})

B(0, 0)

and

C(2, 0)

[Q52 · Apr · 2017]

Centroid is the plain average; the incentre weights by the opposite side lengths $a,b,c$

The centroid

G

is the unweighted average of the three vertices,

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

. The incentre is *not* that average — it is the side-length-weighted average

\dfrac{aA+bB+cC}{a+b+c}

, where

a,b,c

are the lengths of the sides opposite vertices

A,B,C

. Using equal weights for the incentre (or pairing a side with its adjacent vertex) is the usual mistake; the two centres coincide only for an equilateral triangle.

Drill 2 more on centroid, incentre, circumcentre

Concept 3 of 4

Constructing a triangle: vertices, medians, altitudes

Intuition

Many questions give partial data — midpoints, a median, an altitude — and ask for a vertex or a side's equation. Use midpoint and section relations to recover vertices, and slope/perpendicularity for altitudes and bisectors.

Definition

Recover vertices from midpoints: if $M$ is the midpoint of $BC$ , then $B+C=2M$ . An altitude from a vertex is perpendicular to the opposite side (use negative-reciprocal slope through the vertex). Special triangles: an equilateral/isosceles condition fixes the third vertex (often via rotation or equal-distance). The third vertex of an equilateral triangle on a given base has irrational coordinates in general.

Vertex from a midpoint

B+C=2M\qquad C=2M-B

Worked example

Two vertices of a triangle are

B(-5,-1)

and

C(9,3)

. Find the midpoint of

BC

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3LinesMODERATE

Triangle

ABC

has

A(3,5)

. The mid-points of sides

AB

and

AC

are at

(-1,2)

and

(6,4)

respectively. What are the coordinates of the centroid of the triangle

ABC

[Q62 · Apr · 2024]

An altitude is perpendicular to the opposite side — use the negative-reciprocal slope

The altitude from a vertex is perpendicular to the opposite side, so its slope is the negative reciprocal of that side's slope (not the same slope, which would be parallel, and not the side's own slope). To recover a vertex from a midpoint, use

B+C=2M\Rightarrow C=2M-B

— i.e.

2M-B

, not

M-B

; the factor of

2

is essential because

M

is the *average* of

B

and

C

Drill 10 more on constructing a triangle: vertices, medians, altitudes

Concept 4 of 4

Parallelograms, squares and diagonals

Intuition

Quadrilateral problems lean on two facts: in a parallelogram the diagonals bisect each other (so the fourth vertex is

D=A+C-B

), and the diagonals' intersection is their common midpoint. Areas come from the cross-product of adjacent side vectors.

Definition

**Parallelogram $ABCD$ :** diagonals bisect each other ⇒ $A+C=B+D$ , so $D=A+C-B$ ; the diagonals meet at the midpoint of either. Area $=|\,\vec{AB}\times\vec{AD}\,|=|x_1y_2-x_2y_1|$ for the side vectors. A square/rectangle from two given parallel sides uses the perpendicular distance for the side length.

Parallelogram: fourth vertex and area

A+C=B+D\qquad D=A+C-B\qquad \text{Area}=|x_1y_2-x_2y_1|

Worked example

Three consecutive vertices of a parallelogram are

A(1,2),B(4,3),C(6,6)

. Find

D

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4LinesEASY

A parallelogram has three consecutive vertices

(-3, 4)

(0, -4)

and

(5, 2)

. The fourth vertex is

[Q57 · Apr · 2021]

In parallelogram $ABCD$ the diagonals are $AC$ and $BD$ : $A+C=B+D$

The diagonals of

ABCD

join opposite vertices —

AC

and

BD

— and they bisect each other, so the midpoints match:

A+C=B+D

, giving

D=A+C-B

. The slip is pairing adjacent vertices (e.g. computing

A+B

); the order of the labels around the parallelogram tells you which pairs are diagonals. Get the pairing wrong and the 'fourth vertex' is misplaced.

Drill 12 more on parallelograms, squares and diagonals

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 (4)

Area of a triangle and collinearity
Area of a triangle from vertices
$\text{Area}=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$
Centroid, incentre, circumcentre
Centroid and incentre
$G=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)\qquad I=\dfrac{a\,A+b\,B+c\,C}{a+b+c}$
Constructing a triangle: vertices, medians, altitudes
Vertex from a midpoint
$B+C=2M\qquad C=2M-B$
Parallelograms, squares and diagonals
Parallelogram: fourth vertex and area
$A+C=B+D\qquad D=A+C-B\qquad \text{Area}=|x_1y_2-x_2y_1|$

Watch out for (4)

Don't forget the $\tfrac12$ and the absolute value — and collinearity is area $=0$
→ Area of a triangle and collinearity
Centroid is the plain average; the incentre weights by the opposite side lengths $a,b,c$
→ Centroid, incentre, circumcentre
An altitude is perpendicular to the opposite side — use the negative-reciprocal slope
→ Constructing a triangle: vertices, medians, altitudes
In parallelogram $ABCD$ the diagonals are $AC$ and $BD$ : $A+C=B+D$
→ Parallelograms, squares and diagonals

Drill every past-year question on this subtopic

32 questions from the bank — paginated, with cart and Word-export support.

Drill the 32 questions

Drill every past-year question on this subtopic

Related notes