Basic Concepts of Coordinate Geometry

Q: What is the difference between abscissa and ordinate?

The abscissa is the x-coordinate - the signed perpendicular distance of a point from the y-axis. The ordinate is the y-coordinate - the signed perpendicular distance from the x-axis. For a point P(3, -5), the abscissa is 3 and the ordinate is -5.

Q: How is external division different from internal division?

In internal division the point P lies between A and B on the segment. In external division P lies on the line AB but outside the segment. Algebraically the two formulas differ only in a sign - the internal formula has plus in numerator and denominator, the external has minus.

Q: What is the Euler line?

In any triangle (except equilateral), the orthocentre H, centroid G, and circumcentre O are collinear. This line is called the Euler line. The centroid G divides HO in the ratio 2:1 measured from H.

Q: When do we shift the origin instead of rotating axes?

Shift the origin to kill first-degree terms. Rotate the axes to kill the xy term (align the principal axes of a conic with the new coordinate axes). For a general second-degree curve both are usually needed - shift first, rotate second.

Q: What is the equation of a locus?

The equation of a locus is the algebraic relation between x and y that a moving point (x, y) satisfies if and only if it obeys the given geometric condition. Every point of the locus satisfies the equation; every solution of the equation is a point of the locus.

MathsStraight LinesFor JEE aspirants

Coordinate geometry (also called analytical geometry) is the branch of mathematics that uses algebraic equations to describe points, lines, and curves. In the two-dimensional plane, every point is fixed by an ordered pair $(x, y)$ measured against two perpendicular axes. The basic concepts of coordinate geometry - distance formula, section formula, centroid, incentre, circumcentre, orthocentre, area of a triangle, locus, and change of axes - form the foundation on which all of JEE Main and Advanced coordinate-geometry problems are built.

Key Formulas - Quick Reference

Distance: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Section (internal, ratio $m : n$ ): $(\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})$
Section (external, ratio $m : n$ ): $(\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n})$
Midpoint: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Centroid: $G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$
Incentre: $I = (\frac{a x _{1} + b x _{2} + c x _{3}}{a + b + c}, \frac{a y _{1} + b y _{2} + c y _{3}}{a + b + c})$
Area of triangle: $Δ = \frac{1}{2} x_{1} y_{1} 1 x_{2} y_{2} 1 x_{3} y_{3} 1$
Translation ( $O \to O^{'} (h, k)$ ): $x = X + h, y = Y + k$
Rotation (through angle $θ$ ): $x = X cos θ - Y sin θ, y = X sin θ + Y cos θ$

1. Rectangular Cartesian Coordinate System

Two perpendicular number lines - the $x$ -axis (horizontal) and the $y$ -axis (vertical) - meet at a point $O$ called the origin. Any point $P$ in the plane is located by an ordered pair $(x, y)$ , where $x$ (the abscissa) is the signed distance from the $y$ -axis and $y$ (the ordinate) is the signed distance from the $x$ -axis.

The two axes divide the plane into four regions called quadrants, numbered anti-clockwise starting from the top-right:

Figure 1. The rectangular Cartesian plane. For

P (x, y)

the abscissa

x

is the signed distance from the

y

-axis and the ordinate

y

is the signed distance from the

x

-axis. Quadrants are numbered anti-clockwise.

2. Distance Between Two Points

The distance between $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ is

A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} .

The formula is a direct application of Pythagoras' theorem to the right triangle whose legs are horizontal and vertical distances between the two points.

Figure 2. The distance formula is Pythagoras' theorem in disguise:

A B^{2} = ∣ x_{2} - x_{1} ∣^{2} + ∣ y_{2} - y_{1} ∣^{2}

, so

A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Note. Distance is always non-negative, so the order of subtraction doesn't matter:

(x_{2} - x_{1})^{2} = (x_{1} - x_{2})^{2}

. Distance of

P (x, y)

from the origin is

x^{2} + y^{2}

Using distance to identify quadrilaterals

Square: all four sides equal and both diagonals equal.
Rhombus (not a square): four sides equal but diagonals unequal.
Rectangle (not a square): opposite sides equal and diagonals equal.
Parallelogram (not a rectangle): opposite sides equal but diagonals unequal.
Equilateral triangle: all three sides equal.
Isosceles triangle: two sides equal.

In every parallelogram (and its special cases), the diagonals bisect each other, which gives another way to check.

Solved Example 1

Find

x

if the distance between

(x, - 1)

and

(3, 2)

5

Solution:

Using the distance formula, $(x - 3)^{2} + (- 1 - 2)^{2} = 5$ .

Squaring: $(x - 3)^{2} + 9 = 25 \Rightarrow (x - 3)^{2} = 16 \Rightarrow x - 3 = \pm 4$ .

So $x = 7$ or $x = - 1$ .

Solved Example 2

Find the circumcentre of the triangle with vertices

(8, 6)

(8, - 2)

, and

(2, - 2)

, and its circumradius.

Solution:

Let the circumcentre be $C (x, y)$ . Then $C$ is equidistant from all three vertices.

From $C A = C B$ : $(x - 8)^{2} + (y - 6)^{2} = (x - 8)^{2} + (y + 2)^{2}$ , giving $y = 2$ .

From $C B = C D$ : $(x - 8)^{2} + (y + 2)^{2} = (x - 2)^{2} + (y + 2)^{2}$ , giving $x = 5$ .

So $C = (5, 2)$ and circumradius $= (5 - 8)^{2} + (2 - 6)^{2} = 9 + 16 = 5$ .

3. Section Formula

Internal division

If $P (x, y)$ divides the line segment joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m : n$ , then

x = \frac{m x _{2} + n x _{1}}{m + n}, y = \frac{m y _{2} + n y _{1}}{m + n} .

External division

If $P$ divides $A B$ externally in the ratio $m : n$ (so $P$ lies on the line $A B$ outside the segment), then

x = \frac{m x _{2} - n x _{1}}{m - n}, y = \frac{m y _{2} - n y _{1}}{m - n} .

Figure 3. Internal versus external division.

P

lies between

A

and

B

with

A P : P B = m : n

;

Q

lies on the extension with

A Q : QB = m : n

. Only the sign in the denominator changes:

m + n

becomes

m - n

Midpoint

Setting $m = n = 1$ in the internal formula gives the midpoint:

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}) .

Sign convention. If

\frac{m}{n} > 0

the division is internal; if

\frac{m}{n} < 0

it is external. Some textbooks use one formula with a signed ratio.

Harmonic conjugate points

If $P$ divides $A B$ internally in the ratio $m : n$ and $Q$ divides $A B$ externally in the same ratio $m : n$ , then $P$ and $Q$ are called harmonic conjugates of each other with respect to $A$ and $B$ . In that case

\frac{2}{A B} = \frac{1}{A P} + \frac{1}{A Q},

so $A P$ , $A B$ , $A Q$ are in harmonic progression (H.P.).

Solved Example 3

Find the coordinates of the point dividing the segment joining

(6, 3)

and

(- 4, 5)

in the ratio

3 : 2

(i) internally and (ii) externally.

Solution:

(i) Internal: $x = \frac{3 ( - 4 ) + 2 ( 6 )}{3 + 2} = \frac{- 12 + 12}{5} = 0$ and $y = \frac{3 ( 5 ) + 2 ( 3 )}{5} = \frac{21}{5}$ . So $P = (0, \frac{21}{5})$ .

(ii) External: $x = \frac{3 ( - 4 ) - 2 ( 6 )}{3 - 2} = - 24$ and $y = \frac{3 ( 5 ) - 2 ( 3 )}{1} = 9$ . So $P = (- 24, 9)$ .

Solved Example 4

Find the points that trisect the segment joining

A (1, - 2)

and

B (- 3, 4)

Solution:

Let $P$ and $Q$ be the points of trisection with $A P = P Q = QB$ . Then $P$ divides $A B$ in the ratio $1 : 2$ (internally) and $Q$ in the ratio $2 : 1$ (internally).

$P = (\frac{1 ( - 3 ) + 2 ( 1 )}{3}, \frac{1 ( 4 ) + 2 ( - 2 )}{3}) = (- \frac{1}{3}, 0)$ .

$Q = (\frac{2 ( - 3 ) + 1 ( 1 )}{3}, \frac{2 ( 4 ) + 1 ( - 2 )}{3}) = (- \frac{5}{3}, 2)$ .

Solved Example 5

Three vertices of a parallelogram taken in order are

(- 1, 0)

(3, 1)

, and

(2, 2)

. Find the fourth vertex.

Solution:

Let the fourth vertex be $D (x, y)$ so the vertices in order are $A, B, C, D$ . In a parallelogram, diagonals $A C$ and $B D$ share the same midpoint.

Midpoint of $A C = (\frac{- 1 + 2}{2}, \frac{0 + 2}{2}) = (\frac{1}{2}, 1)$ .

Midpoint of $B D = (\frac{3 + x}{2}, \frac{1 + y}{2})$ .

Equating: $3 + x = 1 \Rightarrow x = - 2$ and $1 + y = 2 \Rightarrow y = 1$ . So $D = (- 2, 1)$ .

4. Special Points of a Triangle

Let $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ , $C (x_{3}, y_{3})$ be the vertices of a triangle with side lengths $a = B C$ , $b = C A$ , $c = A B$ .

Centroid $G$

The centroid is the intersection of the three medians (lines from each vertex to the midpoint of the opposite side). It divides each median in the ratio $2 : 1$ from the vertex.

G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3}) .

Figure 4. The centroid

G

is the common point of the three medians and divides each of them in the ratio

2 : 1

from the vertex, so

G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})

Incentre $I$

The incentre is the intersection of the three internal angle bisectors and the centre of the inscribed circle.

I = (\frac{a x _{1} + b x _{2} + c x _{3}}{a + b + c}, \frac{a y _{1} + b y _{2} + c y _{3}}{a + b + c}) .

Figure 5. The incentre

I

is where the three internal angle bisectors meet and is the centre of the incircle. Its coordinates are the side-weighted average

I = (\frac{a x _{1} + b x _{2} + c x _{3}}{a + b + c}, \frac{a y _{1} + b y _{2} + c y _{3}}{a + b + c})

, and the inradius is

r = \frac{Δ}{s}

Excentres $I_{1}, I_{2}, I_{3}$

Each excentre is the intersection of one internal angle bisector and the two external angle bisectors from the other two vertices. It is the centre of one of the three escribed (excircles).

I_{1} = (\frac{- a x _{1} + b x _{2} + c x _{3}}{- a + b + c}, \frac{- a y _{1} + b y _{2} + c y _{3}}{- a + b + c}) (opposite to A) .

$I_{2}$ (opposite to $B$ ) and $I_{3}$ (opposite to $C$ ) follow the same pattern - flip the sign of $b$ or $c$ respectively in both numerator and denominator.

Circumcentre $O$

The circumcentre is the intersection of the perpendicular bisectors of the three sides and the centre of the circumscribed circle. It is equidistant from the three vertices; the common distance is the circumradius $R$ .

Practical method: Let $O = (x, y)$ . Solve $O A^{2} = O B^{2}$ and $O B^{2} = O C^{2}$ - two linear equations in $x, y$ .

Orthocentre $H$

The orthocentre is the intersection of the three altitudes (perpendiculars from each vertex to the opposite side).

Practical method: Find the equations of any two altitudes and solve them simultaneously.

Key properties (memorise)

Euler line: The orthocentre $H$ , centroid $G$ , and circumcentre $O$ are always collinear, and $G$ divides $H O$ in the ratio $2 : 1$ (from $H$ ).
The incentre $I$ divides each internal angle bisector in the ratio $(b + c) : a$ (and cyclic).
Incentre and the excentre on the same angle bisector are harmonic conjugates with respect to the two feet of the bisector.
Isosceles triangle: $G$ , $I$ , $O$ , $H$ all lie on the axis of symmetry.
Equilateral triangle: $G$ , $I$ , $O$ , $H$ all coincide.
Right-angled triangle: orthocentre is at the right-angle vertex; circumcentre is the midpoint of the hypotenuse.
Obtuse-angled triangle: circumcentre and orthocentre both lie outside the triangle.

Figure 6. The Euler line. In every non-equilateral triangle the orthocentre

H

, centroid

G

and circumcentre

O

are collinear, with

H G : GO = 2 : 1

Solved Example 6

Find the centroid and the incentre of the triangle with vertices

A (0, 6)

B (8, 12)

C (8, 0)

Solution:

Centroid: $G = (\frac{0 + 8 + 8}{3}, \frac{6 + 12 + 0}{3}) = (\frac{16}{3}, 6)$ .

Side lengths. $a = B C = (8 - 8)^{2} + (12 - 0)^{2} = 12$ . $b = C A = (8 - 0)^{2} + (0 - 6)^{2} = 10$ . $c = A B = (0 - 8)^{2} + (6 - 12)^{2} = 10$ .

Incentre: $I_{x} = \frac{12 ( 0 ) + 10 ( 8 ) + 10 ( 8 )}{12 + 10 + 10} = \frac{160}{32} = 5$ . $I_{y} = \frac{12 ( 6 ) + 10 ( 12 ) + 10 ( 0 )}{32} = \frac{192}{32} = 6$ . So $I = (5, 6)$ .

Solved Example 7

Find the orthocentre of the triangle with vertices

A (0, 0)

B (1, 4)

C (3, 0)

Solution:

Side $A C$ lies on the $x$ -axis, so the altitude from $B$ is the vertical line $x = 1$ . Hence the orthocentre has $x$ -coordinate $1$ .

Altitude from $A (0, 0)$ is perpendicular to $B C$ . Slope of $B C = \frac{0 - 4}{3 - 1} = - 2$ , so altitude slope $= \frac{1}{2}$ , giving $y = \frac{x}{2}$ .

At $x = 1$ : $y = \frac{1}{2}$ . So the orthocentre is $H = (1, \frac{1}{2})$ .

5. Area of a Triangle and Polygon

Area of a triangle

The area of triangle $A B C$ with vertices $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ , $C (x_{3}, y_{3})$ is

Δ = \frac{1}{2} x_{1} y_{1} 1 x_{2} y_{2} 1 x_{3} y_{3} 1 = \frac{1}{2} x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) .

Without absolute value, the determinant is positive when the vertices are taken in anti-clockwise order and negative when clockwise. The sign carries useful information (orientation), so problems that ask for a signed area drop the modulus.

Figure 7. Orientation decides the sign. The determinant

\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

is positive for an anti-clockwise labelling and negative for a clockwise one; the area is its modulus.

Area of an $n$ -sided polygon (shoelace formula)

For a polygon with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), \dots, (x_{n}, y_{n})$ taken in order,

Area = \frac{1}{2} x_{1} x_{2} y_{1} y_{2} + x_{2} x_{3} y_{2} y_{3} + \dots + x_{n} x_{1} y_{n} y_{1} .

Solved Example 8

A (3, 4)

and

B (5, - 2)

and

P

satisfies

P A = P B

with area of

△ P A B = 10

, find

P

Solution:

Let $P = (x, y)$ . $P A = P B \Rightarrow (x - 3)^{2} + (y - 4)^{2} = (x - 5)^{2} + (y + 2)^{2}$ .

Expanding: $- 6 x - 8 y + 25 = - 10 x + 4 y + 29$ , giving $x - 3 y - 1 = 0$ .

Area $= \frac{1}{2} ∣ x (4 + 2) + 3 (- 2 - y) + 5 (y - 4) ∣ = \frac{1}{2} ∣6 x + 2 y - 26∣ = 10$ .

So $6 x + 2 y - 26 = \pm 20$ , i.e., $3 x + y = 23$ or $3 x + y = 3$ .

Solving with $x - 3 y = 1$ : first pair gives $(7, 2)$ ; second gives $(1, 0)$ . Both are valid.

6. Collinearity of Three Points

Three points $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ , $C (x_{3}, y_{3})$ are collinear if any one of these holds:

Equal slopes: slope of $A B$ = slope of $B C$ , i.e., $\frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{y _{3} - y _{2}}{x _{3} - x _{2}}$ .
Zero area: $x_{1} y_{1} 1 x_{2} y_{2} 1 x_{3} y_{3} 1 = 0$ .
Distance test: $A B + B C = A C$ (or $∣ A B - B C ∣ = A C$ ) - one of the three points lies between the other two on the line.
Section test: one of the three points divides the segment joining the other two in some real ratio.

In practice the determinant test is fastest; use the distance test only when order matters.

Solved Example 9

Show that

(1, 1)

(2, 3)

, and

(3, 5)

are collinear.

Solution:

Slope of $A B = \frac{3 - 1}{2 - 1} = 2$ . Slope of $B C = \frac{5 - 3}{3 - 2} = 2$ . Slopes equal and both lines pass through $B$ , so $A, B, C$ lie on one line.

7. Locus and Equation of a Locus

The locus of a moving point is the set of all positions the point can take when it moves according to a given geometric condition. The equation of the locus is the algebraic equation in $x$ and $y$ that every point of the locus satisfies (and no other point does).

Working rule to find a locus

Let the coordinates of the moving point be $P (h, k)$ .
Write the given geometric condition and express it in terms of $h, k$ (and any given constants).
Eliminate any parameters using the given constraints, leaving a relation between $h$ and $k$ alone.
Replace $h$ by $x$ and $k$ by $y$ - that is the equation of the locus.

Solved Example 10

A point

P

has coordinates

(\frac{4 cos θ + 3}{5}, \frac{4 sin θ}{5})

. Find its locus as

θ

varies.

Solution:

Let $P = (h, k)$ . Then $cos θ = \frac{5 h - 3}{4}$ and $sin θ = \frac{5 k}{4}$ .

Using $cos^{2} θ + sin^{2} θ = 1$ : $\frac{( 5 h - 3 ) ^{2} + ( 5 k ) ^{2}}{16} = 1$ , i.e., $(5 h - 3)^{2} + 25 k^{2} = 16$ .

Replacing $(h, k)$ by $(x, y)$ : $(5 x - 3)^{2} + 25 y^{2} = 16$ . This is a circle in disguise (centre $(3/5, 0)$ , radius $4/5$ ).

Solved Example 11

Find the locus of a point which moves such that its distance from

(2, 0)

is always twice its distance from

(- 1, 0)

Solution:

Let $P (h, k)$ be the moving point. Given $(h - 2)^{2} + k^{2} = 2 (h + 1)^{2} + k^{2}$ .

Squaring: $(h - 2)^{2} + k^{2} = 4 [(h + 1)^{2} + k^{2}]$ , i.e., $h^{2} - 4 h + 4 + k^{2} = 4 h^{2} + 8 h + 4 + 4 k^{2}$ .

Simplifying: $3 h^{2} + 12 h + 3 k^{2} = 0 \Rightarrow h^{2} + 4 h + k^{2} = 0$ .

Locus: $x^{2} + y^{2} + 4 x = 0$ , a circle with centre $(- 2, 0)$ and radius $2$ .

Figure 8. A locus made visible. Every point

P

with

P A = 2 P B

lies on the circle

x^{2} + y^{2} + 4 x = 0

, centre

(- 2, 0)

and radius

2

: the geometric condition becomes an algebraic equation.

8. Transformation of Coordinates

Sometimes a problem becomes far simpler if we shift or rotate the axes to a more convenient position. Two standard transformations are used.

Translation of axes (shifting the origin)

Shift the origin from $O (0, 0)$ to $O^{'} (h, k)$ without changing the directions of the axes. If the old coordinates of a point are $(x, y)$ and the new coordinates (in the shifted system) are $(X, Y)$ , then

x = X + h, y = Y + k or equivalently X = x - h, Y = y - k .

Figure 9. Translation of axes. The directions of the axes are unchanged, only the origin moves to

O^{'} (h, k)

, so

x = X + h

and

y = Y + k

The equation of any curve in the old system becomes a new equation in $X, Y$ by substitution.

Solved Example 12

Transform the equation

x^{2} + y^{2} - 4 x + 6 y - 3 = 0

to a form free of the first-degree terms.

Solution:

Substitute $x = X + h$ , $y = Y + k$ and choose $h, k$ so that the coefficients of $X$ and $Y$ vanish.

The expression becomes $X^{2} + Y^{2} + (2 h - 4) X + (2 k + 6) Y + (h^{2} + k^{2} - 4 h + 6 k - 3) = 0$ .

Set $2 h - 4 = 0 \Rightarrow h = 2$ and $2 k + 6 = 0 \Rightarrow k = - 3$ . The constant becomes $4 + 9 - 8 - 18 - 3 = - 16$ .

Transformed equation: $X^{2} + Y^{2} = 16$ , a circle of radius $4$ centred at the new origin $(2, - 3)$ .

Rotation of axes

Rotate the axes about the origin through an angle $θ$ (anti-clockwise). If a point has old coordinates $(x, y)$ and new coordinates $(X, Y)$ , then

x = X cos θ - Y sin θ, y = X sin θ + Y cos θ,

and the inverse relations are

X = x cos θ + y sin θ, Y = - x sin θ + y cos θ .

Figure 10. Rotation of axes through

θ

. The point does not move, only the frame does:

x = X cos θ - Y sin θ

and

y = X sin θ + Y cos θ

Compact table (memorise).

	$X$	$Y$
$x$	$cos θ$	$- sin θ$
$y$	$sin θ$	$cos θ$

Read a row as:

x = X cos θ - Y sin θ

. Read a column (with signs flipped on the off-diagonal) for the inverse.

Removal of the $x y$ term (application of rotation)

A general second-degree curve $a x^{2} + 2 h x y + b y^{2} + \dots = 0$ can be rotated so that the $x y$ term vanishes in the new coordinates. Choose $θ$ satisfying

tan 2 θ = \frac{2 h}{a - b} (if a \neq = b), θ = \frac{π}{4} (if a = b) .

This trick reduces conics to their standard axis-aligned form and is a favourite JEE Advanced tool.

Solved Example 13

Remove the

x y

term from

x^{2} + 2 x y + y^{2} - 4 = 0

by a suitable rotation of axes.

Solution:

Here $a = b = 1$ , so we take $θ = \frac{π}{4}$ . Substituting $x = \frac{X - Y}{2}$ and $y = \frac{X + Y}{2}$ :

$x + y = \frac{2 X}{2} = X 2$ , so $(x + y)^{2} = 2 X^{2}$ . The original equation is $(x + y)^{2} = 4$ , giving $2 X^{2} = 4 \Rightarrow X^{2} = 2$ .

The curve is the pair of parallel lines $X = \pm 2$ - clean and axis-aligned in the rotated system.

Common Mistakes to Avoid

Watch out

Section formula, external division: the denominator is $m - n$ , not $m + n$ . If $m = n$ externally, the formula is undefined (the two points would coincide with the point at infinity).
Incentre coefficients: the weights $a, b, c$ are the lengths of the sides opposite to $A, B, C$ respectively - not the sides adjacent to each vertex.
Area determinant: always take the modulus for a physical area; only drop the modulus when the problem asks for signed (oriented) area.
Collinearity via slopes: if any two of the three points have the same $x$ -coordinate, the slope is undefined - use the determinant test instead.
Rotation formulas: $x = X cos θ - Y sin θ$ (mind the minus sign on $sin$ ). Mixing it up with $cos$ is the most common exam slip.
Locus: after eliminating parameters, remember to replace $(h, k)$ with $(x, y)$ ; leaving $h, k$ in the final equation is technically wrong.

Frequently Asked Questions

Q1. What is the difference between abscissa and ordinate?

The abscissa is the $x$ -coordinate - the signed perpendicular distance of a point from the $y$ -axis. The ordinate is the $y$ -coordinate - the signed perpendicular distance from the $x$ -axis. For a point $P (3, - 5)$ , the abscissa is $3$ and the ordinate is $- 5$ .

Q2. How is external division different from internal division?

In internal division the point $P$ lies between $A$ and $B$ on the segment. In external division $P$ lies on the line $A B$ but outside the segment. Algebraically the two formulas differ only in a sign - the internal formula has $+$ in numerator and denominator, the external has $-$ .

Q3. Which triangle has all four special points coinciding?

An equilateral triangle. In an equilateral triangle the centroid, incentre, circumcentre, and orthocentre are all the same point. Any isosceles triangle has these four points collinear (on the axis of symmetry) but not coincident.

Q4. Where is the orthocentre of a right-angled triangle?

At the vertex of the right angle. The two legs are themselves altitudes (each is perpendicular to the other), so the three altitudes meet at the right-angle vertex. The circumcentre in a right triangle is the midpoint of the hypotenuse.

Q5. What is the Euler line?

In any triangle (except equilateral, where they coincide), the orthocentre $H$ , centroid $G$ , and circumcentre $O$ are collinear. This line is called the Euler line. The centroid $G$ divides $H O$ in the ratio $2 : 1$ measured from $H$ .

Q6. Can the area of a triangle come out negative from the determinant formula?

Yes - if the vertices are listed clockwise. The determinant $\frac{1}{2} x_{1} y_{1} 1 x_{2} y_{2} 1 x_{3} y_{3} 1$ (without the absolute value bars) is a signed area: positive for anti-clockwise, negative for clockwise. The physical area is the absolute value.

Q7. When do we shift the origin instead of rotating axes?

Shift the origin to kill first-degree terms (make the curve pass through the new origin, or centre a circle at the new origin). Rotate the axes to kill the $x y$ term (align the principal axes of a conic with the new coordinate axes). For a general second-degree curve both are usually needed - shift first, rotate second.

Q8. What is the equation of a locus, in one line?

The equation of a locus is the algebraic relation between $x$ and $y$ that a moving point $(x, y)$ satisfies if and only if it obeys the given geometric condition. Every point of the locus satisfies the equation; every solution of the equation is a point of the locus.

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1. Rectangular Cartesian Coordinate System

2. Distance Between Two Points

Using distance to identify quadrilaterals

3. Section Formula

Internal division

External division

Midpoint

Harmonic conjugate points

4. Special Points of a Triangle

Centroid G

Incentre I

Excentres I1​,I2​,I3​

Circumcentre O

Orthocentre H

Key properties (memorise)

5. Area of a Triangle and Polygon

Area of a triangle

Area of an n-sided polygon (shoelace formula)

6. Collinearity of Three Points

7. Locus and Equation of a Locus

Working rule to find a locus

8. Transformation of Coordinates

Translation of axes (shifting the origin)

Rotation of axes

Removal of the xy term (application of rotation)

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between abscissa and ordinate?

Q2. How is external division different from internal division?

Q3. Which triangle has all four special points coinciding?

Q4. Where is the orthocentre of a right-angled triangle?

Q5. What is the Euler line?

Q6. Can the area of a triangle come out negative from the determinant formula?

Q7. When do we shift the origin instead of rotating axes?

Q8. What is the equation of a locus, in one line?

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Centroid $G$

Incentre $I$

Excentres $I_{1}, I_{2}, I_{3}$

Circumcentre $O$

Orthocentre $H$

Area of an $n$ -sided polygon (shoelace formula)

Removal of the $x y$ term (application of rotation)