Chapter 7: Coordinate Geometry
1. The Cartesian Plane — Basic Terms
- The Cartesian plane is formed by two perpendicular number lines: the x-axis (horizontal) and the y-axis (vertical), intersecting at the origin O(0, 0).
- A point is written as an ordered pair P(x, y), where x is the abscissa (distance from the y-axis) and y is the ordinate (distance from the x-axis).
- Points on the axes: any point on the x-axis has coordinates (x, 0); any point on the y-axis has coordinates (0, y).
- Distance of P(x, y) from the x-axis = |y|; distance from the y-axis = |x|.
2. Distance Formula
Distance between A(x1, y1) and B(x2, y2):
AB = √[(x2 − x1)2 + (y2 − y1)2]
Distance of P(x, y) from the origin: OP = √(x2 + y2)
The formula follows from the Pythagoras theorem: the horizontal leg of the right triangle is (x2 − x1) and the vertical leg is (y2 − y1).
Note: Distance is always non-negative. AB = BA, so the order of points does not matter.
3. Applications of the Distance Formula
(a) Checking collinearity
Three points A, B, C are collinear if the sum of two of the distances equals the third, i.e. AB + BC = AC (with B between A and C). If no such relation holds, the points form a triangle.
(b) Type of triangle
- Isosceles: exactly two sides equal.
- Equilateral: all three sides equal.
- Right-angled: the square of the longest side equals the sum of squares of the other two (converse of Pythagoras), e.g. AB2 + BC2 = AC2.
- Scalene: all sides unequal.
(c) Type of quadrilateral (vertices taken in order A, B, C, D)
| Quadrilateral | Condition on sides | Condition on diagonals |
|---|---|---|
| Parallelogram | Opposite sides equal: AB = CD, BC = DA | Diagonals unequal (in general); diagonals bisect each other (midpoints of AC and BD coincide) |
| Rectangle | Opposite sides equal | Diagonals equal: AC = BD |
| Rhombus | All four sides equal | Diagonals unequal |
| Square | All four sides equal | Diagonals equal: AC = BD |
Exam tip: To prove a quadrilateral is a square, showing all four sides equal is NOT enough (that only proves a rhombus). You must also show that the diagonals are equal. Always compute all four sides AND both diagonals, then write the conclusion in words.
4. Section Formula (Internal Division)
The point P which divides the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2 is:
P = ( (m1x2 + m2x1) / (m1 + m2) , (m1y2 + m2y1) / (m1 + m2) )
Memory aid: the ratio part m1 (nearer to A) multiplies the FAR point B, and m2 multiplies the near point A — “cross-multiply” the ratio with the coordinates.
Midpoint Formula (special case m1 : m2 = 1 : 1)
Midpoint of AB = ( (x1 + x2)/2 , (y1 + y2)/2 )
Points of Trisection
Points P and Q trisect AB if AP = PQ = QB. Then P divides AB in the ratio 1 : 2 and Q divides AB in the ratio 2 : 1. Apply the section formula twice (or find Q as the midpoint of PB).
5. Finding an Unknown Ratio — the k : 1 Method
When a point P is known to divide AB in some ratio, assume the ratio is k : 1. Then
P = ( (kx2 + x1)/(k + 1) , (ky2 + y1)/(k + 1) )
Equate one coordinate of P to solve for k. Useful facts:
- x-axis divides AB: put the y-coordinate of P equal to 0 ⇒ ratio = −y1 : y2 in magnitude, i.e. k = −y1/y2.
- y-axis divides AB: put the x-coordinate of P equal to 0 ⇒ k = −x1/x2.
- After finding k from one coordinate, substitute back to find the other coordinate of the point of division.
6. Parallelogram Problems Using Midpoints
In a parallelogram ABCD (vertices in order), the diagonals bisect each other. Therefore:
Midpoint of AC = Midpoint of BD
(x1 + x3)/2 = (x2 + x4)/2 and (y1 + y3)/2 = (y2 + y4)/2
This is the standard method to find a missing vertex of a parallelogram.
7. Common Board Question Patterns and Stepwise Methods
- Point on an axis equidistant from two points: Take the point as (x, 0) [x-axis] or (0, y) [y-axis]. Equate the two squared distances, simplify (the x2/y2 terms cancel), solve the linear equation.
- Find x/y if distance is given: Square both sides of the distance equation first to remove the square root; expect a quadratic with possibly two valid answers — give both.
- Prove points form a specific figure: Compute all sides (and diagonals if needed) → compare → state the geometric conclusion in a full sentence.
- Ratio in which a point/line divides a segment: Use the k : 1 method; verify with the second coordinate.
- Point equidistant from two given points (locus-type): Let the point be (x, y), equate squared distances, obtain a linear relation in x and y.
- Circle-based: If points lie on a circle with a given centre, their distances from the centre (radii) are equal.
Exam tips:
- Always write the formula first, then substitute — formula marks are awarded even if arithmetic slips.
- Leave answers in exact surd form (e.g. 2√5) unless a decimal is asked for.
- While squaring distances, be careful with signs: (−2 − 3)2 = 25, not −25.
- In section-formula answers, fractional coordinates like (0, −7/3) are perfectly acceptable — do not force decimals.
- 1-mark questions frequently test: distance from origin, midpoint, coordinates of a point on an axis, and distance of a point from an axis.
Deleted from the Rationalised Syllabus (NOT to be studied)
- Area of a triangle from coordinates — the formula ½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| and all questions based on it (including collinearity via area = 0) are deleted.
- Section formula is restricted to internal division only — external division is not in the syllabus.
- Retained: distance formula, section formula (internal), midpoint as a special case, checking collinearity and types of triangles/quadrilaterals using distances, and points on the axes.