Chapter 12 — Coordinate Geometry

To describe where something is, we need a reference. On a street we say "third house on the left after the temple"; on a chessboard we say "e4". Coordinate geometry does this for every point of a flat surface using two perpendicular number lines, the x-axis (horizontal) and the y-axis (vertical), meeting at the origin O. Any point P is fixed by two numbers: its distance from the y-axis measured along the x-direction, called the x-coordinate or abscissa, and its distance from the x-axis measured along the y-direction, the y-coordinate or ordinate. We write P(x, y); the order matters, so (3, 5) and (5, 3) are different points. This is the Cartesian system, named after René Descartes, who saw in the seventeenth century that a curve could be described by an equation once points were named by numbers.

The axes divide the plane into four quadrants, numbered anticlockwise from the upper right: in the first both coordinates are positive, in the second x is negative and y positive, in the third both are negative, in the fourth x is positive and y negative. A point on the x-axis has y = 0, so it looks like (a, 0); a point on the y-axis has x = 0, so it looks like (0, b). The origin is (0, 0).

Some quick facts follow from the picture. The distance of P(x, y) from the y-axis is |x| and from the x-axis is |y|. The point (x, 0) is the foot of the perpendicular from P to the x-axis. Points with the same y-coordinate lie on a horizontal line; with the same x-coordinate, on a vertical line. The reflection of (x, y) in the x-axis is (x, −y), in the y-axis is (−x, y), and in the origin is (−x, −y).

Plotting is the first skill: to plot (−2, 3), move 2 units left of the origin along the x-axis, then 3 units up. To read off a plotted point, drop perpendiculars to both axes. In the examination, plotting is asked with the area of a triangle or the shape of a quadrilateral, and a neat, labelled diagram earns marks even when the question can be answered by formula alone.

Everything in this chapter rests on one idea: because points are numbers, questions about lengths, ratios, areas and directions become questions about numbers, and can be answered by calculation rather than construction. The first such question is the distance between two points.

Two points on the x-axis, A(x₁, 0) and B(x₂, 0), are |x₂ − x₁| apart, the difference of their abscissae. Two points on a vertical line, (a, y₁) and (a, y₂), are |y₂ − y₁| apart. For two general points P(x₁, y₁) and Q(x₂, y₂), draw the horizontal through P and the vertical through Q; they meet at R(x₂, y₁), and triangle PRQ is right-angled at R. PR = |x₂ − x₁| and RQ = |y₂ − y₁|, so by Pythagoras

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

This is the distance formula. Since the differences are squared, their signs and order do not matter, and PQ = QP. The distance of P(x, y) from the origin is the special case OP = √(x² + y²).

Examples. The distance between (2, 3) and (4, 1): √[(4 − 2)² + (1 − 3)²] = √(4 + 4) = √8 = 2√2. Between (−5, 7) and (−1, 3): √[(−1 + 5)² + (3 − 7)²] = √(16 + 16) = 4√2. Between (a, b) and (−a, −b): √(4a² + 4b²) = 2√(a² + b²). From the origin to (−6, 8): √(36 + 64) = 10.

Finding an unknown coordinate from a distance. Find the values of y for which the distance between P(2, −3) and Q(10, y) is 10 units. √[(10 − 2)² + (y + 3)²] = 10, so 64 + (y + 3)² = 100, (y + 3)² = 36, y + 3 = ±6, y = 3 or y = −9. Two answers, both valid.

Distance in a real situation. A town B is 36 km east and 15 km north of town A. Taking A as origin, B is (36, 15), and AB = √(1296 + 225) = √1521 = 39 km. Two friends at (3, 4) and (6, 7) in a classroom grid are √(9 + 9) = 3√2 ≈ 4.24 units apart.

A point equidistant from two points. Find the point on the x-axis equidistant from (2, −5) and (−2, 9). A point on the x-axis is (x, 0). Set the squares of the distances equal: (x − 2)² + 25 = (x + 2)² + 81, so x² − 4x + 4 + 25 = x² + 4x + 4 + 81, −8x = 56, x = −7. The point is (−7, 0). Working with squared distances avoids square roots, and is the standard habit.

A frequent error is to write the formula as √[(x₂ − y₂)² + …], mixing x with y. Say it as "difference of x's squared plus difference of y's squared, then root". Another is to forget the square root at the end or to leave √8 unsimplified; write 2√2.

Given three points, the distance formula gives the three sides of the triangle they form, and the sides classify the triangle. Equal sides mean isosceles (two) or equilateral (three); if the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled, by the converse of Pythagoras; and if the longest side equals the sum of the other two, the "triangle" is degenerate — the points are collinear.

Isosceles. Show that the points (7, 10), (−2, 5), (3, −4) form an isosceles right triangle. AB² = (−2 − 7)² + (5 − 10)² = 81 + 25 = 106. BC² = (3 + 2)² + (−4 − 5)² = 25 + 81 = 106. CA² = (7 − 3)² + (10 + 4)² = 16 + 196 = 212. AB = BC, so it is isosceles; and AB² + BC² = 212 = CA², so the angle at B is a right angle. An isosceles right triangle.

Equilateral. Show that (0, 0), (3, √3), (3, −√3) are the vertices of an equilateral triangle. OA² = 9 + 3 = 12, OB² = 9 + 3 = 12, AB² = 0 + (2√3)² = 12. All sides 2√3. Equilateral.

Right-angled. Do (3, 2), (−2, −3), (2, 3) form a right triangle? AB² = 25 + 25 = 50, BC² = 16 + 36 = 52, CA² = 1 + 1 = 2. Since 50 + 2 = 52, yes; the right angle is at A, opposite the longest side BC.

Collinear (no triangle). Are (1, 5), (2, 3), (−2, −11) collinear? AB = √(1 + 4) = √5, BC = √(16 + 196) = √212, AC = √(9 + 256) = √265. Since √5 + √212 ≠ √265 (approximately 2.24 + 14.56 = 16.80 versus 16.28), they are not collinear. For (1, 3), (2, 5), (3, 7): AB = √5, BC = √5, AC = √20 = 2√5 = AB + BC, so collinear. A cleaner collinearity test uses area or slope, covered later, but the distance test is what this part of the chapter uses.

Working with squares. Always compare squared distances; roots are only needed for the final side length. The right-angle check needs only squares, and the equal-side check needs only squares. Only the collinearity check by distances needs roots, which is why it is clumsy and why the area test is preferred.

Finding a vertex. If (x, y) is equidistant from (3, 6) and (−3, 4), find a relation between x and y. (x − 3)² + (y − 6)² = (x + 3)² + (y − 4)², so −6x − 12y + 45 = 6x − 8y + 25, so 12x + 4y = 20, that is, 3x + y = 5. This line is the perpendicular bisector of the segment joining the two points.

In the examination, name the vertices A, B, C, compute the three squared sides in a column, and then state the classification with its reason in one sentence.

Four points in order form a quadrilateral, and its four sides and two diagonals, all found by the distance formula, decide what kind it is.

Square. Show that (1, 7), (4, 2), (−1, −1), (−4, 4) are the vertices of a square. AB² = 9 + 25 = 34, BC² = 25 + 9 = 34, CD² = 9 + 25 = 34, DA² = 25 + 9 = 34: all sides equal. AC² = 4 + 64 = 68, BD² = 64 + 4 = 68: diagonals equal. All sides equal and diagonals equal, so it is a square. (Equal sides alone would give a rhombus; equal diagonals are needed to exclude that.)

Rhombus and its area. Show that (3, 0), (4, 5), (−1, 4), (−2, −1) form a rhombus and find its area. AB² = 1 + 25 = 26, BC² = 25 + 1 = 26, CD² = 1 + 25 = 26, DA² = 25 + 1 = 26. AC² = 16 + 16 = 32, BD² = 36 + 36 = 72. Sides equal, diagonals unequal: a rhombus, not a square. Area = ½ × d₁ × d₂ = ½ × √32 × √72 = ½ × 4√2 × 6√2 = ½ × 48 = 24 square units.

Rectangle. (−1, −2), (1, 0), (−1, 2), (−3, 0): AB² = 4 + 4 = 8, BC² = 4 + 4 = 8, CD² = 8, DA² = 8, and AC² = 0 + 16 = 16, BD² = 16 + 0 = 16. All sides equal and diagonals equal — actually a square. For a rectangle that is not a square, take (5, 6), (1, 5), (2, 1), (6, 2): AB² = 17, BC² = 17, CD² = 17, DA² = 17 and AC² = 9 + 25 = 34, BD² = 25 + 9 = 34 — again a square. A true rectangle example: (0, 0), (4, 0), (4, 3), (0, 3): sides 4, 3, 4, 3, diagonals both 5.

Parallelogram. Are (1, 2), (4, 3), (6, 6), (3, 5) the vertices of a parallelogram? AB² = 9 + 1 = 10, CD² = 9 + 1 = 10, BC² = 4 + 9 = 13, DA² = 4 + 9 = 13. Opposite sides equal, so yes. Alternatively, the midpoint of AC is (7/2, 4) and of BD is (7/2, 4): the diagonals bisect each other, which is the neater test once the midpoint formula is known.

Two cautions. First, the order of the vertices matters — A, B, C, D must go around the figure; taking them in the wrong order gives a crossed figure and wrong "sides". Plot the points first. Second, equal sides do not make a square; check the diagonals. The examination's classic trap is a rhombus presented as if it might be a square.

Suppose a point P lies on the segment AB and divides it so that AP : PB = m : n. Where is P? Let A = (x₁, y₁), B = (x₂, y₂), P = (x, y). Draw perpendiculars from A, P, B to the x-axis, meeting it at A', P', B'. The lines are parallel, so they cut every transversal in the same ratio: A'P' : P'B' = AP : PB = m : n. Now A'P' = x − x₁ and P'B' = x₂ − x, so (x − x₁)/(x₂ − x) = m/n, which gives n(x − x₁) = m(x₂ − x), nx + mx = mx₂ + nx₁, and

x = (mx₂ + nx₁)/(m + n), and by the same argument with the y-axis, y = (my₂ + ny₁)/(m + n).

This is the section formula for internal division. The pattern is: multiply the far coordinate by the near part of the ratio — m goes with B's coordinates, n with A's — and divide by the total. A mnemonic: "cross-multiply the ratio with the endpoints".

Examples. The point dividing the segment from (4, −3) to (8, 5) in the ratio 3 : 1: x = (3 × 8 + 1 × 4)/4 = 28/4 = 7, y = (3 × 5 + 1 × (−3))/4 = 12/4 = 3. The point is (7, 3). The point dividing (−1, 7) to (4, −3) in the ratio 2 : 3: x = (2 × 4 + 3 × (−1))/5 = 5/5 = 1, y = (2 × (−3) + 3 × 7)/5 = 15/5 = 3. The point is (1, 3).

Points of trisection. The points that divide AB into three equal parts divide it in the ratios 1 : 2 and 2 : 1. For A(2, −2) and B(−7, 4): the first point is ((1 × (−7) + 2 × 2)/3, (1 × 4 + 2 × (−2))/3) = (−1, 0), and the second is ((2 × (−7) + 1 × 2)/3, (2 × 4 + 1 × (−2))/3) = (−4, 2). Check that (−1, 0) is the midpoint of A and (−4, 2): ((2 − 4)/2, (−2 + 2)/2) = (−1, 0). Yes.

Dividing into n equal parts. The points dividing (−2, 2) to (2, 8) into four equal parts are found with ratios 1 : 3, 2 : 2 (the midpoint), 3 : 1: (−1, 7/2), (0, 5), (1, 13/2).

Ratio notation. "P divides AB in the ratio 3 : 1" means AP : PB = 3 : 1, so P is nearer B. Reversing the ratio reverses the point. If the ratio is given as k : 1, the formula becomes x = (kx₂ + x₁)/(k + 1), useful when k is to be found.

In the examination, write the formula with m, n, x₁, y₁, x₂, y₂ identified, substitute, simplify, and give the point. The derivation by similar triangles is a possible four-mark question; know the figure and the parallel-lines argument.

The midpoint M of AB divides it in the ratio 1 : 1, so putting m = n = 1 in the section formula gives

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

the average of the coordinates. The midpoint of (2, 3) and (6, 7) is (4, 5); of (−3, 5) and (7, −1) is (2, 2); of (a, b) and (−a, −b) is the origin.

Finding an endpoint. If the midpoint of AB is (2, 5) and A is (−1, 3), find B. (−1 + x)/2 = 2 gives x = 5; (3 + y)/2 = 5 gives y = 7. B = (5, 7).

Parallelogram tests and missing vertices. The diagonals of a parallelogram bisect each other, so they have the same midpoint. If (1, 2), (4, y), (x, 6), (3, 5) are the vertices of a parallelogram taken in order, then the midpoint of the diagonal from (1, 2) to (x, 6) equals that of the diagonal from (4, y) to (3, 5): ((1 + x)/2, 4) = (7/2, (y + 5)/2). So 1 + x = 7, x = 6, and y + 5 = 8, y = 3. The fourth vertex D of a parallelogram ABCD with A(1, 2), B(4, 3), C(6, 6): midpoint of AC = (7/2, 4) = midpoint of BD = ((4 + x)/2, (3 + y)/2), so D = (3, 5).

Median and centroid. The median from a vertex goes to the midpoint of the opposite side. The three medians meet at the centroid G, which divides each median in the ratio 2 : 1 from the vertex. For a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), the midpoint D of BC is ((x₂ + x₃)/2, (y₂ + y₃)/2), and G divides AD in 2 : 1: G = ((2 × (x₂ + x₃)/2 + 1 × x₁)/3, …) = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3). The centroid is the average of the vertices. For (3, −5), (−7, 4), (10, −2): G = (6/3, −3/3) = (2, −1). If the centroid of a triangle with vertices (a, b), (b, c), (c, a) is the origin, then a + b + c = 0.

Length of a median. For A(7, −3), B(5, 3), C(3, −1), the midpoint of BC is D(4, 1), and AD = √[(7 − 4)² + (−3 − 1)²] = √(9 + 16) = 5. Medians are found by midpoint then distance.

Circle centre. The centre of a circle is the midpoint of any diameter. If (2, 3) and (−6, 7) are the ends of a diameter, the centre is (−2, 5) and the radius is half the diameter: ½√(64 + 16) = ½ × √80 = 2√5. If the centre is (2, −3) and one end of a diameter is (1, 4), the other end is (3, −10).

The midpoint formula is the single most used tool of the chapter; it appears inside the centroid, inside the parallelogram test, inside medians and inside circle problems. Know it as "average of the ends".

The section formula can be run backwards: given the dividing point, find the ratio. The trick is to call the ratio k : 1, so that only one unknown appears.

A given point. In what ratio does (−4, 6) divide the segment joining A(−6, 10) and B(3, −8)? Let the ratio be k : 1. Then the x-coordinate is (3k − 6)/(k + 1) = −4, so 3k − 6 = −4k − 4, 7k = 2, k = 2/7. Check with y: (−8k + 10)/(k + 1) = (−16/7 + 10)/(9/7) = (54/7)/(9/7) = 6. Correct. The ratio is 2 : 7. If the y-check had failed, the point would not be on the segment at all.

The x-axis as divider. Find the ratio in which the x-axis divides the segment joining (1, −5) and (−4, 5), and the point of division. A point on the x-axis has y = 0. With ratio k : 1, y = (5k − 5)/(k + 1) = 0, so k = 1. The ratio is 1 : 1 and the point is the midpoint, (−3/2, 0). In general, the x-axis divides AB in the ratio −y₁ : y₂ and the y-axis in the ratio −x₁ : x₂; the sign tells whether the division is internal (positive) or external.

The y-axis as divider. The y-axis divides (5, −6) to (−1, −4): ratio −x₁ : x₂ = −5 : −1 = 5 : 1. The point: y = (5 × (−4) + 1 × (−6))/6 = −26/6 = −13/3, so (0, −13/3).

A given line as divider. Find the ratio in which the line 2x + y − 4 = 0 divides the segment joining (2, −2) and (3, 7). Let the point be P dividing in k : 1: P = ((3k + 2)/(k + 1), (7k − 2)/(k + 1)). P lies on the line, so 2(3k + 2) + (7k − 2) − 4(k + 1) = 0, that is, 6k + 4 + 7k − 2 − 4k − 4 = 0, 9k = 2, k = 2/9. The ratio is 2 : 9.

Unknown coordinate from a ratio. If (x, 2) divides the segment from (−3, 5) to (7, −1) in some ratio, find x and the ratio. Using y: (−k + 5)/(k + 1) = 2 gives −k + 5 = 2k + 2, k = 1. So it is the midpoint, and x = (7 − 3)/2 = 2.

Checking that a point lies on a segment. Does (3, 5) lie on the segment from (1, 3) to (5, 7)? With k : 1, x gives (5k + 1)/(k + 1) = 3, 5k + 1 = 3k + 3, k = 1; y gives (7k + 3)/(k + 1) = 5, k = 1. Consistent, so yes, and it is the midpoint. If the two values of k had differed, the point would be off the segment.

The k : 1 device keeps the algebra linear. The examination's typical two-mark version is the x-axis or y-axis case; the four-mark version is the line as divider.

Given three vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃), the area can be found without measuring any height. Drop perpendiculars from A, B, C to the x-axis. The triangle's area is the area of the trapezium under AB, plus the trapezium under BC, minus the trapezium under AC (when the points are arranged suitably). A trapezium with parallel sides p and q and width w has area ½(p + q)w. Adding and subtracting the three trapezia and simplifying gives

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

The absolute value is taken because the expression may come out negative depending on the order of the vertices; area is always positive. The pattern inside is cyclic: each x is multiplied by the difference of the other two y's, in the order 2−3, 3−1, 1−2.

Example 1. Area of the triangle with vertices (1, −1), (−4, 6), (−3, −5): ½ |1(6 + 5) + (−4)(−5 + 1) + (−3)(−1 − 6)| = ½ |11 + 16 + 21| = ½ × 48 = 24 square units.

Example 2. Vertices (5, 2), (4, 7), (7, −4): ½ |5(7 + 4) + 4(−4 − 2) + 7(2 − 7)| = ½ |55 − 24 − 35| = ½ |−4| = 2 square units. The negative sign inside shows the vertices were taken clockwise; the area is 2.

Example 3 — checking a right triangle's area. (0, 0), (4, 0), (0, 3): ½ |0 + 4(3 − 0) + 0| = 6, which agrees with ½ × 4 × 3.

Area of a quadrilateral. Split it by a diagonal into two triangles and add. For (−4, −2), (−3, −5), (3, −2), (2, 3): triangle ABC has area ½ |−4(−5 + 2) + (−3)(−2 + 2) + 3(−2 + 5)| = ½ |12 + 0 + 9| = 21/2; triangle ACD has area ½ |−4(−2 − 3) + 3(3 + 2) + 2(−2 + 2)| = ½ |20 + 15 + 0| = 35/2; total 28 square units.

Finding an unknown from an area. If the area of the triangle with vertices (k, 0), (4, 0), (0, 2) is 4 square units, find k. ½ |k(0 − 2) + 4(2 − 0) + 0| = 4, so |−2k + 8| = 8, so −2k + 8 = ±8, k = 0 or k = 8. Two positions of the vertex give the same area, one on each side of (4, 0).

Heron's formula as a check. For (1, −1), (−4, 6), (−3, −5) the sides are √74, √122, √32; Heron's formula would give 24 as well, but with much more work. The coordinate formula is the efficient route whenever vertices are known.

Write the formula, then substitute in the cyclic order, keeping the signs of the coordinates inside brackets. The most frequent error is a sign slip in y₂ − y₃ when a coordinate is negative; write (6 − (−5)) = 11 explicitly.

Three points are collinear when they lie on one straight line, and in that case the "triangle" they form has zero area. So the area formula gives a clean test:

A, B, C are collinear ⇔ x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) = 0

Unlike the distance test, this needs no square roots and no guess about which distance is longest.

Example 1. Are (1, 5), (2, 3), (−2, −11) collinear? 1(3 + 11) + 2(−11 − 5) + (−2)(5 − 3) = 14 − 32 − 4 = −22 ≠ 0. Not collinear.

Example 2. Are (−1.5, 3), (6, −2), (−3, 4) collinear? −1.5(−2 − 4) + 6(4 − 3) + (−3)(3 + 2) = 9 + 6 − 15 = 0. Collinear.

Finding an unknown for collinearity. Find k if (7, −2), (5, 1), (3, k) are collinear. 7(1 − k) + 5(k + 2) + 3(−2 − 1) = 0, so 7 − 7k + 5k + 10 − 9 = 0, −2k + 8 = 0, k = 4. Find x if (x, −1), (2, 1), (4, 5) are collinear: x(1 − 5) + 2(5 + 1) + 4(−1 − 1) = 0, −4x + 12 − 8 = 0, x = 1.

Find a relation between x and y if (x, y), (1, 2), (7, 0) are collinear. x(2 − 0) + 1(0 − y) + 7(y − 2) = 0, so 2x − y + 7y − 14 = 0, x + 3y − 7 = 0. This is the equation of the line through (1, 2) and (7, 0); collinearity with a variable point produces the line's equation, a result the Intermediate course takes up in detail.

Which three of four points are collinear. Given (1, 2), (3, 4), (5, 6), (2, −1): the first three satisfy 1(4 − 6) + 3(6 − 2) + 5(2 − 4) = −2 + 12 − 10 = 0, so they are collinear; the fourth is off that line since with (1, 2), (3, 4), (2, −1): 1(4 + 1) + 3(−1 − 2) + 2(2 − 4) = 5 − 9 − 4 = −8 ≠ 0.

Why the test works. If A, B, C are on one line, the trapezia under AB and BC exactly make up the trapezium under AC, so the difference is zero. Conversely, if the area is zero the height from C to AB is zero, so C is on line AB.

Three ways to test collinearity. By distances: AB + BC = AC (roots needed). By area: the expression equals zero (cleanest). By slope, from the next topic: slope of AB = slope of BC (also clean; fails only for vertical lines, where the area test still works). The examination accepts any, but the area test is the one named in this chapter and the safest to present.

The slope (or gradient) of a line measures how steeply it rises. For a line through P(x₁, y₁) and Q(x₂, y₂), the slope is the rise divided by the run:

m = (y₂ − y₁)/(x₂ − x₁), with x₁ ≠ x₂.

It is the same whichever point is called P: reversing both differences leaves the quotient unchanged. It is also the same for any two points on the line, because all the right triangles formed with horizontal and vertical legs are similar.

Sign and size. A positive slope means the line rises to the right; a negative slope means it falls to the right. Slope 0 means a horizontal line (y₁ = y₂). A vertical line (x₁ = x₂) has no slope — the run is zero and the division is undefined; we say the slope is not defined. A slope of 1 is a line at 45°, slope 2 is steeper, slope 1/2 is gentler.

Examples. Slope through (4, −8) and (5, −2): (−2 + 8)/(5 − 4) = 6. Through (0, 0) and (√3, 3): 3/√3 = √3. Through (2, 3) and (4, 3): 0, horizontal. Through (3, 1) and (3, 7): undefined, vertical. Through (−2, 5) and (4, −7): (−7 − 5)/(4 + 2) = −2.

Slope and angle. If the line makes angle θ with the positive x-axis, then m = tan θ. A slope of √3 means θ = 60°; a slope of 1 means 45°; a slope of 1/√3 means 30°. (This connects to trigonometry and is used more in the Intermediate course.)

Parallel lines have equal slopes. Two lines making the same angle with the x-axis are parallel. The line through (1, 2) and (3, 6) has slope 2; the line through (0, −1) and (2, 3) has slope 2; they are parallel. The sides AB and DC of the quadrilateral (1, 2), (4, 3), (6, 6), (3, 5) have slopes 1/3 and 1/3, and BC and AD have slopes 3/2 and 3/2: both pairs of opposite sides are parallel, so it is a parallelogram — a third proof of what distances and midpoints showed earlier.

Perpendicular lines have slopes whose product is −1. If one line has slope m, a perpendicular line has slope −1/m (for non-vertical, non-horizontal lines). The line through (2, 3) and (4, 7) has slope 2; a perpendicular line has slope −1/2. In triangle (3, 2), (−2, −3), (2, 3): slope of AB = 1, slope of AC = −1, product −1, so the angle at A is a right angle — confirming the Pythagoras check made earlier.

Collinearity by slope. Three points are collinear if the slope of AB equals the slope of BC (with B common). For (1, 3), (2, 5), (3, 7): slopes 2 and 2, collinear. For (1, 5), (2, 3), (−2, −11): slopes −2 and 14/4 = 7/2, not collinear.

The Class 10 chapter stops at finding slopes and using them for parallel, perpendicular and collinear; the equation of a line y = mx + c belongs to the next course, but the slope defined here is exactly its m.

Examination problems often need two or three of the chapter's formulas together. The following are worked in full.

Problem 1. Find the coordinates of the point which divides the line segment joining (−1, 7) and (4, −3) in the ratio 2 : 3, and find the distance of this point from the origin. Section formula: x = (2 × 4 + 3 × (−1))/5 = 1, y = (2 × (−3) + 3 × 7)/5 = 3. The point is (1, 3), and its distance from the origin is √(1 + 9) = √10.

Problem 2. Show that the points A(1, 2), B(5, 4), C(3, 8), D(−1, 6) are the vertices of a square, and find its area. AB² = 16 + 4 = 20, BC² = 4 + 16 = 20, CD² = 16 + 4 = 20, DA² = 4 + 16 = 20; AC² = 4 + 36 = 40, BD² = 36 + 4 = 40. Equal sides and equal diagonals: a square, of side √20 and area 20 square units. Check by the triangle formula: area of ABC = ½ |1(4 − 8) + 5(8 − 2) + 3(2 − 4)| = ½ |−4 + 30 − 6| = 10, and doubling for the square gives 20.

Problem 3. The vertices of a triangle are A(4, 6), B(1, 5), C(7, 2). A line is drawn to intersect AB and AC at D and E respectively such that AD/AB = AE/AC = 1/4. Find the area of triangle ADE and compare it with the area of ABC. D divides AB in 1 : 3: D = ((1 × 1 + 3 × 4)/4, (1 × 5 + 3 × 6)/4) = (13/4, 23/4). E divides AC in 1 : 3: E = ((7 + 12)/4, (2 + 18)/4) = (19/4, 5). Area of ADE = ½ |4(23/4 − 5) + 13/4(5 − 6) + 19/4(6 − 23/4)| = ½ |3 − 13/4 + 19/16| = ½ |(48 − 52 + 19)/16| = 15/32. Area of ABC = ½ |4(5 − 2) + 1(2 − 6) + 7(6 − 5)| = ½ |12 − 4 + 7| = 15/2. The ratio is (15/32)/(15/2) = 1/16 = (1/4)², as similar triangles predict.

Problem 4. Find the centre of a circle passing through (6, −6), (3, −7), (3, 3). Let the centre be (x, y); it is equidistant from all three. (x − 6)² + (y + 6)² = (x − 3)² + (y + 7)² gives −12x + 36 + 12y + 36 = −6x + 9 + 14y + 49, so −6x − 2y = −14, 3x + y = 7. And (x − 3)² + (y + 7)² = (x − 3)² + (y − 3)² gives 14y + 49 = −6y + 9, 20y = −40, y = −2. Then x = 3. The centre is (3, −2) and the radius is √(9 + 16) = 5.

Problem 5. If A(−2, 1), B(a, 0), C(4, b), D(1, 2) are the vertices of a parallelogram ABCD, find a and b. Midpoint of AC = midpoint of BD: ((−2 + 4)/2, (1 + b)/2) = ((a + 1)/2, 1). So a + 1 = 2, a = 1, and 1 + b = 2, b = 1.

Problem 6. Find the point on the y-axis equidistant from (−5, −2) and (3, 2). Let it be (0, y): 25 + (y + 2)² = 9 + (y − 2)², so 25 + 4y + 4 = 9 − 4y + 4, 8y = −16, y = −2. The point is (0, −2).

Each problem starts by asking which quantity is wanted — a point, a length, an area, a ratio — and choosing the formula that produces it. Then the other formulas supply the inputs.

The chapter has five formulas. For points (x₁, y₁) and (x₂, y₂): the distance √[(x₂ − x₁)² + (y₂ − y₁)²]; the section point ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)); the midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2); the slope (y₂ − y₁)/(x₂ − x₁). For three points: the area ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| and the centroid (average of the three vertices).

One-mark questions ask for the distance of a point from the origin or an axis, the midpoint of two points, the quadrant of a point, the slope of a segment, or the centroid. Answer with the formula and the value.

Two-mark questions ask for the distance between two points, a point dividing a segment in a ratio, the ratio in which an axis divides a segment, the area of a triangle, or a value of k for collinearity. Write the formula, substitute with the signs visible, and simplify.

Four-mark questions ask to show that four points form a square, rhombus, rectangle or parallelogram (with the reason for each property); to find the area of a quadrilateral; to find a vertex or the centre of a circle from equidistance; to find the ratio in which a line divides a segment; or to derive the section formula or the distance formula. The derivations are set regularly: know the figure, the similar-triangle or Pythagoras argument, and the final statement.

The usual errors. Mixing x with y in the distance formula. Attaching m to the wrong endpoint in the section formula (m goes with the second point). Forgetting the absolute value in the area, or reporting a negative area. A sign slip when a coordinate is negative — write (6 − (−5)) in full. Calling a rhombus a square without checking the diagonals. Taking the vertices of a quadrilateral out of order. Writing "slope undefined" as "slope zero" for a vertical line. In ratio problems, forgetting to verify the second coordinate.

Presentation. A labelled sketch first, even rough. Name the points. State which formula is being used and why. Keep squared distances in a table. End with the conclusion in words: "Since all four sides are equal and the diagonals are equal, ABCD is a square."

Where this leads. Intermediate coordinate geometry begins with the equation of a straight line, y = mx + c, whose m is this chapter's slope and whose c is where it crosses the y-axis, then circles, parabolas and the rest of the conic sections, all handled by the same idea: a point is a pair of numbers, so geometry is algebra. The distance and section formulas of this chapter are used, unchanged, in three dimensions and in physics.