Chapter 7 — Coordinate Geometry

To locate a point in a plane we draw two perpendicular number lines meeting at a point O called the origin. The horizontal line is the x-axis and the vertical line the y-axis. Any point P is described by an ordered pair (x, y): x, the abscissa, is the perpendicular distance of P from the y-axis, taken positive to the right of it and negative to the left; y, the ordinate, is the perpendicular distance of P from the x-axis, positive above and negative below. The pair is ordered: (3, 5) and (5, 3) are different points.

The axes divide the plane into four quadrants, numbered anticlockwise from the upper right. In the first quadrant both coordinates are positive, (+, +); in the second (−, +); in the third (−, −); in the fourth (+, −). A point on the x-axis has ordinate 0 and is written (x, 0); a point on the y-axis has abscissa 0 and is written (0, y). The origin is (0, 0).

Two further facts are used throughout the chapter. First, the distance of a point (x, y) from the y-axis is |x| and from the x-axis is |y|, the absolute values, since distance is never negative. Second, if two points have the same ordinate, the segment joining them is parallel to the x-axis and its length is the difference of the abscissae; if they have the same abscissa, the segment is parallel to the y-axis and its length is the difference of the ordinates. So the points (2, 3) and (7, 3) are 5 units apart, and (4, −1) and (4, 6) are 7 units apart.

Plotting is the first skill to secure. To plot (−3, 2), start at O, move 3 units to the left along the x-axis and then 2 units up. To plot (0, −4), stay on the y-axis and go 4 units down. When you plot a set of points and join them in order, you get a polygon whose shape you can begin to guess; the formulas of this chapter then confirm the guess with certainty.

Coordinate geometry converts every geometric statement into an algebraic one. A point becomes a pair of numbers; a length becomes an expression under a square root; the mid-point of a segment becomes an average; the area of a triangle becomes a determinant-like sum. The examination tests exactly these conversions.

Suppose a town B is 36 km east and 15 km north of a town A. How far apart are they? The eastward and northward displacements form the legs of a right triangle whose hypotenuse is the direct distance, so AB = √(36² + 15²) = √(1296 + 225) = √1521 = 39 km. The distance formula is nothing but this reasoning written for any two points.

Derivation. Let P(x₁, y₁) and Q(x₂, y₂) be two points. Draw PR and QS perpendicular to the x-axis, and draw PT perpendicular to QS, meeting it at T. Then PT is parallel to the x-axis, so PT = RS = OS − OR = x₂ − x₁. Also QT = QS − TS = QS − PR = y₂ − y₁. In right triangle PTQ, by the Pythagoras theorem, PQ² = PT² + QT² = (x₂ − x₁)² + (y₂ − y₁)². Hence

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²], the distance formula.

The derivation assumed the points in the first quadrant, but since the squares (x₂ − x₁)² and (x₁ − x₂)² are equal, the formula holds for points anywhere in the plane and in either order. The distance of a point P(x, y) from the origin is a special case: OP = √(x² + y²).

Points of care. Subtract coordinates of the same kind: x from x, y from y. Keep the signs inside the brackets, for instance for (−2, 3) and (4, −5) the differences are 4 − (−2) = 6 and −5 − 3 = −8, giving √(36 + 64) = 10. Leave the answer in simplest surd form, such as 2√5, unless a decimal is asked. And remember that the formula gives the length of the segment, so the answer is positive.

The distance formula is the workhorse of the chapter. In the next topics it is used to show that points are collinear, to classify triangles and quadrilaterals, to find unknown coordinates from a given distance, and to locate points equidistant from given points.

Three points are collinear when they lie on one straight line. With the distance formula this becomes a test: compute the three distances AB, BC and AC; if the sum of the two smaller ones equals the largest, say AB + BC = AC, then B lies on the segment AC and the points are collinear. If the sum of any two is greater than the third, the points form a triangle (triangle inequality).

Once we know the three points form a triangle, its type is read from the side lengths. If two sides are equal the triangle is isosceles; if all three are equal it is equilateral; if all three differ it is scalene. It is right-angled if the square of the largest side equals the sum of the squares of the other two, by the converse of the Pythagoras theorem. A triangle can be both isosceles and right-angled.

Worked example: are the points A(1, 5), B(2, 3) and C(−2, −11) collinear? AB = √(1 + 4) = √5 ≈ 2.24; BC = √(16 + 196) = √212 ≈ 14.56; AC = √(9 + 256) = √265 ≈ 16.28. Now AB + BC ≈ 16.80, which is not equal to AC, so the points are not collinear.

Worked example: show that A(3, 2), B(−2, −3) and C(2, 3) form a right triangle. AB² = 25 + 25 = 50; BC² = 16 + 36 = 52; AC² = 1 + 1 = 2. Since AB² + AC² = 52 = BC², the triangle is right-angled at A.

Worked example: show that (0, 0), (5, 5), (−5, 5) form an isosceles right triangle. The distances from the origin are both √50 = 5√2, and the third side is 10. Since (5√2)² + (5√2)² = 100 = 10², the triangle is isosceles and right-angled at the origin.

A practical tip: compare squares of distances rather than the distances themselves. Squares are integers when the coordinates are integers, so equality can be checked exactly, and the Pythagoras test needs squares anyway. Only take square roots when the question asks for the actual length. In the answer always state the conclusion in words, for instance since AB = AC, the triangle is isosceles.

Four points taken in order form a quadrilateral, and its type can be decided from the lengths of its four sides and two diagonals. The distinguishing properties are:

• Parallelogram: both pairs of opposite sides equal; diagonals are generally unequal.
• Rectangle: opposite sides equal and the diagonals equal.
• Rhombus: all four sides equal; diagonals unequal.
• Square: all four sides equal and the diagonals equal.

Because the diagonals of a parallelogram bisect each other, an alternative test is that the mid-points of the two diagonals coincide; this uses the mid-point formula of a later topic. But the distance approach alone is enough for all examination questions of this kind.

Worked example: show that A(1, 7), B(4, 2), C(−1, −1), D(−4, 4) are the vertices of a square. AB² = 9 + 25 = 34; BC² = 25 + 9 = 34; CD² = 9 + 25 = 34; DA² = 25 + 9 = 34. So all four sides equal √34. Diagonals: AC² = 4 + 64 = 68 and BD² = 64 + 4 = 68, equal. Four equal sides and equal diagonals make ABCD a square.

Worked example: do the points (−1, −2), (1, 0), (−1, 2), (−3, 0) form a square? Sides: each is √(4 + 4) = 2√2. Diagonals: from (−1, −2) to (−1, 2) is 4; from (1, 0) to (−3, 0) is 4. Equal sides and equal diagonals, so yes.

Worked example: A(−3, 5), B(3, 1), C(0, 3), D(−1, −4). AB = √(36 + 16) = √52, BC = √(9 + 4) = √13, CD = √(1 + 49) = √50, DA = √(4 + 81) = √85; no two sides equal. Check whether A, B, C are collinear: AC = √(9 + 4) = √13, and AC + CB = 2√13 = √52 = AB. So C lies on AB and the four points do not form a quadrilateral at all. This shows why one must be alert: verify that the points really form a quadrilateral before naming it.

When the vertices are given in order, the sides are AB, BC, CD, DA and the diagonals AC and BD. If a question gives the points without order, plot them roughly to decide the order.

Many questions reverse the distance formula: a distance is known and a coordinate is unknown. The unknown is found by writing the distance formula, squaring both sides to remove the root, and solving the resulting equation, which is usually linear or quadratic.

Point on an axis. A point on the x-axis has the form (x, 0); on the y-axis, (0, y). Suppose we want the point on the x-axis equidistant from A(2, −5) and B(−2, 9). Let it be P(x, 0). Then PA = PB gives PA² = PB²: (x − 2)² + 25 = (x + 2)² + 81. Expanding, x² − 4x + 4 + 25 = x² + 4x + 4 + 81, so −8x = 56 and x = −7. The point is (−7, 0). Notice that the x² terms cancel because both distances are from the same variable point; this always happens in equidistance problems, leaving a linear equation.

Unknown coordinate with given distance. Find y if the distance between P(2, −3) and Q(10, y) is 10. Then (10 − 2)² + (y + 3)² = 100, so 64 + (y + 3)² = 100, (y + 3)² = 36, y + 3 = ±6, giving y = 3 or y = −9. Two answers are possible because a circle of radius 10 about P meets the vertical line x = 10 in two points.

Relation between coordinates. If P(x, y) is equidistant from A(3, 6) and B(−3, 4), find the relation between x and y. PA² = PB²: (x − 3)² + (y − 6)² = (x + 3)² + (y − 4)². Expanding and cancelling squares, −6x − 12y + 45 = 6x − 8y + 25, so 12x + 4y = 20, that is, 3x + y = 5. This is the equation of the perpendicular bisector of AB, the set of all points equidistant from A and B.

Circumcentre. The point equidistant from three vertices of a triangle is its circumcentre. To find the point P(x, y) equidistant from A(6, −6), B(3, −7) and C(3, 3), set PA² = PB² and PB² = PC², giving two linear equations. PA² = PB² gives −12x + 12y + 72 = −6x + 14y + 58, so 6x + 2y = 14, i.e. 3x + y = 7. PB² = PC² gives 14y + 58 = −6y + 18, so 20y = −40 and y = −2; then x = 3. The point is (3, −2).

Always finish by checking the answer with a direct distance computation; it takes a minute and catches sign mistakes.

Suppose a telephone company wants to place a relay tower on the line between two towns A and B so that it is twice as far from A as from B. Where exactly should it stand? Questions of this kind ask for the point that divides a segment in a given ratio, and the section formula answers them.

Theorem. The coordinates of the point P(x, y) which divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ are

x = (m₁x₂ + m₂x₁)/(m₁ + m₂), y = (m₁y₂ + m₂y₁)/(m₁ + m₂).

Derivation. Let P divide AB so that AP : PB = m₁ : m₂. Draw AR, PS and BT perpendicular to the x-axis. Draw AQ ⊥ PS and PC ⊥ BT. Then AQ = RS = x − x₁, PC = ST = x₂ − x, PQ = PS − QS = PS − AR = y − y₁, and BC = BT − CT = BT − PS = y₂ − y.

In ΔAQP and ΔPCB, ∠PAQ = ∠BPC (corresponding angles, since AQ ∥ PC) and ∠AQP = ∠PCB = 90°. So ΔAQP ~ ΔPCB by AA, and therefore AQ/PC = PQ/BC = AP/PB = m₁/m₂.

From AQ/PC = m₁/m₂: (x − x₁)/(x₂ − x) = m₁/m₂, so m₂x − m₂x₁ = m₁x₂ − m₁x, giving x(m₁ + m₂) = m₁x₂ + m₂x₁, hence the formula for x. From PQ/BC = m₁/m₂: (y − y₁)/(y₂ − y) = m₁/m₂, which gives the formula for y in the same way.

Memory aid: the ratio m₁ : m₂ is written between the points, and each coordinate is found by cross-multiplying, m₁ with the far point B and m₂ with the near point A, then dividing by the sum. If the ratio is given as k : 1, the formulas simplify to x = (kx₂ + x₁)/(k + 1), y = (ky₂ + y₁)/(k + 1), a form convenient for finding an unknown ratio.

The derivation, with its figure, is a standard four-mark question. The key steps are the two similar right triangles and the ratio of their corresponding sides.

The section formula is applied in three directions: to find the dividing point when the ratio is known, to find the ratio when the dividing point is known, and to find an endpoint when the dividing point and the ratio are known. Each is a direct substitution followed by simple algebra.

Finding the ratio. In what ratio does the point (−4, 6) divide the segment joining A(−6, 10) and B(3, −8)? Let the ratio be k : 1. Then −4 = (3k − 6)/(k + 1), so −4k − 4 = 3k − 6, giving 7k = 2 and k = 2/7. The ratio is 2 : 7. Check with the y-coordinate: (−8 × 2 + 10 × 7)/9 = (−16 + 70)/9 = 6, correct. Always verify with the other coordinate; if it fails, the point is not on the line at all.

Points of trisection. The points that divide a segment 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 divides in 1 : 2, giving ((−7 + 4)/3, (4 − 4)/3) = (−1, 0); the second divides in 2 : 1, giving ((−14 + 2)/3, (8 − 2)/3) = (−4, 2). Alternatively, the second point is the mid-point of the first point and B.

Dividing into n equal parts. To divide AB into four equal parts, the three dividing points divide it in ratios 1 : 3, 1 : 1 and 3 : 1. For A(−2, 2) and B(2, 8): 1 : 3 gives (−1, 7/2); 1 : 1 gives (0, 5); 3 : 1 gives (1, 13/2).

Finding an endpoint. If (2, 3) divides AB in ratio 1 : 2 with A(1, 1), find B(x, y). Then 2 = (x + 2)/3, so x = 4, and 3 = (y + 2)/3, so y = 7. B is (4, 7).

Vertices of a parallelogram. If three vertices are known and the fourth is (x, y), use the fact that the diagonals bisect each other: the mid-point of AC equals the mid-point of BD. For A(1, 2), B(4, y), C(x, 6), D(3, 5): mid-point of AC = ((1 + x)/2, 4) and mid-point of BD = (7/2, (y + 5)/2). So 1 + x = 7, x = 6, and y + 5 = 8, y = 3.

In a word problem such as flower beds on a lawn or a flag on a running track, translate the description into coordinates first, then apply the formula, then translate the answer back into words.

The most used special case of the section formula is the mid-point, which divides a segment in the ratio 1 : 1. Putting m₁ = m₂ = 1 in the section formula gives

Mid-point of A(x₁, y₁) and B(x₂, y₂) = ((x₁ + x₂)/2, (y₁ + y₂)/2).

In words, the mid-point has coordinates equal to the averages of the corresponding coordinates of the endpoints. This is easy to remember and easy to apply mentally: the mid-point of (3, −5) and (−7, 9) is ((3 − 7)/2, (−5 + 9)/2) = (−2, 2).

Uses of the mid-point formula in this chapter:

• Finding an endpoint. If the mid-point of AB is M(2, −3) and A is (−1, 4), then B = (2 × 2 − (−1), 2 × (−3) − 4) = (5, −10), since each coordinate of B is twice the mid-point coordinate minus the coordinate of A.
• Diagonals of a parallelogram. Because they bisect each other, the mid-point of one diagonal equals the mid-point of the other. This is the quickest way to find a missing vertex or to prove that four given points form a parallelogram.
• Medians of a triangle. A median joins a vertex to the mid-point of the opposite side; its length is found by the distance formula once the mid-point is known.
• Centre of a circle. The centre is the mid-point of any diameter. If the ends of a diameter are (2, −3) and (−6, 7), the centre is (−2, 2).
• Mid-point theorem in coordinates. The segment joining the mid-points of two sides of a triangle is half the third side; this can be verified by distances.

Worked example: find the length of the median from A in the triangle A(4, 2), B(6, 5), C(1, 4). The mid-point of BC is D((6 + 1)/2, (5 + 4)/2) = (7/2, 9/2). AD = √[(7/2 − 4)² + (9/2 − 2)²] = √[1/4 + 25/4] = √(26/4) = √26/2 units.

Worked example: if (1, 2), (4, y), (x, 6) and (3, 5) are vertices of a parallelogram taken in order, the mid-point of the diagonal joining (1, 2) and (x, 6) is ((1 + x)/2, 4) and the mid-point of the diagonal joining (4, y) and (3, 5) is (7/2, (y + 5)/2). Equating gives x = 6, y = 3.

A frequently asked question is: in what ratio does the x-axis (or the y-axis, or a given line) divide the segment joining two points? The method is the section formula in the k : 1 form combined with the property of the axis.

Division by the x-axis. Every point on the x-axis has y = 0. Let the x-axis meet AB at P, dividing it in ratio k : 1. Then the y-coordinate of P is (ky₂ + y₁)/(k + 1) = 0, so ky₂ + y₁ = 0 and k = −y₁/y₂. For A(1, −5) and B(−4, 5): k = 5/5 = 1, so the x-axis bisects AB, in ratio 1 : 1, and the point of division is ((−4 + 1)/2, 0) = (−3/2, 0).

Division by the y-axis. Every point on the y-axis has x = 0, so k = −x₁/x₂. For A(5, −6) and B(−1, −4): k = 5/1 = 5, ratio 5 : 1, and the point is (0, (5 × (−4) + (−6))/6) = (0, −26/6) = (0, −13/3).

The sign of k has a meaning. If k comes out positive, the axis cuts the segment between A and B (internal division), which happens when A and B are on opposite sides of the axis. If k comes out negative, the points are on the same side of the axis and the axis meets the line AB extended, outside the segment; the syllabus deals with internal division, so state that the axis does not cut the segment between the points.

Division by a given line. Find the ratio in which the line 2x + 3y − 5 = 0 divides the segment joining (8, −9) and (2, 1). Let the ratio be k : 1. The dividing point is ((2k + 8)/(k + 1), (k − 9)/(k + 1)). It lies on the line, so 2(2k + 8) + 3(k − 9) − 5(k + 1) = 0, that is, 4k + 16 + 3k − 27 − 5k − 5 = 0, so 2k = 16 and k = 8. The ratio is 8 : 1, and the point is ((16 + 8)/9, (8 − 9)/9) = (8/3, −1/9).

A neat shortcut for a line L(x, y) = 0: the ratio in which it divides AB is −L(x₁, y₁) : L(x₂, y₂). For the example, L(8, −9) = 16 − 27 − 5 = −16 and L(2, 1) = 4 + 3 − 5 = 2, giving 16 : 2 = 8 : 1, the same answer. Use the substitution method in the examination and the shortcut to check.

A median of a triangle joins a vertex to the mid-point of the opposite side. The three medians of any triangle meet at a single point called the centroid, and this point divides each median in the ratio 2 : 1 measured from the vertex. In coordinates the centroid has a beautifully simple form.

Theorem. The centroid of the triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) is

G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3).

Derivation. Let D be the mid-point of BC: D = ((x₂ + x₃)/2, (y₂ + y₃)/2). The centroid G lies on the median AD and divides it in ratio 2 : 1 from A, so by the section formula with m₁ = 2, m₂ = 1:
x = [2 × (x₂ + x₃)/2 + 1 × x₁]/(2 + 1) = (x₁ + x₂ + x₃)/3,
y = [2 × (y₂ + y₃)/2 + 1 × y₁]/3 = (y₁ + y₂ + y₃)/3.
Repeating the computation on the medians from B and from C gives the same point, which proves that the three medians are concurrent at G.

In words, the centroid is the average of the three vertices. It is the balance point of a triangular sheet of uniform thickness: a cardboard triangle balances on a pin placed at its centroid.

Worked example: the centroid of the triangle with vertices (3, −5), (−7, 4) and (10, −2) is ((3 − 7 + 10)/3, (−5 + 4 − 2)/3) = (2, −1).

Worked example: two vertices of a triangle are (1, 2) and (3, 5) and its centroid is (−1, 4). Find the third vertex (x, y). Then (1 + 3 + x)/3 = −1 gives x = −7, and (2 + 5 + y)/3 = 4 gives y = 5. The third vertex is (−7, 5).

Worked example: verify that the centroid divides the median in 2 : 1 for A(4, 2), B(6, 5), C(1, 4). G = (11/3, 11/3). D, the mid-point of BC, is (7/2, 9/2). The point dividing AD in 2 : 1 from A is ((2 × 7/2 + 4)/3, (2 × 9/2 + 2)/3) = (11/3, 11/3) = G, as required.

Note the difference from the circumcentre, which is equidistant from the vertices and is found by solving distance equations; the centroid is found by averaging. Do not confuse the two.

In earlier classes you found the area of a triangle as ½ × base × height, or by Heron's formula from the three sides. When the vertices are given as coordinates, a direct formula is available.

Theorem. The area of the triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) is

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

Derivation. Draw perpendiculars AP, BQ, CR from the vertices to the x-axis. The area of the triangle equals the area of trapezium ABQP plus the area of trapezium APRC minus the area of trapezium BQRC (for the configuration in which A lies between B and C horizontally, with A highest). The area of a trapezium is ½ × (sum of parallel sides) × distance between them. So
ar(ABQP) = ½(BQ + AP) × QP = ½(y₂ + y₁)(x₁ − x₂),
ar(APRC) = ½(AP + CR) × PR = ½(y₁ + y₃)(x₃ − x₁),
ar(BQRC) = ½(BQ + CR) × QR = ½(y₂ + y₃)(x₃ − x₂).
Adding the first two and subtracting the third, then expanding and collecting terms, every product of the form x_iy_i cancels and what remains is ½[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)]. The expression can be negative for another order of the vertices, so we take the absolute value to get the area.

Memory aid: write the vertices in a cyclic pattern. Each x is multiplied by the difference of the y's of the other two vertices, taken in the order next, then next-next: x₁ with (y₂ − y₃), x₂ with (y₃ − y₁), x₃ with (y₁ − y₂).

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

Worked example: area of the triangle with vertices (5, 2), (4, 7), (7, −4). Area = ½ |5(7 + 4) + 4(−4 − 2) + 7(2 − 7)| = ½ |55 − 24 − 35| = ½ |−4| = 2 square units. Here the bracket came out negative; the absolute value makes the area positive.

Units: area is in square units of the coordinate grid. If a question fixes 1 unit = 1 cm, answer in cm².

If three points lie on a straight line, the triangle they form collapses to a segment and has zero area. Conversely, if the area formula gives zero, the points cannot form a triangle and so they are collinear. This gives a second test for collinearity, quicker than the distance method because it involves no square roots.

Test. Points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are collinear if and only if x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) = 0.

Worked example: are (1, 5), (2, 3), (−2, −11) collinear? Compute 1(3 + 11) + 2(−11 − 5) + (−2)(5 − 3) = 14 − 32 − 4 = −22 ≠ 0. Not collinear; indeed the triangle has area 11 square units.

Worked example: show that (1, −1), (5, 2), (9, 5) are collinear. Compute 1(2 − 5) + 5(5 + 1) + 9(−1 − 2) = −3 + 30 − 27 = 0. Collinear.

Finding an unknown for collinearity. Find k if the points (7, −2), (5, 1), (3, k) are collinear. Set 7(1 − k) + 5(k + 2) + 3(−2 − 1) = 0, that is, 7 − 7k + 5k + 10 − 9 = 0, so −2k + 8 = 0 and k = 4.

Worked example: find the value of k for which (8, 1), (k, −4), (2, −5) are collinear. 8(−4 + 5) + k(−5 − 1) + 2(1 + 4) = 8 − 6k + 10 = 0, so k = 3.

Comparison of the two tests. The distance test requires three square roots and a check that two of them add to the third, which can fail on rounding; the area test is exact arithmetic with integers. In the examination, use the area test unless the question specifically asks you to use the distance formula. State the principle clearly: the area of the triangle formed by the points is zero, hence the points are collinear.

The area test also underlies the equation of a line through two points: a variable point (x, y) is on the line through A and B exactly when x(y₁ − y₂) + x₁(y₂ − y) + x₂(y − y₁) = 0. You will use this form in the Intermediate course; here it is enough to know that collinearity means zero area.

A quadrilateral has no single area formula in coordinates in this course, but every quadrilateral can be split by a diagonal into two triangles, and the area of each triangle is found by the formula of the previous topics. The area of the quadrilateral is their sum.

Method. For quadrilateral ABCD with vertices in order, draw the diagonal AC. Then ar(ABCD) = ar(ABC) + ar(ACD). The vertices must be taken in order around the figure, otherwise the diagonal chosen may lie outside and the two triangles overlap.

Worked example: find the area of the quadrilateral with vertices A(−4, −2), B(−3, −5), C(3, −2), D(2, 3). ar(ABC) = ½ |−4(−5 + 2) + (−3)(−2 + 2) + 3(−2 + 5)| = ½ |12 + 0 + 9| = 21/2. ar(ACD) = ½ |−4(−2 − 3) + 3(3 + 2) + 2(−2 + 2)| = ½ |20 + 15 + 0| = 35/2. Total area = 21/2 + 35/2 = 28 square units.

Worked example: the vertices of a quadrilateral are (−5, 7), (−4, −5), (−1, −6), (4, 5). With diagonal from (−5, 7) to (−1, −6): first triangle ½ |−5(−5 + 6) + (−4)(−6 − 7) + (−1)(7 + 5)| = ½ |−5 + 52 − 12| = 35/2; second triangle ½ |−5(−6 − 5) + (−1)(5 − 7) + 4(7 + 6)| = ½ |55 + 2 + 52| = 109/2. Total = 144/2 = 72 square units.

Combined problems. Examination questions often chain the formulas. Typical patterns:

• Find the area of the triangle formed by joining the mid-points of the sides of a triangle, and compare with the whole. For A(0, −1), B(2, 1), C(0, 3), the mid-points are (1, 0), (1, 2), (0, 1); the small triangle has area ½ |1(2 − 1) + 1(1 − 0) + 0| = 1, and the big triangle has area ½ |0 + 2(3 + 1) + 0| = 4. The ratio is 1 : 4, as the mid-point theorem predicts.
• Find the area of a triangle whose vertices are given in terms of a variable, and hence find the variable from a given area.
• Use the area of a triangle to find the length of an altitude: since area = ½ × base × height, height = 2 × area ÷ base, with the base found by the distance formula.
• Find the ratio in which a line divides a segment and then the area of the triangle so formed.

In every combined problem write each formula you use by name before substituting, keep fractions exact, and give the final answer with its unit.