Chapter 13 — Coordinate Geometry Optional

The main chapter teaches five formulas one at a time. The optional exercise sets problems in which the formulas must be chosen and combined: a circle's centre needs the distance formula three times and a pair of linear equations; a parallelogram with unknown vertices needs the midpoint formula on both diagonals; a triangle cut by a line needs the section formula for two points and then the area formula twice; a locus problem needs a squared-distance equation simplified into a line. The arithmetic is not harder than in the main exercises, but there is more of it, and the order of operations must be planned.

Three habits make the difference. First, sketch. A rough plot of the given points shows which vertex is which, whether a quadrilateral's vertices are in order, which side is the base, and roughly where the answer should lie — a centre inside the triangle, a fourth vertex opposite the second. Second, work with squares. Equating PA² = PB² instead of PA = PB removes every square root; the x² and y² terms cancel and a linear equation remains. Third, name the formula before using it. "By the section formula with m : n = 1 : 3", "by the midpoint formula", "by the area formula" — the examiner is looking for the choice as much as the computation.

The families of problem in this exercise: (1) centre of a circle through three points; (2) parallelogram vertices and the fourth vertex; (3) areas of quadrilaterals; (4) a line cutting two sides of a triangle in the same ratio, and the ratio of areas; (5) altitude from area; (6) a line dividing a segment; (7) loci — the relation between x and y under a condition; (8) medians, centroid and their proofs; (9) mixed problems. The chapter takes them in this order.

A note on the last step. In many of these problems the answer must be checked: a centre must be equidistant from all three points, not just two; a fourth vertex must make both pairs of opposite sides equal; a ratio found from the x-coordinate must agree with the y-coordinate. The check is quick and is part of a complete answer. The optional exercise is also where the difference between "the points are the vertices of a rhombus" and "the points are the vertices of a square" is tested: the extra check on the diagonals is expected, and its absence loses a mark.

Every point of a circle is at the same distance, the radius, from the centre. So if a circle passes through A, B and C, its centre O(x, y) satisfies OA = OB = OC, and two equations OA² = OB² and OB² = OC² determine x and y.

Worked example. Find the centre of the circle passing through (6, −6), (3, −7) and (3, 3). Let O(x, y). OA² = OB²: (x − 6)² + (y + 6)² = (x − 3)² + (y + 7)². Expanding, x² − 12x + 36 + y² + 12y + 36 = x² − 6x + 9 + y² + 14y + 49. The x² and y² cancel, leaving −12x + 12y + 72 = −6x + 14y + 58, so −6x − 2y = −14, that is, 3x + y = 7 … (1). OB² = OC²: (x − 3)² + (y + 7)² = (x − 3)² + (y − 3)². The (x − 3)² cancels, so y² + 14y + 49 = y² − 6y + 9, 20y = −40, y = −2. Then from (1), x = 3. The centre is (3, −2). The radius is OA = √[(6 − 3)² + (−6 + 2)²] = √(9 + 16) = 5. Check with C: OC² = 0 + 25 = 25. Correct.

Notice that in the second equation x cancelled entirely because B and C have the same x-coordinate; the centre lies on the horizontal line halfway between them, y = (−7 + 3)/2 = −2. Whenever two of the points share a coordinate, one equation is trivial; use it first.

Second example. Find the centre of the circle through (1, 1), (2, −1), (3, 2). OA² = OB²: (x − 1)² + (y − 1)² = (x − 2)² + (y + 1)², so −2x − 2y + 2 = −4x + 2y + 5, 2x − 4y = 3. OB² = OC²: (x − 2)² + (y + 1)² = (x − 3)² + (y − 2)², so −4x + 2y + 5 = −6x − 4y + 13, 2x + 6y = 8, x + 3y = 4. Solving: from the second, x = 4 − 3y; substitute: 8 − 6y − 4y = 3, y = 1/2, x = 5/2. Centre (5/2, 1/2), radius √[(1 − 5/2)² + (1 − 1/2)²] = √(9/4 + 1/4) = √(10/4) = √10/2.

Related: the circumcentre of a triangle. The centre of the circle through the three vertices of a triangle is its circumcentre, and the same method finds it. For a right triangle the circumcentre is the midpoint of the hypotenuse: for (0, 0), (6, 0), (0, 8) it is (3, 4), at distance 5 from each vertex.

Finding a point on an axis equidistant from three points is impossible in general; but a point equidistant from two points on an axis, or a point on a given line equidistant from two points, is a common two-mark variant: find the point on the y-axis equidistant from (−5, −2) and (3, 2): (0, y) with 25 + (y + 2)² = 9 + (y − 2)², 8y = −16, y = −2, point (0, −2).

Write both equations in full, show the cancellation, solve the linear pair, state the centre and the radius, and verify with the third point. That is the complete four-mark answer.

In a parallelogram ABCD (vertices in order), the diagonals AC and BD bisect each other, so they have the same midpoint. This single fact solves every problem about unknown vertices.

Two unknowns among four vertices. If (1, 2), (4, y), (x, 6), (3, 5) are the vertices of a parallelogram in order, find x and y. Midpoint of AC: ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4). Midpoint of BD: ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2). Equate: (1 + x)/2 = 7/2 gives x = 6; (y + 5)/2 = 4 gives y = 3. The vertices are (1, 2), (4, 3), (6, 6), (3, 5).

The fourth vertex. Three vertices of a parallelogram ABCD are A(1, 2), B(4, 3), C(6, 6). Find D. Midpoint of AC = (7/2, 4). If D = (x, y), midpoint of BD = ((4 + x)/2, (3 + y)/2). So 4 + x = 7, x = 3; 3 + y = 8, y = 5. D = (3, 5). A shortcut: D = A + C − B = (1 + 6 − 4, 2 + 6 − 3) = (3, 5), because the diagonals share a midpoint means A + C = B + D as coordinate sums.

Which vertex is the fourth? If three points are given without saying which is opposite which, there are three possible fourth vertices. For A(−2, −1), B(1, 0), C(4, 3): if AC is a diagonal, D = A + C − B = (1, 2); if AB is a diagonal, D = A + B − C = (−5, −4); if BC is a diagonal, D = B + C − A = (7, 4). The problem usually says "ABCD in order", which fixes D = A + C − B.

Confirming with sides. For (1, 2), (4, 3), (6, 6), (3, 5): AB² = 9 + 1 = 10, CD² = 9 + 1 = 10, BC² = 4 + 9 = 13, DA² = 4 + 9 = 13. Opposite sides equal — a parallelogram confirmed. Since AC² = 25 + 16 = 41 and BD² = 1 + 4 = 5 are unequal, it is not a rectangle.

A parallelogram from vectors of position. The relation A + C = B + D is a coordinate way of saying that the displacement from A to B equals the displacement from D to C: B − A = (3, 1) and C − D = (3, 1). Checking the displacement of one pair of opposite sides is the fastest verification of a parallelogram, and it is the same as checking equal slopes and equal lengths at once.

Area of the parallelogram. Twice the area of triangle ABC: 2 × ½ |1(3 − 6) + 4(6 − 2) + 6(2 − 3)| = |−3 + 16 − 6| = 7 square units. The parallelogram (1, 2), (4, 3), (6, 6), (3, 5) has area 7.

The examination's usual form is the two-unknowns problem; the four-mark form adds "and find the area" or "show that it is not a rectangle".

A quadrilateral ABCD is split by the diagonal AC into triangles ABC and ACD, and its area is the sum of theirs. Either diagonal may be used; choose the one that gives simpler numbers, and take the vertices in order so the two triangles do not overlap.

Example 1. Find the area of the quadrilateral with vertices (−4, −2), (−3, −5), (3, −2), (2, 3), taken in order. Triangle ABC with A(−4, −2), B(−3, −5), C(3, −2): ½ |−4(−5 + 2) + (−3)(−2 + 2) + 3(−2 + 5)| = ½ |12 + 0 + 9| = 21/2. Triangle ACD with A(−4, −2), C(3, −2), D(2, 3): ½ |−4(−2 − 3) + 3(3 + 2) + 2(−2 + 2)| = ½ |20 + 15 + 0| = 35/2. Total 21/2 + 35/2 = 56/2 = 28 square units.

Example 2. Vertices (1, 2), (6, 2), (5, 3), (3, 4). Triangle ABC: ½ |1(2 − 3) + 6(3 − 2) + 5(2 − 2)| = ½ |−1 + 6 + 0| = 5/2. Triangle ACD: ½ |1(3 − 4) + 5(4 − 2) + 3(2 − 3)| = ½ |−1 + 10 − 3| = 3. Total 11/2 square units.

Example 3 — a rhombus by diagonals. For (3, 0), (4, 5), (−1, 4), (−2, −1), the diagonals have lengths AC = √(16 + 16) = 4√2 and BD = √(36 + 36) = 6√2, so the area is ½ × 4√2 × 6√2 = 24. By triangles: ABC = ½ |3(5 − 4) + 4(4 − 0) + (−1)(0 − 5)| = ½ |3 + 16 + 5| = 12, and ACD = 12 by symmetry; total 24. Both methods agree.

Example 4 — a field as a triangle with a fence. Students of a school are standing in rows and columns in their playground. A, B, C, D are at (3, 1), (6, 4), (1, 6), (5, 7) on a grid; find the area of the quadrilateral ABDC (note the order: A, B, D, C go round the figure). Triangle ABD: ½ |3(4 − 7) + 6(7 − 1) + 5(1 − 4)| = ½ |−9 + 36 − 15| = 6. Triangle ADC: ½ |3(7 − 6) + 5(6 − 1) + 1(1 − 7)| = ½ |3 + 25 − 6| = 11. Total 17 square units.

When the order is wrong. If the vertices are not taken around the figure, the "diagonal" is actually a side and the two triangles overlap; the sum then exceeds the true area. Always sketch to fix the order. For the points of Example 1, the order A, B, C, D goes anticlockwise around the shape.

A general formula. For a quadrilateral in order, the area is ½ |(x₁y₂ − x₂y₁) + (x₂y₃ − x₃y₂) + (x₃y₄ − x₄y₃) + (x₄y₁ − x₁y₄)|, the "shoelace" pattern, which is exactly the sum of the two triangle formulas. It is not required, but it explains why the triangle method works for any polygon: keep adding triangles from one vertex.

The vertices of triangle ABC are A(4, 6), B(1, 5), C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of triangle ADE and compare it with the area of triangle ABC.

Step 1 — locate D and E. AD/AB = 1/4 means AD : DB = 1 : 3, so D divides AB in the ratio 1 : 3. By the section formula, D = ((1 × 1 + 3 × 4)/4, (1 × 5 + 3 × 6)/4) = (13/4, 23/4). Similarly E divides AC in 1 : 3: E = ((1 × 7 + 3 × 4)/4, (1 × 2 + 3 × 6)/4) = (19/4, 20/4) = (19/4, 5).

Step 2 — area of ADE. ½ |4(23/4 − 5) + 13/4(5 − 6) + 19/4(6 − 23/4)| = ½ |4 × 3/4 + 13/4 × (−1) + 19/4 × 1/4| = ½ |3 − 13/4 + 19/16| = ½ |(48 − 52 + 19)/16| = ½ × 15/16 = 15/32 square units.

Step 3 — area of ABC. ½ |4(5 − 2) + 1(2 − 6) + 7(6 − 5)| = ½ |12 − 4 + 7| = 15/2 square units.

Step 4 — compare. (15/32) ÷ (15/2) = 2/32 = 1/16. The area of ADE is one-sixteenth of the area of ABC.

Why 1/16. Since AD/AB = AE/AC and angle A is common, triangles ADE and ABC are similar (SAS similarity) with ratio 1 : 4, and the areas of similar triangles are in the ratio of the squares of corresponding sides: (1/4)² = 1/16. The coordinate calculation confirms the theorem of the Similar Triangles chapter, and the examination often asks for exactly this comparison.

Variant with ratio 1 : 2. Same triangle, AD/AB = AE/AC = 1/3. D divides AB in 1 : 2: D = ((1 + 8)/3, (5 + 12)/3) = (3, 17/3). E = ((7 + 8)/3, (2 + 12)/3) = (5, 14/3). Area ADE = ½ |4(17/3 − 14/3) + 3(14/3 − 6) + 5(6 − 17/3)| = ½ |4 − 4 + 5/3| = 5/6. Ratio to 15/2: (5/6)/(15/2) = 1/9 = (1/3)². Again the square of the ratio.

A related problem. If D and E are the midpoints of AB and AC, then DE is parallel to BC and half its length, and area ADE = ¼ area ABC. With the same triangle, D = (5/2, 11/2), E = (11/2, 4), and area ADE = ½ |4(11/2 − 4) + 5/2(4 − 6) + 11/2(6 − 11/2)| = ½ |6 − 5 + 11/4| = 15/8 = ¼ × 15/2. Confirmed.

The fractions are the only difficulty. Keep every coordinate as a fraction with a common denominator, compute each bracket separately, and combine at the end. Write the ratio of areas as a single fraction and state its relation to the ratio of sides.

The area of a triangle is ½ × base × height. If the area is known from the coordinate formula and the base is known from the distance formula, the height follows — and the height is the perpendicular distance from the opposite vertex to the line of the base. This is the Class 10 way to find such a distance without the point-to-line formula of the Intermediate course.

Example 1. Find the length of the altitude from A in the triangle A(1, −1), B(−4, 6), C(−3, −5). Area = ½ |1(6 + 5) + (−4)(−5 + 1) + (−3)(−1 − 6)| = ½ |11 + 16 + 21| = 24. Base BC = √[(−3 + 4)² + (−5 − 6)²] = √(1 + 121) = √122. Then ½ × √122 × h = 24, so h = 48/√122 = 48√122/122 = 24√122/61 ≈ 4.35 units.

Example 2. The vertices of a triangle are (2, 3), (4, 7), (6, 1). Find the altitude on the side joining (4, 7) and (6, 1). Area = ½ |2(7 − 1) + 4(1 − 3) + 6(3 − 7)| = ½ |12 − 8 − 24| = 10. Base = √(4 + 36) = √40 = 2√10. h = 2 × 10 / (2√10) = 10/√10 = √10 units.

Example 3 — a right triangle check. For (0, 0), (4, 0), (0, 3), the altitude on the hypotenuse: area 6, hypotenuse 5, h = 12/5 = 2.4, which agrees with the standard result (product of legs)/(hypotenuse).

Distance of a point from a line through two points. The same method gives the distance of any point P from the line through Q and R: form triangle PQR, compute its area, divide twice the area by QR. Distance of (3, 4) from the line through (0, 0) and (6, 0): area = ½ |3(0 − 0) + 0 + 6(4 − 0)| = 12, QR = 6, distance = 24/6 = 4 — obviously right, since the line is the x-axis.

Area from base and height in reverse. If a triangle with base along the x-axis from (−2, 0) to (5, 0) has area 14, its third vertex has |y| = 4; it lies on y = 4 or y = −4. This is the same idea as finding k for a given area: the unknown is a height.

Finding k from a given area. For what k is the area of the triangle (k, 0), (4, 0), (0, 2) equal to 4? ½ |k(0 − 2) + 4(2 − 0) + 0| = 4, |8 − 2k| = 8, so k = 0 or k = 8. The base is on the x-axis with height 2, so the base must be 4 units long: (4 − k) = ±4.

The examination phrases this as "find the length of the altitude", "find the height of the triangle on the base BC" or "find the distance of A from BC". All three are the same computation: area, base, divide.

A line and a segment cross at a point that divides the segment in some ratio. To find it, call the ratio k : 1, write the point of division by the section formula, and require it to satisfy the line's equation.

Example 1. Find the ratio in which the line 2x + y − 4 = 0 divides the segment joining A(2, −2) and B(3, 7). Let P divide AB in k : 1: P = ((3k + 2)/(k + 1), (7k − 2)/(k + 1)). P is on the line, so 2(3k + 2)/(k + 1) + (7k − 2)/(k + 1) − 4 = 0. Multiply by (k + 1): 6k + 4 + 7k − 2 − 4k − 4 = 0, 9k − 2 = 0, k = 2/9. The ratio is 2 : 9, and P = ((6/9 + 2)/(11/9), (14/9 − 2)/(11/9)) = ((24/9)/(11/9), (−4/9)/(11/9)) = (24/11, −4/11). Check: 2(24/11) − 4/11 − 4 = (48 − 4 − 44)/11 = 0.

Example 2. In what ratio does the line x − y − 2 = 0 divide the segment joining (3, −1) and (8, 9)? P = ((8k + 3)/(k + 1), (9k − 1)/(k + 1)). On the line: (8k + 3) − (9k − 1) − 2(k + 1) = 0, 8k + 3 − 9k + 1 − 2k − 2 = 0, −3k + 2 = 0, k = 2/3. Ratio 2 : 3; P = ((16/3 + 3)/(5/3), (6 − 1)/(5/3)) = (5, 3). Check: 5 − 3 − 2 = 0.

Example 3 — the axes as lines. The x-axis is the line y = 0, so it divides AB in the ratio −y₁ : y₂. For (1, −5) and (−4, 5): 5 : 5 = 1 : 1. The y-axis is x = 0 and divides AB in −x₁ : x₂. For (5, −6) and (−1, −4): −5 : −1 = 5 : 1. A positive ratio means the crossing is between A and B (internal division); a negative ratio means the line meets AB produced, outside the segment.

Example 4 — a given point on the segment. Find the ratio in which (−4, 6) divides the segment from (−6, 10) to (3, −8), and confirm the point is on the segment. From x: (3k − 6)/(k + 1) = −4, 3k − 6 = −4k − 4, k = 2/7. From y: (−8k + 10)/(k + 1) = 6, −8k + 10 = 6k + 6, k = 2/7. Same k, so the point lies on the segment and divides it in 2 : 7.

Example 5 — the relation between x and y for a point on a segment. If P(x, y) divides (2, 3) to (6, 11) in some ratio k : 1, then x = (6k + 2)/(k + 1) and y = (11k + 3)/(k + 1). Eliminating k: from the first, k = (x − 2)/(6 − x); substituting into the second and simplifying gives y = 2x − 1, the line through the two points. Every point of the segment satisfies this, which is why the collinearity condition gives the line's equation.

The device k : 1 is chosen so that there is one unknown; if the ratio comes out as a fraction p/q, report it as p : q. If k is negative, the line does not cut the segment between its ends, and the examination expects that observation.

When a point P(x, y) is described by a condition rather than by numbers — "equidistant from A and B", "forming a triangle of area 5 with A and B", "collinear with A and B" — the condition becomes an equation in x and y, and that equation is the locus of P: the set of all positions P can occupy. In Class 10 these loci are straight lines; the Intermediate course adds circles and other curves.

Equidistant from two points. Find a relation between x and y such that (x, y) is equidistant from (3, 6) and (−3, 4). PA² = PB²: (x − 3)² + (y − 6)² = (x + 3)² + (y − 4)². Expanding, x² − 6x + 9 + y² − 12y + 36 = x² + 6x + 9 + y² − 8y + 16, so −6x − 12y + 45 = 6x − 8y + 25, −12x − 4y + 20 = 0, 3x + y − 5 = 0. This is the perpendicular bisector of AB. Check: the midpoint of AB, (0, 5), satisfies 0 + 5 − 5 = 0; and the slope of AB is (4 − 6)/(−3 − 3) = 1/3 while the line 3x + y = 5 has slope −3, and (1/3)(−3) = −1, so they are perpendicular.

Equidistant from (7, 1) and (3, 5). (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)²: −14x + 49 − 2y + 1 = −6x + 9 − 10y + 25, −8x + 8y + 16 = 0, x − y − 2 = 0. Verify with the midpoint (5, 3): 5 − 3 − 2 = 0.

Collinear with two points. The relation for (x, y) collinear with (1, 2) and (7, 0): x(2 − 0) + 1(0 − y) + 7(y − 2) = 0, 2x − y + 7y − 14 = 0, x + 3y − 7 = 0. This is the line AB itself.

Triangle of fixed area. Find the locus of (x, y) such that the triangle with (1, 0) and (0, 1) has area 1. ½ |x(0 − 1) + 1(1 − y) + 0| = 1, so |−x + 1 − y| = 2, x + y − 1 = ±2. Two lines: x + y = 3 and x + y = −1, both parallel to AB (slope −1) and at the right distance on either side.

A point at a given distance from a fixed point. If (x, y) is at distance 5 from the origin, x² + y² = 25 — a circle, not a line. This is beyond the Class 10 syllabus but is the natural next step and appears in the Intermediate course as the equation of a circle.

Sum of squared distances. If PA² + PB² = 20 with A(1, 0), B(−1, 0): (x − 1)² + y² + (x + 1)² + y² = 20, 2x² + 2y² + 2 = 20, x² + y² = 9. Again a circle.

The technique is uniform: write the condition in coordinates, expand, cancel, simplify to lowest integer coefficients, and verify with one point that obviously satisfies the condition (the midpoint for equidistance, one of the two points for collinearity). The examination asks for "the relation between x and y"; give it as an equation in the form ax + by + c = 0 with integer coefficients.

A median joins a vertex to the midpoint of the opposite side. The three medians of any triangle pass through one point, the centroid G, and G divides each median in the ratio 2 : 1 measured from the vertex. In coordinates, G is the average of the three vertices.

Proof that the centroid is the average. Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃). The midpoint of BC is D((x₂ + x₃)/2, (y₂ + y₃)/2). The point dividing AD in the ratio 2 : 1 is, by the section formula, ((2 × (x₂ + x₃)/2 + 1 × x₁)/3, (2 × (y₂ + y₃)/2 + 1 × y₁)/3) = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3). By symmetry the point dividing BE in 2 : 1 (E the midpoint of CA) and the point dividing CF in 2 : 1 are the same expression. So all three medians pass through this point, and it divides each in 2 : 1. This proves both the concurrence of the medians and the 2 : 1 property in one stroke, and is a standard four-mark derivation.

Example 1. Find the centroid of the triangle with vertices (3, −5), (−7, 4), (10, −2). G = ((3 − 7 + 10)/3, (−5 + 4 − 2)/3) = (2, −1).

Example 2. The centroid of a triangle is (2, 3), and two vertices are (1, 2) and (5, 1). Find the third vertex. (1 + 5 + x)/3 = 2 gives x = 0; (2 + 1 + y)/3 = 3 gives y = 6. The third vertex is (0, 6).

Example 3 — a relation from the centroid. If the centroid of the triangle with vertices (a, b), (b, c), (c, a) is the origin, show that a + b + c = 0. The centroid is ((a + b + c)/3, (b + c + a)/3) = (0, 0), so a + b + c = 0. Then also a³ + b³ + c³ = 3abc, from the identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Example 4 — length of a median. A(7, −3), B(5, 3), C(3, −1). Midpoint of BC is D(4, 1). AD = √[(7 − 4)² + (−3 − 1)²] = √(9 + 16) = 5. The centroid G divides AD in 2 : 1: G = ((2 × 4 + 7)/3, (2 × 1 − 3)/3) = (5, −1/3), and directly G = ((7 + 5 + 3)/3, (−3 + 3 − 1)/3) = (5, −1/3). Consistent.

Example 5 — a median divides the triangle into two equal areas. For A(4, 6), B(1, 5), C(7, 2), D = midpoint of BC = (4, 7/2). Area ABD = ½ |4(5 − 7/2) + 1(7/2 − 6) + 4(6 − 5)| = ½ |6 − 5/2 + 4| = 15/4. Area ADC = ½ |4(7/2 − 2) + 4(2 − 6) + 7(6 − 7/2)| = ½ |6 − 16 + 35/2| = 15/4. Equal halves of the total 15/2, as the theory says: same height, equal bases.

Example 6 — the three triangles at the centroid. Triangles GAB, GBC, GCA have equal areas, each one-third of ABC. With A(4, 6), B(1, 5), C(7, 2), G = (4, 13/3): area GAB = ½ |4(5 − 13/3) + 1(13/3 − 6) + 4(6 − 5)| = ½ |8/3 − 5/3 + 4| = 5/2 = (15/2)/3. The other two are the same by the same computation.

Medians, centroid, and their areas are the examination's favourite proof-plus-computation questions: know the section-formula derivation and be able to verify any claimed property numerically.

Several optional problems ask for a point that is constrained to a line or an axis and also satisfies a distance condition. The line supplies one relation, the distance another, and together they fix the point.

A point on the x-axis at a given distance from a point. Find the points on the x-axis at distance 5 from (2, 4). Let the point be (x, 0): (x − 2)² + 16 = 25, (x − 2)² = 9, x = 5 or x = −1. Two points: (5, 0) and (−1, 0). A circle of radius 5 about (2, 4) cuts the x-axis twice.

A point on the y-axis equidistant from two points. (0, y) equidistant from (−5, −2) and (3, 2): 25 + (y + 2)² = 9 + (y − 2)², 25 + 4y + 4 = 9 − 4y + 4, 8y = −16, y = −2. The point is (0, −2).

A point on a given line equidistant from two points. Find the point on the line x + y = 4 equidistant from (2, 3) and (6, 1). Let the point be (x, 4 − x): (x − 2)² + (1 − x)² = (x − 6)² + (3 − x)². Expanding, x² − 4x + 4 + 1 − 2x + x² = x² − 12x + 36 + 9 − 6x + x², so −6x + 5 = −18x + 45, 12x = 40, x = 10/3, y = 2/3. The point is (10/3, 2/3).

A point on a segment at a given distance from one end. The point on the segment from (2, 3) to (8, 11) at distance 5 from (2, 3): the whole segment has length √(36 + 64) = 10, so the point is the midpoint, (5, 7). In general, if the required distance is d and the segment length L, the point divides the segment in the ratio d : (L − d).

A point on the line y = x equidistant from the origin and (4, 2). (t, t): 2t² = (t − 4)² + (t − 2)² = 2t² − 12t + 20, so 12t = 20, t = 5/3. Point (5/3, 5/3).

Where a perpendicular bisector meets an axis. The perpendicular bisector of the segment joining (1, 5) and (4, 6) cuts the y-axis where? Equidistant point on the y-axis: (0, y) with 1 + (y − 5)² = 16 + (y − 6)², 1 + y² − 10y + 25 = 16 + y² − 12y + 36, 2y = 26, y = 13. The point is (0, 13). Alternatively the bisector's equation from PA² = PB² is 6x + 2y − 26 = 0, i.e. 3x + y = 13, which meets x = 0 at y = 13.

A point whose distance from the origin equals its distance from (4, 4). x² + y² = (x − 4)² + (y − 4)² gives 8x + 8y = 32, x + y = 4 — the perpendicular bisector again.

In each case the unknown is reduced to one variable by the line condition, and then a squared-distance equation gives a linear or quadratic equation in that variable. Two solutions are common when the condition is a fixed distance (a circle meets a line twice); one solution when the condition is equidistance (a line meets a line once).

Each of the following is a complete answer of the kind that scores fully. The commentary in brackets notes the mark-earning steps.

Problem 1. Show that A(−3, 2), B(−5, −5), C(2, −3), D(4, 4) are the vertices of a rhombus, and find its area. [Sketch; name the vertices.] AB² = 4 + 49 = 53, BC² = 49 + 4 = 53, CD² = 4 + 49 = 53, DA² = 49 + 4 = 53. [All sides equal — rhombus.] AC² = 25 + 25 = 50, BD² = 81 + 81 = 162. [Diagonals unequal — not a square.] Area = ½ × √50 × √162 = ½ × 5√2 × 9√2 = ½ × 90 = 45 square units. [Formula named, answer with units.]

Problem 2. Find the coordinates of the point that divides the segment joining (−1, 7) and (4, −3) in the ratio 2 : 3, and find the area of the triangle formed by this point with the origin and (4, −3). [Section formula.] P = ((8 − 3)/5, (−6 + 21)/5) = (1, 3). [Area formula.] Area of O(0, 0), P(1, 3), B(4, −3) = ½ |0 + 1(−3 − 0) + 4(0 − 3)| = ½ |−3 − 12| = 15/2 square units.

Problem 3. If A(−2, 1), B(a, 0), C(4, b), D(1, 2) are the vertices of a parallelogram ABCD, find a and b, and the length of the sides. [Midpoints of diagonals.] ((−2 + 4)/2, (1 + b)/2) = ((a + 1)/2, (0 + 2)/2), so a + 1 = 2, a = 1; 1 + b = 2, b = 1. [Distances.] AB = √(9 + 1) = √10, BC = √(9 + 1) = √10. All sides √10 — in fact a rhombus. AC² = 36 + 0 = 36, BD² = 0 + 4 = 4; unequal, so a rhombus, not a square.

Problem 4. Find the point on the x-axis equidistant from (2, −5) and (−2, 9), and the distance of this point from each. [Squared distances.] (x − 2)² + 25 = (x + 2)² + 81, −8x = 56, x = −7. Point (−7, 0). Distance = √(81 + 25) = √106.

Problem 5. Prove that the points (a, b + c), (b, c + a), (c, a + b) are collinear. [Area expression.] a[(c + a) − (a + b)] + b[(a + b) − (b + c)] + c[(b + c) − (c + a)] = a(c − b) + b(a − c) + c(b − a) = ac − ab + ab − bc + bc − ac = 0. Zero area, hence collinear. [Alternatively, all three lie on the line x + y = a + b + c.]

Problem 6. Find the ratio in which the point P(−1, y) lying on the segment joining A(−3, 10) and B(6, −8) divides it, and find y. [k : 1 from x.] (6k − 3)/(k + 1) = −1, 6k − 3 = −k − 1, k = 2/7. Ratio 2 : 7. [Then y.] y = (2 × (−8) + 7 × 10)/9 = (−16 + 70)/9 = 6.

Problem 7. The line segment joining A(2, 1) and B(5, −8) is trisected at P and Q. If P lies on the line 2x − y + k = 0, find k. [Trisection: P divides AB in 1 : 2.] P = ((5 + 4)/3, (−8 + 2)/3) = (3, −2). [Substitute.] 6 + 2 + k = 0, k = −8.

Every problem here has been solved by choosing the right formula, writing it, substituting with the signs visible, and closing with a stated conclusion. That is the whole of the examination's expectation.

Choosing the tool. Ask what is wanted. A length → distance formula. A point on a segment → section or midpoint formula. A ratio → section formula with k : 1. An area → area formula, or ½ d₁d₂ for a rhombus. A height or a distance from a line → area divided by base. A centre → equidistance, squared. A relation between x and y → the condition written in coordinates and simplified. Parallel or perpendicular → slopes. Collinear → area expression zero or equal slopes.

Keeping the algebra clean. Square distances before equating; the x² and y² terms cancel and a linear equation remains. In the section formula, m multiplies the second point's coordinates and n the first's; check by imagining m = 0, which must give the first point. In the area formula, keep the cyclic order (1, 2, 3) and write each y-difference in a bracket with the sign shown. Simplify surds: √50 = 5√2, √162 = 9√2, √122 stays as it is.

Order of vertices. For any quadrilateral, plot first; a wrong order gives a crossed figure and a wrong area or a false "not a parallelogram". For a parallelogram ABCD the diagonals are AC and BD, never AB and CD.

Checking. A centre must be equidistant from all three points; a ratio must agree from both coordinates; a locus must be satisfied by an obvious point (the midpoint for a perpendicular bisector); a fourth vertex must make opposite sides equal; an area ratio for similar triangles must be the square of the side ratio. Each check is one line.

Common errors. Writing √[(x₂ − x₁) + (y₂ − y₁)]² without squaring separately. Using PA = PB with roots and getting stuck. Attaching m to the wrong endpoint. Dropping the absolute value in the area and reporting a negative area. Confusing the midpoint of a diagonal with the centroid. Declaring a square after checking only sides. Taking AD/AB = 1/4 as the ratio AD : DB instead of 1 : 3. Forgetting the units, or giving a distance as an unsimplified fraction with a surd in the denominator.

Time. Two-mark questions: three minutes. Four-mark: eight minutes, of which one goes to the sketch and one to the check.

What comes next. The Intermediate course begins with the equation of a straight line in several forms — slope-intercept y = mx + c, point-slope, two-point, intercept form — and the distance of a point from a line, |ax₁ + by₁ + c|/√(a² + b²), which replaces the area-divided-by-base method of this chapter. It goes on to pairs of lines, circles, parabolas, ellipses and hyperbolas, and to three-dimensional coordinates. Every one of these uses the distance and section formulas unchanged, and the habit of turning a geometric condition into an equation, which is precisely what the optional exercise has practised.