High School Maths · Year 2 · Unit 15 · Lesson 4 of 5

Combined Loci and Regions: Shade Exactly Where the Rules Allow

About 15 minutes · combining two or more loci · "and" means the overlap · solid and dashed boundaries · loci on a coordinate grid: circles, perpendicular bisectors and y = x · testing whether a point is in a region · areas of regions in terms of π · no calculator · exit quiz

Hook

On a grid in metres, a tree T is at (3, 3), a well W at (2, 7) and a gate G at (6, 3). A treasure is buried less than 3 m from the tree, and nearer to the well than to the gate.

Where should you dig?

Draw each locus. "Less than 3 m from T" is the inside of a circle of radius 3 m round T. "Nearer to W than to G" is W's side of the perpendicular bisector of WG. The treasure is where the two regions overlap: the part of the circle on W's side of the line. Both boundaries are dashed, because points on them do not count.

Inside the circle AND on W's side of the bisector of WG.
Learn

Master the essentials

Every point on y = x (first quadrant) is the same distance from both axes.

One rule at a time

Draw the locus for each rule in turn, accurately, leaving any construction arcs.

Then decide which side of each line or curve you need

"And" means overlap

A point that must obey several rules lies where ALL the regions overlap.

Shade only the overlap

Solid or dashed

A solid boundary means points on it count ("at most", "at least"). A dashed boundary means they do not ("less than", "more than").

"At most 3 m": solid circle

Nearer to A than to B

Draw the perpendicular bisector of AB. Shade the side that contains A.

The bisector itself is dashed

Distance on a grid

Use Pythagoras with the differences in x and y.

(1, 2) to (4, 6): √(3² + 4²) = 5

Bisector on a grid

It passes through the midpoint of AB and is perpendicular to AB. If AB is horizontal, it is a vertical line x = ….

A(0, 2), B(4, 2): x = 2

Equidistant from the axes

The same distance from the x-axis and the y-axis: the line y = x (and y = −x in the other quadrants).

Nearer to the x-axis: below y = x in the first quadrant

Test a point

A point is in the region only if it passes EVERY rule. Compare squared distances to avoid square roots.

Check one point inside your shading

Worked examples

See it done

1. On a grid, A is (1, 1) and B is (9, 1). Find the equation of the locus of points equidistant from A and B. Is C (4, 6) nearer to A or to B?

  1. AB is horizontal, so the perpendicular bisector is vertical.
  2. It passes through the midpoint of AB: ((1 + 9) ÷ 2, 1) = (5, 1). So the locus is the line x = 5.
  3. C has x = 4, which is on A's side of x = 5, so C is nearer to A.
  4. Check: CA² = 3² + 5² = 34 and CB² = 5² + 5² = 50. 34 < 50 ✓.

x = 5; C is nearer to A

2. On a grid, P is (4, 3), A is (2, 6) and B is (6, 6). Describe the region of points less than 3 units from P and nearer to A than to B, and find its area in terms of π.

  1. Less than 3 units from P: inside a dashed circle, centre P, radius 3.
  2. Nearer to A than to B: left of the perpendicular bisector of AB, which is x = 4.
  3. The line x = 4 passes through P, the centre, so the region is the left half of the circle.
  4. Area = ½ × π × 3² = 4.5π square units.

The left half of the circle: 4.5π square units

3. Which of the points (5, 2), (2, 4), (5, 1) and (4, 2) lies in the region that is at most 5 units from O (0, 0) and nearer to the x-axis than to the y-axis?

  1. At most 5 units from O: x² + y² must be at most 25.
  2. (5, 2): 25 + 4 = 29, too far. (5, 1): 25 + 1 = 26, too far.
  3. Nearer to the x-axis: the distance to the x-axis (y) must be less than the distance to the y-axis (x). (2, 4) fails: 4 is more than 2.
  4. (4, 2): 16 + 4 = 20 ≤ 25 and 2 < 4. It passes both tests.

(4, 2)

Common mistake: Shading the wrong side, or drawing the wrong kind of line. After shading, test one point inside your shading against every rule. And match the boundary to the words: "less than" needs a dashed line, "at most" a solid one.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Draw each locus on a sketch of the grid, then test points against every rule. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.A is (1, 2) and B is (7, 2). Which line is the locus of points equidistant from A and B?

Not quite. This is the line through A and B.

Solution

  1. AB is horizontal, so the perpendicular bisector is a vertical line.
  2. It passes through the midpoint: (1 + 7) ÷ 2 = 4.
  3. The locus is x = 4.

Answer: C. x = 4.

Not quite. 3 is half of the distance 6. Add it to the x-coordinate of A: 1 + 3 = 4.

Solution

  1. AB is horizontal, so the perpendicular bisector is a vertical line.
  2. It passes through the midpoint: (1 + 7) ÷ 2 = 4.
  3. The locus is x = 4.

Answer: C. x = 4.

Correct.

Solution

  1. AB is horizontal, so the perpendicular bisector is a vertical line.
  2. It passes through the midpoint: (1 + 7) ÷ 2 = 4.
  3. The locus is x = 4.

Not quite. This adds the x-coordinates but forgets to halve: (1 + 7) ÷ 2 = 4.

Solution

  1. AB is horizontal, so the perpendicular bisector is a vertical line.
  2. It passes through the midpoint: (1 + 7) ÷ 2 = 4.
  3. The locus is x = 4.

Answer: C. x = 4.

2.Which point is less than 5 units from the origin O?

Not quite. This point is √32 units from O: 4² + 4² = 32, which is more than 25.

Solution

  1. Distance² from O = x² + y². It must be less than 5² = 25.
  2. (2, 4): 2² + 4² = 4 + 16 = 20 < 25 ✓.
  3. The others give 32, 25 and 26: none is less than 25.

Answer: C. (2, 4).

Not quite. This point is exactly 5 units from O. "Less than" leaves it out.

Solution

  1. Distance² from O = x² + y². It must be less than 5² = 25.
  2. (2, 4): 2² + 4² = 4 + 16 = 20 < 25 ✓.
  3. The others give 32, 25 and 26: none is less than 25.

Answer: C. (2, 4).

Correct.

Solution

  1. Distance² from O = x² + y². It must be less than 5² = 25.
  2. (2, 4): 2² + 4² = 4 + 16 = 20 < 25 ✓.
  3. The others give 32, 25 and 26: none is less than 25.

Not quite. 5² + 1² = 26, which is more than 25, so it is just over 5 units away.

Solution

  1. Distance² from O = x² + y². It must be less than 5² = 25.
  2. (2, 4): 2² + 4² = 4 + 16 = 20 < 25 ✓.
  3. The others give 32, 25 and 26: none is less than 25.

Answer: C. (2, 4).

3.Which line is the locus of points in the first quadrant that are the same distance from the x-axis as from the y-axis?

Correct.

Solution

  1. A point (x, y) is y units from the x-axis and x units from the y-axis.
  2. Equal distances: y = x.

Not quite. This line is in the second and fourth quadrants.

Solution

  1. A point (x, y) is y units from the x-axis and x units from the y-axis.
  2. Equal distances: y = x.

Answer: A. y = x.

Not quite. This is the y-axis itself. Its points are 0 units from the y-axis.

Solution

  1. A point (x, y) is y units from the x-axis and x units from the y-axis.
  2. Equal distances: y = x.

Answer: A. y = x.

Not quite. These points are twice as far from the x-axis as from the y-axis.

Solution

  1. A point (x, y) is y units from the x-axis and x units from the y-axis.
  2. Equal distances: y = x.

Answer: A. y = x.

Exam standard (Several steps, like real exam questions.)

4.On the grid, A is (2, 6), B is (6, 2) and C is (5, 3). A solid circle has centre C and radius 3. The dashed line is y = x. The shaded region is inside the circle and on the same side of y = x as A. Which description gives the shaded region?

Not quite. The circle is solid, so points on it are included: "at most", not "less than".

Solution

  1. The circle has centre C and radius 3. It is solid, so the rule is "at most 3 units from C".
  2. y = x is the perpendicular bisector of AB (it passes through the midpoint (4, 4), at right angles to AB).
  3. The shading is on A's side of it: "nearer to A than to B".

Answer: B. At most 3 units from C, and nearer to A than to B.

Correct.

Solution

  1. The circle has centre C and radius 3. It is solid, so the rule is "at most 3 units from C".
  2. y = x is the perpendicular bisector of AB (it passes through the midpoint (4, 4), at right angles to AB).
  3. The shading is on A's side of it: "nearer to A than to B".

Not quite. The shading is on A's side of the line y = x, not B's.

Solution

  1. The circle has centre C and radius 3. It is solid, so the rule is "at most 3 units from C".
  2. y = x is the perpendicular bisector of AB (it passes through the midpoint (4, 4), at right angles to AB).
  3. The shading is on A's side of it: "nearer to A than to B".

Answer: B. At most 3 units from C, and nearer to A than to B.

Not quite. The shading is inside the circle, so the points are at most 3 units from C.

Solution

  1. The circle has centre C and radius 3. It is solid, so the rule is "at most 3 units from C".
  2. y = x is the perpendicular bisector of AB (it passes through the midpoint (4, 4), at right angles to AB).
  3. The shading is on A's side of it: "nearer to A than to B".

Answer: B. At most 3 units from C, and nearer to A than to B.

5.A is (0, 6), B is (6, 0) and C is (2, 2). Which point is nearer to A than to B, and less than 4 units from C?

Not quite. This point is nearer to B: it is below the line y = x.

Solution

  1. The perpendicular bisector of AB is y = x. Nearer to A means above it (y > x).
  2. (1, 4): 4 > 1 ✓. Distance² to C = 1² + 2² = 5, less than 4² = 16 ✓.
  3. So (1, 4) obeys both rules.

Answer: C. (1, 4).

Not quite. This point is nearer to A, but it is 5 units from C.

Solution

  1. The perpendicular bisector of AB is y = x. Nearer to A means above it (y > x).
  2. (1, 4): 4 > 1 ✓. Distance² to C = 1² + 2² = 5, less than 4² = 16 ✓.
  3. So (1, 4) obeys both rules.

Answer: C. (1, 4).

Correct.

Solution

  1. The perpendicular bisector of AB is y = x. Nearer to A means above it (y > x).
  2. (1, 4): 4 > 1 ✓. Distance² to C = 1² + 2² = 5, less than 4² = 16 ✓.
  3. So (1, 4) obeys both rules.

Not quite. This point is on y = x, the perpendicular bisector of AB: it is the same distance from A and B.

Solution

  1. The perpendicular bisector of AB is y = x. Nearer to A means above it (y > x).
  2. (1, 4): 4 > 1 ✓. Distance² to C = 1² + 2² = 5, less than 4² = 16 ✓.
  3. So (1, 4) obeys both rules.

Answer: C. (1, 4).

6.A park must be at most 2 km from a school S and nearer to a hospital H than to S. The table shows the distances of four possible sites. Which site obeys both rules?

SiteDistance from school S (km)Distance from hospital H (km)
Site 11.82.3
Site 22.41.1
Site 31.91.6
Site 422

Not quite. Site 1 is nearer to the school (1.8 km) than to the hospital (2.3 km).

Solution

  1. Rule 1: the distance from S is 2 km or less. Sites 1, 3 and 4 pass.
  2. Rule 2: the distance from H is less than the distance from S.
  3. Site 3: 1.6 < 1.9 ✓. It is the only site that passes both.

Answer: D. Site 3.

Not quite. Site 2 is 2.4 km from the school, more than 2 km.

Solution

  1. Rule 1: the distance from S is 2 km or less. Sites 1, 3 and 4 pass.
  2. Rule 2: the distance from H is less than the distance from S.
  3. Site 3: 1.6 < 1.9 ✓. It is the only site that passes both.

Answer: D. Site 3.

Not quite. Site 4 is the same distance from both, so it is not nearer to the hospital.

Solution

  1. Rule 1: the distance from S is 2 km or less. Sites 1, 3 and 4 pass.
  2. Rule 2: the distance from H is less than the distance from S.
  3. Site 3: 1.6 < 1.9 ✓. It is the only site that passes both.

Answer: D. Site 3.

Correct.

Solution

  1. Rule 1: the distance from S is 2 km or less. Sites 1, 3 and 4 pass.
  2. Rule 2: the distance from H is less than the distance from S.
  3. Site 3: 1.6 < 1.9 ✓. It is the only site that passes both.

7.Find the area of the region of points in the first quadrant that are at most 6 units from the origin and nearer to the x-axis than to the y-axis. Leave your answer in terms of π.

Not quite. This is the whole quarter circle. Only the part below y = x is nearer to the x-axis.

Solution

  1. Nearer to the x-axis than to the y-axis means below the line y = x.
  2. So the region is the slice between the x-axis and y = x: 45° of a circle of radius 6.
  3. Area = 45360 × π × 6² = 36π ÷ 8 = 4.5π.

Answer: D. 4.5π square units.

Not quite. This is half a circle. The region is a 45° slice.

Solution

  1. Nearer to the x-axis than to the y-axis means below the line y = x.
  2. So the region is the slice between the x-axis and y = x: 45° of a circle of radius 6.
  3. Area = 45360 × π × 6² = 36π ÷ 8 = 4.5π.

Answer: D. 4.5π square units.

Not quite. This uses 2πr, which is a length. Area needs πr².

Solution

  1. Nearer to the x-axis than to the y-axis means below the line y = x.
  2. So the region is the slice between the x-axis and y = x: 45° of a circle of radius 6.
  3. Area = 45360 × π × 6² = 36π ÷ 8 = 4.5π.

Answer: D. 4.5π square units.

Correct.

Solution

  1. Nearer to the x-axis than to the y-axis means below the line y = x.
  2. So the region is the slice between the x-axis and y = x: 45° of a circle of radius 6.
  3. Area = 45360 × π × 6² = 36π ÷ 8 = 4.5π.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A is (2, 1) and B is (6, 5). Which working gives the equation of the perpendicular bisector of AB?

Not quite. This uses the gradient of AB itself. That gives the line AB, not the bisector.

Solution

  1. Midpoint: ((2 + 6) ÷ 2, (1 + 5) ÷ 2) = (4, 3).
  2. Gradient of AB = (5 − 1) ÷ (6 − 2) = 1. A perpendicular line has gradient −1.
  3. y = −x + c through (4, 3): 3 = −4 + c, so c = 7. The bisector is y = −x + 7.

Answer: B. Midpoint (4, 3); AB has gradient 1, so the bisector has gradient −1: y = −x + 7.

Correct.

Solution

  1. Midpoint: ((2 + 6) ÷ 2, (1 + 5) ÷ 2) = (4, 3).
  2. Gradient of AB = (5 − 1) ÷ (6 − 2) = 1. A perpendicular line has gradient −1.
  3. y = −x + c through (4, 3): 3 = −4 + c, so c = 7. The bisector is y = −x + 7.

Not quite. This uses the y-coordinate of the midpoint as the intercept. Substitute (4, 3) into y = −x + c.

Solution

  1. Midpoint: ((2 + 6) ÷ 2, (1 + 5) ÷ 2) = (4, 3).
  2. Gradient of AB = (5 − 1) ÷ (6 − 2) = 1. A perpendicular line has gradient −1.
  3. y = −x + c through (4, 3): 3 = −4 + c, so c = 7. The bisector is y = −x + 7.

Answer: B. Midpoint (4, 3); AB has gradient 1, so the bisector has gradient −1: y = −x + 7.

Not quite. This halves the differences (4 and 4) instead of finding the average of the coordinates.

Solution

  1. Midpoint: ((2 + 6) ÷ 2, (1 + 5) ÷ 2) = (4, 3).
  2. Gradient of AB = (5 − 1) ÷ (6 − 2) = 1. A perpendicular line has gradient −1.
  3. y = −x + c through (4, 3): 3 = −4 + c, so c = 7. The bisector is y = −x + 7.

Answer: B. Midpoint (4, 3); AB has gradient 1, so the bisector has gradient −1: y = −x + 7.

9.The point (a, 3) is exactly 5 units from the point (1, 0), and a is positive. Find a.

Not quite. This is a − 1. Add 1 to get a.

Solution

  1. (a − 1)² + 3² = 5², so (a − 1)² = 25 − 9 = 16.
  2. a − 1 = 4 or a − 1 = −4.
  3. a is positive, so a = 5.

Answer: D. a = 5.

Not quite. This is the negative solution, but a is positive.

Solution

  1. (a − 1)² + 3² = 5², so (a − 1)² = 25 − 9 = 16.
  2. a − 1 = 4 or a − 1 = −4.
  3. a is positive, so a = 5.

Answer: D. a = 5.

Not quite. This forgets the square root: (a − 1)² = 16, so a − 1 = 4.

Solution

  1. (a − 1)² + 3² = 5², so (a − 1)² = 25 − 9 = 16.
  2. a − 1 = 4 or a − 1 = −4.
  3. a is positive, so a = 5.

Answer: D. a = 5.

Correct.

Solution

  1. (a − 1)² + 3² = 5², so (a − 1)² = 25 − 9 = 16.
  2. a − 1 = 4 or a − 1 = −4.
  3. a is positive, so a = 5.

10.A is (0, 0) and B is (5, 0). Which statement about the region of points less than 2 units from A and less than 2 units from B is true?

Correct.

Solution

  1. The two regions are the insides of circles of radius 2 round A and round B.
  2. For them to overlap, 2 + 2 would have to be more than AB = 5.
  3. 4 < 5, so the circles do not meet and the region is empty.

Not quite. Points in that strip are more than 2 units from at least one of A and B.

Solution

  1. The two regions are the insides of circles of radius 2 round A and round B.
  2. For them to overlap, 2 + 2 would have to be more than AB = 5.
  3. 4 < 5, so the circles do not meet and the region is empty.

Answer: A. There are no such points, because 2 + 2 is less than 5.

Not quite. (2.5, 0) is 2.5 units from A and from B, which is more than 2.

Solution

  1. The two regions are the insides of circles of radius 2 round A and round B.
  2. For them to overlap, 2 + 2 would have to be more than AB = 5.
  3. 4 < 5, so the circles do not meet and the region is empty.

Answer: A. There are no such points, because 2 + 2 is less than 5.

Not quite. That is the locus of points equidistant from A and B, a different rule.

Solution

  1. The two regions are the insides of circles of radius 2 round A and round B.
  2. For them to overlap, 2 + 2 would have to be more than AB = 5.
  3. 4 < 5, so the circles do not meet and the region is empty.

Answer: A. There are no such points, because 2 + 2 is less than 5.

Apply

Plan a new shop

On a grid in kilometres, a school S is at (2, 2) and a park K is at (8, 2). A new shop must be nearer to S than to K, and at most 4 km from K. Answer each question, then compare with the model.

1. Find the equation of the locus of points equidistant from S and K.

2. Which part of the line y = 2 is in the region?

3. Is the point (4.5, 4) in the region? Show your working.

  1. SK is horizontal; its midpoint is (5, 2). The perpendicular bisector is x = 5.
  1. Nearer to S: x < 5. At most 4 km from K along y = 2: x from 8 − 4 = 4 to 12.
  2. Both: 4 ≤ x < 5 (x = 5 is not included).
  1. x = 4.5 < 5, so it is nearer to S ✓.
  2. Distance² to K = 3.5² + 2² = 12.25 + 4 = 16.25, which is more than 4² = 16.
  3. No: it is just over 4 km from K.

Next: the Lesson 4 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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