Applications and Mathematical Modelling

Introduction

Geometry is the study of shapes, sizes, positions, distances, angles, and spatial relationships. Many geometric problems involve two or more unknown quantities connected by mathematical rules.

For example, we might need to determine:

• The length and width of a rectangle.

• The angles of a triangle.

• The coordinates where two lines intersect.

• The dimensions of a shape from its perimeter and area.

• The point where a line intersects a circle.

• The coordinates of the vertices of a geometric figure.

• The dimensions of a three-dimensional object.

Systems of equations provide a systematic way to solve these problems by combining different geometric relationships.

In coordinate geometry, equations can represent lines, circles, and other curves. Solving a system allows us to identify points shared by two or more geometric objects.

In measurement problems, equations can describe relationships between lengths, widths, heights, perimeters, areas, and volumes.

By constructing appropriate systems, solving them, and interpreting the results geometrically, we can solve problems that would otherwise be difficult to approach.

Why Are Systems Useful in Geometry?

A single geometric condition often does not provide enough information to determine every unknown.

Suppose a rectangle has a perimeter of 40 cm.

Its length and width satisfy:

2l + 2w = 40

or:

l + w = 20

Many rectangles satisfy this condition.

Examples include:

• Length 12 cm and width 8 cm.

• Length 15 cm and width 5 cm.

• Length 11 cm and width 9 cm.

The perimeter alone does not determine a unique rectangle.

However, suppose we also know that the length is 4 cm greater than the width.

Then:

l − w = 4

We now have a system:

l + w = 20

l − w = 4

Solving this system determines the dimensions uniquely.

This illustrates an important idea:

Each independent geometric condition provides information that can help determine unknown quantities.

Constructing Systems from Geometric Relationships

A reliable approach to geometric modelling involves six steps.

1. Identify the unknown lengths, angles, or coordinates.

2. Define variables and include units where appropriate.

3. Identify relevant geometric properties or formulas.

4. Construct equations from the given conditions.

5. Solve the equations and check all possible solutions.

6. Verify that the results satisfy both the equations and the geometric restrictions.

Useful geometric relationships

Some geometric relationships produce linear equations, while others produce nonlinear equations involving squares or products of variables.

Systems of equations can be used in both situations, although the solving methods may differ.

Application: Finding the Dimensions of a Rectangle

A rectangular garden has a perimeter of 52 m.

Its length is 6 m greater than its width.

Find the dimensions.

Step 1: Define variables

l = length of the garden, in metres

w = width of the garden, in metres

Step 2: Construct the perimeter equation

P = 2l + 2w

52 = 2l + 2w

Divide by 2:

l + w = 26

Step 3: Construct the second equation

The length is 6 m greater than the width.

l − w = 6

Step 4: Solve by elimination

l + w = 26

l − w = 6

Add:

2l = 32

l = 16

Substitute:

16 + w = 26

w = 10

Step 5: Verify

Perimeter:

2(16) + 2(10) = 52 m

Difference:

16 − 10 = 6 m

Both conditions are satisfied.

Answer: The garden is 16 m long and 10 m wide.

Geometric interpretation

The solution determines a unique rectangle satisfying both the perimeter and dimension conditions.

Its area is:

A = lw

A = 16(10)

A = 160 m²

The system provides dimensions that can be used for further calculations involving fencing, landscaping, or construction materials.

Application: Rectangle Dimensions from Area and Perimeter

Not all geometric systems are linear.

Suppose a rectangle has:

• Perimeter = 34 cm.

• Area = 60 cm².

Find its dimensions.

Step 1: Define variables

l = length, in cm

w = width, in cm

Step 2: Construct equations

Perimeter:

2l + 2w = 34

Simplify:

l + w = 17

Area:

lw = 60

Step 3: Substitute

From the perimeter equation:

l = 17 − w

Substitute into the area equation:

(17 − w)w = 60

17w − w² = 60

Rearrange:

w² − 17w + 60 = 0

Step 4: Factor

(w − 5)(w − 12) = 0

Therefore:

w = 5 or w = 12

If w = 5:

l = 12

If w = 12:

l = 5

These describe the same rectangle with its dimensions interchanged.

Answer: The rectangle measures 12 cm by 5 cm.

Verify

Perimeter:

2(12) + 2(5) = 34 cm

Area:

12(5) = 60 cm²

Important observation

The system contains a linear equation and a nonlinear equation.

Substitution transforms the system into a quadratic equation.

This demonstrates how systems of equations connect algebra and geometry.

Application: Finding Unknown Angles

Geometric angle relationships can be represented using systems of equations.

Worked Example: Complementary Angles

Two angles are complementary.

The larger angle is 18° greater than the smaller angle.

Find both angles.

Step 1: Define variables

x = larger angle, in degrees

y = smaller angle, in degrees

Step 2: Construct equations

Complementary angles add to 90°.

x + y = 90

The larger angle exceeds the smaller by 18°.

x − y = 18

Step 3: Solve

Add the equations:

2x = 108

x = 54

Then:

y = 90 − 54

y = 36

Step 4: Verify

54° + 36° = 90°

54° − 36° = 18°

Answer: The angles are 54° and 36°.

Application: Angles in a Triangle

The three interior angles of a triangle add to 180°.

Suppose:

• The second angle is 20° greater than the first.

• The third angle is twice the first.

Find all three angles.

Step 1: Define variables

x = first angle

y = second angle

z = third angle

Step 2: Construct equations

Angle sum:

x + y + z = 180

Second angle relationship:

y = x + 20

Third angle relationship:

z = 2x

Step 3: Substitute

x + (x + 20) + 2x = 180

4x + 20 = 180

4x = 160

x = 40

Then:

y = 40 + 20 = 60

z = 2(40) = 80

Step 4: Verify

40° + 60° + 80° = 180°

60° − 40° = 20°

80° = 2(40°)

Answer: The triangle's angles are 40°, 60°, and 80°.

Geometric interpretation

All three angles are less than 90°, so the triangle is acute.

This example shows how a three-variable system can be solved efficiently using relationships between the variables.

Coordinate Geometry: Intersections of Straight Lines

In coordinate geometry, a straight line can be represented by an equation such as:

y = mx + b

where:

• m = gradient or slope.

• b = y-intercept.

• x and y = coordinates of points on the line.

When two lines intersect, the intersection point satisfies both equations simultaneously.

Therefore, finding the intersection of two lines is equivalent to solving a system of equations.

Worked Example: Two Intersecting Lines

Find the intersection of:

y = 2x + 1

y = −x + 7

Step 1: Set the expressions equal

At the intersection, both equations have the same x-value and y-value.

2x + 1 = −x + 7

Step 2: Solve for x

3x = 6

x = 2

Step 3: Solve for y

y = 2(2) + 1

y = 5

Intersection: (2, 5).

Step 4: Verify

First equation:

5 = 2(2) + 1

Second equation:

5 = −2 + 7

Both are correct.

Geometric interpretation

The ordered pair (2, 5) represents a point that lies on both lines.

The x-coordinate is 2, and the y-coordinate is 5.

Because the lines have different gradients, they intersect at exactly one point.

Parallel and Coincident Lines

Not every pair of lines has one intersection.

Consider three possibilities.

Example: Parallel Lines

y = 3x + 2

y = 3x − 4

Setting the expressions equal:

3x + 2 = 3x − 4

2 = −4

This is impossible.

Conclusion: No solution.

Example: Coincident Lines

2x + 4y = 12

x + 2y = 6

Multiplying the second equation by 2 produces the first equation.

The equations represent the same line.

Conclusion: Infinitely many solutions.

Every point on that line satisfies both equations.

Application: Finding the Vertices of a Triangle

A triangle is formed by three lines:

Line A: y = x + 1

Line B: y = −x + 7

Line C: y = 2

Find the three vertices of the triangle.

Step 1: Find the intersection of A and B

x + 1 = −x + 7

2x = 6

x = 3

y = 4

First vertex:

P = (3, 4)

Step 2: Find the intersection of A and C

2 = x + 1

x = 1

Second vertex:

Q = (1, 2)

Step 3: Find the intersection of B and C

2 = −x + 7

x = 5

Third vertex:

R = (5, 2)

Step 4: Interpret geometrically

The vertices are:

P = (3, 4)

Q = (1, 2)

R = (5, 2)

Step 5: Calculate the area

The base QR lies horizontally along y = 2.

Its length is:

5 − 1 = 4 units

The vertical height from P to the base is:

4 − 2 = 2 units

Area of a triangle:

A = ½bh

A = ½(4)(2)

A = 4 square units

Verification

Check each vertex in its two corresponding line equations.

For P = (3, 4):

4 = 3 + 1

4 = −3 + 7

For Q = (1, 2):

2 = 1 + 1

2 = 2

For R = (5, 2):

2 = −5 + 7

2 = 2

All three vertices are correct.

Application: Finding the Intersection of a Line and a Circle

Systems can also contain nonlinear equations.

The equation of a circle with centre (h, k) and radius r is:

(x − h)² + (y − k)² = r²

Suppose we want to find where the line:

y = x + 1

intersects the circle:

x² + y² = 25

Step 1: Substitute the line equation

Replace y with x + 1:

x² + (x + 1)² = 25

Step 2: Expand

x² + x² + 2x + 1 = 25

2x² + 2x − 24 = 0

Divide by 2:

x² + x − 12 = 0

Step 3: Factor

(x + 4)(x − 3) = 0

Therefore:

x = −4 or x = 3

Step 4: Find the corresponding y-values

If x = −4:

y = −4 + 1 = −3

If x = 3:

y = 3 + 1 = 4

Intersection points: (−4, −3) and (3, 4).

Step 5: Verify

For (−4, −3):

Circle:

(−4)² + (−3)² = 16 + 9 = 25

Line:

−3 = −4 + 1

For (3, 4):

Circle:

3² + 4² = 9 + 16 = 25

Line:

4 = 3 + 1

Both points satisfy both equations.

How many intersections are possible?

A straight line and a circle can have:

• Two intersection points.

• One intersection point, when the line is tangent.

• No real intersection points.

The number of real solutions of the resulting quadratic equation determines the number of intersections.

Using the discriminant

For:

ax² + bx + c = 0

the discriminant is:

D = b² − 4ac

If:

• D is positive: two distinct real intersections.

• D = 0: one tangent intersection.

• D is negative: no real intersections.

This provides a connection between quadratic equations and coordinate geometry.

Application: Finding an Unknown Point Using the Midpoint

The midpoint of a line segment lies halfway between its endpoints.

For endpoints A(x₁, y₁) and B(x₂, y₂), the midpoint is:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Suppose a line segment has:

A = (2, 4)

M = (5, 7)

Find the coordinates of B.

Step 1: Define the unknown endpoint

B = (x, y)

Step 2: Construct equations

For the x-coordinate:

(2 + x)/2 = 5

For the y-coordinate:

(4 + y)/2 = 7

Step 3: Solve

2 + x = 10

x = 8

4 + y = 14

y = 10

Answer: B = (8, 10).

Verify

Midpoint of (2, 4) and (8, 10):

M = ((2 + 8)/2, (4 + 10)/2)

M = (5, 7)

The calculated endpoint is correct.

Geometric interpretation

The midpoint is exactly halfway between the endpoints in both coordinate directions.

The two coordinate equations can be solved independently because each involves a different unknown.

Application: Perpendicular Lines and Intersection Points

Two nonvertical lines are perpendicular when their gradients multiply to −1.

m₁m₂ = −1

Suppose a line passes through (2, 3) and has gradient 2.

Find the equation of the perpendicular line passing through (4, 1), and determine where the two lines intersect.

Step 1: Find the first line

Using point-slope form:

y − 3 = 2(x − 2)

y = 2x − 1

Step 2: Find the perpendicular gradient

The negative reciprocal of 2 is:

m = −1/2

Step 3: Find the second line

Through (4, 1):

y − 1 = −½(x − 4)

y = −½x + 3

Step 4: Solve the system

2x − 1 = −½x + 3

2.5x = 4

x = 8/5

Step 5: Find y

y = 2(8/5) − 1

y = 16/5 − 5/5

y = 11/5

Intersection: (8/5, 11/5).

Verify

First line:

2(8/5) − 1 = 11/5

Second line:

−½(8/5) + 3 = −4/5 + 15/5 = 11/5

Gradient product:

2(−1/2) = −1

Both the intersection and perpendicularity conditions are satisfied.

Geometric interpretation

The two lines meet at a right angle.

This method is useful when constructing perpendicular paths, locating perpendicular projections, or analysing geometric figures.

Application: Dimensions of a Rectangular Prism

Three-dimensional geometry can involve three unknown dimensions.

A rectangular storage box has:

• Length + width + height = 24 cm.

• Length is twice the width.

• Height is 3 cm greater than the width.

Find the dimensions and volume.

Step 1: Define variables

l = length, in cm

w = width, in cm

h = height, in cm

Step 2: Construct equations

l + w + h = 24

l = 2w

h = w + 3

Step 3: Substitute

2w + w + (w + 3) = 24

4w + 3 = 24

4w = 21

w = 21/4 = 5.25 cm

Step 4: Find the remaining dimensions

l = 2(5.25) = 10.5 cm

h = 5.25 + 3 = 8.25 cm

Step 5: Verify

10.5 + 5.25 + 8.25 = 24 cm

10.5 = 2(5.25)

8.25 = 5.25 + 3

All conditions are satisfied.

Step 6: Calculate volume

V = lwh

V = 10.5 × 5.25 × 8.25

V = 454.78125 cm³

Interpretation

The box measures:

• Length: 10.5 cm.

• Width: 5.25 cm.

• Height: 8.25 cm.

Its volume is approximately 454.78 cm³.

This type of modelling is useful in packaging, storage design, and construction.

Applications in Design, Architecture, and Engineering

Systems of equations are used extensively when geometric conditions must be satisfied simultaneously.

Example: Locating Two Paths

Two straight paths are modelled by:

Path A: y = 0.5x + 2

Path B: y = −x + 8

Their intersection is found by solving:

0.5x + 2 = −x + 8

1.5x = 6

x = 4

Then:

y = 0.5(4) + 2

y = 4

The paths intersect at (4, 4).

If the coordinates are measured in metres from a defined origin, the intersection is 4 m east and 4 m north of that origin, assuming positive x points east and positive y points north.

In a real design, the coordinate system and measurement units must be clearly defined.

Using Technology to Verify Geometric Solutions

Graphing calculators and mathematical software can help visualise geometric systems.

For example, enter:

y = 2x + 1

y = −x + 7

into Desmos Graphing Calculator.

The graph shows the lines intersecting at (2, 5).

For a line-circle system, enter:

x² + y² = 25

y = x + 1

The graph displays two intersection points.

Benefits of technology

Technology can:

• Display geometric relationships visually.

• Estimate intersection coordinates.

• Check algebraic solutions.

• Investigate the effect of changing parameters.

• Help identify the number of intersections.

Limitations

Graphing technology may display approximate values.

A poorly chosen graphing window may hide intersections.

Some intersections may be difficult to identify accurately from the graph alone.

Therefore, algebraic verification remains important.

Evaluating Geometric Solutions

A solution may satisfy an equation but still be inappropriate for the geometric situation.

Positive dimensions

Lengths, widths, heights, radii, and areas cannot be negative in ordinary geometric measurement.

For example, a quadratic equation might produce:

w = 5 or w = −3

If w represents a physical width, the negative solution must be rejected.

Angle restrictions

Interior angles of a nondegenerate triangle must be greater than 0° and less than 180°.

Three proposed angles must also add to 180°.

Coordinate order

The point (3, 5) is different from (5, 3).

Coordinate order must be preserved.

Extraneous solutions

Squaring equations or performing other non-equivalent transformations may introduce solutions that do not satisfy the original system.

Every candidate solution should be checked.

Geometric assumptions

A drawing may not be to scale.

Do not assume two lines are perpendicular or two lengths are equal unless this is given or established mathematically.

Units

Lengths are measured in units such as cm or m.

Areas are measured in square units such as cm² or m².

Volumes are measured in cubic units such as cm³ or m³.

Using incorrect units can make an otherwise correct numerical result incomplete.

Common Mistakes

Using the wrong perimeter formula

For a rectangle:

P = 2l + 2w

not l + w.

Confusing area and perimeter

Area measures the region enclosed by a shape, while perimeter measures the distance around it.

Forgetting angle-sum relationships

Triangle angles add to 180°, while quadrilateral angles add to 360°.

Reversing coordinate values

The point (x, y) must be written in the correct order.

Assuming every system is linear

Area and circle equations often produce nonlinear systems.

Ignoring multiple solutions

A line may intersect a circle at two points.

Both points must be included when valid.

Accepting negative dimensions

Negative lengths or widths are not valid physical dimensions.

Assuming a diagram is drawn to scale

Only use relationships given or justified by geometry.

Misinterpreting parallel lines

Parallel lines with different intercepts have no intersection.

Forgetting to verify intersections

A coordinate must satisfy both equations to be an intersection point.

Using incorrect units

Lengths, areas, and volumes require different types of units.

Relying only on graphing technology

Graphical answers may be approximate or incomplete.

Key Terms

• System of equations: Two or more equations solved simultaneously.

• Coordinate geometry: The study of geometric figures using coordinates and equations.

• Cartesian plane: A coordinate plane with perpendicular x- and y-axes.

• Ordered pair: A pair of coordinates written as (x, y).

• Intersection point: A point shared by two or more geometric objects.

• Gradient: The slope of a straight line.

• Y-intercept: The point where a line crosses the y-axis.

• Parallel lines: Coplanar lines that never intersect.

• Perpendicular lines: Lines that meet at a right angle.

• Coincident lines: Lines that occupy exactly the same position.

• Perimeter: The total distance around a shape.

• Area: The amount of two-dimensional space enclosed by a shape.

• Volume: The amount of three-dimensional space occupied by an object.

• Complementary angles: Two angles whose sum is 90°.

• Supplementary angles: Two angles whose sum is 180°.

• Midpoint: The point halfway between two endpoints.

• Circle: The set of points at a fixed distance from a centre.

• Radius: The distance from the centre of a circle to its boundary.

• Tangent: A line that touches a circle at exactly one point.

• Discriminant: The expression b² − 4ac, which helps determine the number of real roots of a quadratic equation.

• Nonlinear system: A system containing at least one nonlinear equation.

• Extraneous solution: A solution introduced during algebraic manipulation that does not satisfy the original problem.

• Geometric constraint: A condition restricting possible dimensions, angles, or positions.

• Verification: Checking that a solution satisfies the original equations and geometric conditions.

Key Takeaways

• Systems of equations can model lengths, widths, angles, coordinates, and dimensions.

• Geometric properties provide relationships that can be expressed as equations.

• Rectangle problems may involve perimeter, area, and dimension relationships.

• Angle problems can use triangle angle sums and relationships between angles.

• The intersection of two lines is a solution common to both line equations.

• Parallel lines have no intersection, while coincident lines have infinitely many shared points.

• Triangle vertices can be found by solving pairs of line equations.

• A line and a circle may have zero, one, or two real intersection points.

• Midpoint relationships can be used to find unknown coordinates.

• Three-variable systems can determine the dimensions of three-dimensional objects.

• Geometric solutions must satisfy relevant restrictions and use correct units.

• Verification is essential for confirming that calculated values represent valid geometric results.

Check Your Understanding

Recall and Understanding

1. Explain how a system of equations can be used to determine the dimensions of a rectangle.

2. What does the solution to a system of two straight-line equations represent geometrically?

3. Explain the difference between intersecting, parallel, and coincident lines.

4. Why might a line-circle system have two solutions?

5. Why must negative solutions sometimes be rejected in geometric problems?

Rectangle and Angle Problems

6. A rectangle has a perimeter of 48 cm.

Its length is 8 cm greater than its width.

Find its dimensions and area.

7. A rectangle has a perimeter of 30 cm and an area of 54 cm².

Find its dimensions.

8. Two complementary angles differ by 24°.

Find both angles.

9. The second angle of a triangle is 15° greater than the first.

The third angle is twice the first.

Find all three angles.

10. Three angles of a quadrilateral are x, x + 20°, and 2x. The fourth angle is 100°.

Find all four angles.

Coordinate Geometry

11. Find the intersection of:

y = 3x − 2

y = −x + 10

12. Determine the number of solutions:

y = 4x + 3

y = 4x − 7

13. Determine the number of solutions:

2x + 6y = 18

x + 3y = 9

14. Find the vertices of the triangle formed by:

y = x

y = −x + 6

y = 1

15. Find the intersection points of:

x² + y² = 25

y = 3

Applications and Challenge

16. A line segment has endpoint A = (3, 5) and midpoint M = (7, 9).

Find the other endpoint.

17. A line has gradient 3 and passes through (1, 2).

Find the equation of the perpendicular line passing through (4, 5), then determine the intersection of the two lines.

18. A rectangular prism has length, width, and height whose sum is 30 cm.

The length is twice the width.

The height is 6 cm greater than the width.

Find its dimensions and volume.

19. Find the intersection points of:

y = x + 2

x² + y² = 20

20. A rectangular garden has a perimeter of 60 m and an area of 200 m².

Find its dimensions and explain how you verify the result.

Answers to Check Your Understanding

1. Rectangle dimensions

Define variables for length and width.

Use geometric information, such as perimeter and a relationship between the dimensions, to construct two equations.

Solve the system and verify that the dimensions satisfy the original conditions.

2. Intersection of two lines

The solution represents the coordinates of a point lying on both lines.

For example, (3, 5) means both lines pass through x = 3, y = 5.

3. Types of lines

Intersecting lines cross at one point and have one solution.

Distinct parallel lines never meet and have no solution.

Coincident lines represent the same line and have infinitely many solutions.

4. Line-circle intersections

A line can pass through a circle at two different points.

Both intersection points satisfy the line equation and circle equation.

The resulting quadratic equation may therefore have two distinct real solutions.

5. Negative geometric values

Physical dimensions such as lengths, widths, and radii cannot be negative.

Therefore, a negative algebraic solution must be rejected when it represents an invalid dimension.

Negative coordinates, however, can be perfectly valid.

6. Rectangle perimeter

Let:

l = length

w = width

Perimeter:

2l + 2w = 48

l + w = 24

Difference:

l − w = 8

Add:

2l = 32

l = 16

Then:

w = 8

Area:

A = 16(8) = 128 cm²

Answer: Length = 16 cm, width = 8 cm, area = 128 cm².

7. Rectangle area and perimeter

Perimeter:

2l + 2w = 30

l + w = 15

Area:

lw = 54

Substitute l = 15 − w:

(15 − w)w = 54

15w − w² = 54

w² − 15w + 54 = 0

(w − 6)(w − 9) = 0

Therefore:

w = 6 or w = 9

Answer: The rectangle measures 9 cm by 6 cm.

Verification:

2(9) + 2(6) = 30 cm

9(6) = 54 cm²

8. Complementary angles

Let:

x = larger angle

y = smaller angle

Equations:

x + y = 90

x − y = 24

Add:

2x = 114

x = 57

Then:

y = 33

Answer: 57° and 33°.

9. Triangle angles

Let the first angle be x.

Second angle:

x + 15

Third angle:

2x

Angle sum:

x + (x + 15) + 2x = 180

4x + 15 = 180

4x = 165

x = 41.25

Second angle:

56.25°

Third angle:

82.5°

Answer: 41.25°, 56.25°, and 82.5°.

Verification:

41.25 + 56.25 + 82.5 = 180°

10. Quadrilateral angles

Angle sum:

x + (x + 20) + 2x + 100 = 360

4x + 120 = 360

4x = 240

x = 60

Therefore:

First angle = 60°

Second angle = 80°

Third angle = 120°

Fourth angle = 100°

Answer: 60°, 80°, 120°, and 100°.

Verification:

60 + 80 + 120 + 100 = 360°

11. Intersection of two lines

Equations:

y = 3x − 2

y = −x + 10

Set equal:

3x − 2 = −x + 10

4x = 12

x = 3

Then:

y = 3(3) − 2 = 7

Answer: (3, 7).

Verification:

7 = −3 + 10

12. Parallel lines

Equations:

y = 4x + 3

y = 4x − 7

Both lines have gradient 4 but different y-intercepts.

Answer: No solution.

13. Coincident lines

Equations:

2x + 6y = 18

x + 3y = 9

Multiplying the second equation by 2 gives the first.

Answer: Infinitely many solutions.

14. Triangle vertices

Lines:

y = x

y = −x + 6

y = 1

First intersection:

x = −x + 6

2x = 6

x = 3

y = 3

First vertex: (3, 3)

Second intersection:

1 = x

Second vertex: (1, 1)

Third intersection:

1 = −x + 6

x = 5

Third vertex: (5, 1)

Answer: (3, 3), (1, 1), and (5, 1).

15. Line-circle intersections

Circle:

x² + y² = 25

Line:

y = 3

Substitute:

x² + 3² = 25

x² + 9 = 25

x² = 16

x = ±4

Answer: (−4, 3) and (4, 3).

Verification:

16 + 9 = 25 for both points.

16. Unknown endpoint

Given:

A = (3, 5)

M = (7, 9)

Let the other endpoint be B = (x, y).

Midpoint equations:

(3 + x)/2 = 7

(5 + y)/2 = 9

Solve:

3 + x = 14

x = 11

5 + y = 18

y = 13

Answer: B = (11, 13).

17. Perpendicular lines

The first line has gradient 3 and passes through (1, 2).

Using point-slope form:

y − 2 = 3(x − 1)

y = 3x − 1

The perpendicular gradient is:

−1/3

Through (4, 5):

y − 5 = −⅓(x − 4)

y = −x/3 + 19/3

Find the intersection:

3x − 1 = −x/3 + 19/3

Multiply by 3:

9x − 3 = −x + 19

10x = 22

x = 11/5

Substitute:

y = 3(11/5) − 1

y = 28/5

Answer: The perpendicular line is y = −x/3 + 19/3, and the intersection is (11/5, 28/5).

Verification:

3(−1/3) = −1

The gradients confirm that the lines are perpendicular.

18. Rectangular prism

Let:

l = length

w = width

h = height

Equations:

l + w + h = 30

l = 2w

h = w + 6

Substitute:

2w + w + (w + 6) = 30

4w + 6 = 30

4w = 24

w = 6

Then:

l = 12

h = 12

Volume:

V = lwh

V = 12(6)(12)

V = 864 cm³

Answer: Length = 12 cm, width = 6 cm, height = 12 cm, volume = 864 cm³.

19. Line-circle intersections

Given:

y = x + 2

x² + y² = 20

Substitute:

x² + (x + 2)² = 20

Expand:

x² + x² + 4x + 4 = 20

2x² + 4x − 16 = 0

Divide by 2:

x² + 2x − 8 = 0

Factor:

(x + 4)(x − 2) = 0

Therefore:

x = −4 or x = 2

If x = −4:

y = −2

If x = 2:

y = 4

Answer: (−4, −2) and (2, 4).

Verification:

For (−4, −2):

16 + 4 = 20

For (2, 4):

4 + 16 = 20

Both points also satisfy y = x + 2.

20. Rectangular garden

Given:

Perimeter = 60 m

Area = 200 m²

Let:

l = length

w = width

Perimeter:

2l + 2w = 60

l + w = 30

Area:

lw = 200

Substitute:

l = 30 − w

(30 − w)w = 200

30w − w² = 200

w² − 30w + 200 = 0

Factor:

(w − 10)(w − 20) = 0

Therefore:

w = 10 or w = 20

The dimensions are interchangeable.

Answer: The garden measures 20 m by 10 m.

Verification:

Perimeter:

2(20) + 2(10) = 60 m

Area:

20(10) = 200 m²

Both original conditions are satisfied.

Final Summary

Systems of equations provide a powerful connection between algebra and geometry. They allow us to determine unknown dimensions, calculate angles, find intersections, locate vertices, and analyse geometric figures in two and three dimensions.

Some geometric systems are linear, while others involve quadratic equations or additional geometric relationships.

The most important part of solving these problems is constructing equations that accurately represent the geometric conditions.

Once a solution has been obtained, it should be verified using the original equations, relevant geometric properties, and appropriate units.

Successful geometric problem-solving combines algebraic accuracy, visual understanding, and careful interpretation of mathematical results.