Worked Examples
Comprehensive Worked Examples
This page provides detailed, step-by-step solutions to help students understand the problem-solving process.
Example Set 1: Distance and Midpoint
Problem 1: Distance Between Points
Find the distance between points A(3, 7) and B(15, 12).
Solution:
Given: \(A(3, 7)\) and \(B(15, 12)\)
Using the distance formula: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
Substitute the coordinates: \[d = \sqrt{(15 - 3)^2 + (12 - 7)^2}\]
Simplify inside the square root: \[d = \sqrt{12^2 + 5^2}\] \[d = \sqrt{144 + 25}\] \[d = \sqrt{169}\] \[d = 13\]
Answer: The distance is 13 units.
Problem 2: Finding a Missing Coordinate
Point M is the midpoint of line segment AB, where A(-2, 5) and M(3, 4). Find the coordinates of point B.
Solution:
Given: \(A(-2, 5)\), \(M(3, 4)\), and B\((x_2, y_2)\) is unknown
Using the midpoint formula: \[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\]
For the x-coordinate: \[3 = \frac{-2 + x_2}{2}\] \[6 = -2 + x_2\] \[x_2 = 8\]
For the y-coordinate: \[4 = \frac{5 + y_2}{2}\] \[8 = 5 + y_2\] \[y_2 = 3\]
Answer: Point B is at (8, 3).
Example Set 2: Lines and Gradients
Problem 3: Finding the Equation of a Line
Find the equation of the line that passes through points P(2, 5) and Q(8, 17).
Solution:
Step 1: Calculate the gradient \[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{17 - 5}{8 - 2} = \frac{12}{6} = 2\]
Step 2: Use point-slope form with point P(2, 5) \[y - y_1 = m(x - x_1)\] \[y - 5 = 2(x - 2)\]
Step 3: Expand and simplify \[y - 5 = 2x - 4\] \[y = 2x + 1\]
Answer: The equation is \(y = 2x + 1\).
Verification: Check both points satisfy the equation: - For P(2, 5): \(5 = 2(2) + 1 = 5\) ✓ - For Q(8, 17): \(17 = 2(8) + 1 = 17\) ✓
Problem 4: Intersection of Lines
Find the point of intersection of lines \(L_1: y = 2x + 1\) and \(L_2: y = -x + 7\).
Solution:
At the intersection point, both equations are satisfied, so: \[2x + 1 = -x + 7\]
Solve for x: \[2x + x = 7 - 1\] \[3x = 6\] \[x = 2\]
Substitute into \(L_1\) to find y: \[y = 2(2) + 1 = 5\]
Answer: The lines intersect at point (2, 5).
Verification: Check with \(L_2\): \(y = -(2) + 7 = 5\) ✓
Example Set 3: Parallel and Perpendicular Lines
Problem 5: Perpendicular Line Through a Point
Line \(L\) has equation \(y = 4x - 3\). Find the equation of the line perpendicular to \(L\) that passes through the point (8, 5).
Solution:
Step 1: Identify the gradient of \(L\) \[m_L = 4\]
Step 2: Find the gradient of the perpendicular line \[m_{\perp} = -\frac{1}{m_L} = -\frac{1}{4}\]
Step 3: Use point-slope form with (8, 5) \[y - 5 = -\frac{1}{4}(x - 8)\] \[y - 5 = -\frac{1}{4}x + 2\] \[y = -\frac{1}{4}x + 7\]
Answer: The equation is \(y = -\frac{1}{4}x + 7\).
Check: Verify the product of gradients equals -1: \[m_L \times m_{\perp} = 4 \times \left(-\frac{1}{4}\right) = -1\] ✓
Problem 6: Geometric Problem
Points A(2, 3), B(6, 1), and C(4, 5) form a triangle. Show that angle ABC is a right angle.
Solution:
To show angle ABC is a right angle, we need to prove that lines BA and BC are perpendicular.
Step 1: Find the gradient of BA \[m_{BA} = \frac{3 - 1}{2 - 6} = \frac{2}{-4} = -\frac{1}{2}\]
Step 2: Find the gradient of BC \[m_{BC} = \frac{5 - 1}{4 - 6} = \frac{4}{-2} = -2\]
Step 3: Check if lines are perpendicular \[m_{BA} \times m_{BC} = -\frac{1}{2} \times (-2) = 1\]
Since the product of the gradients is 1 (not -1), the lines are NOT perpendicular.
Answer: Angle ABC is not a right angle, as the product of gradients is 1, not -1.
Tips for Success
- Always show your working - partial credit is often awarded
- Check your answers - substitute back into original equations
- Draw diagrams - visual representations help understanding
- Use appropriate formulas - know when to apply each formula
- Be careful with signs - negative numbers need special attention
Practice Strategy
- Work through these examples without looking at the solutions
- Compare your method with the shown solution
- If you made errors, identify where and why
- Try creating similar problems and solving them