IB Mathematics AI – Topic 2

1. Gradient-Intercept Form (Slope-Intercept):

\[ y = mx + c \]

where:

• m: gradient (slope) of the line
• c: y-intercept (where line crosses y-axis)

When to use: When you know the gradient and y-intercept, or want to identify them quickly

2. Point-Gradient Form (Point-Slope):

\[ y - y_1 = m(x - x_1) \]

where:

• m: gradient of the line
• \((x_1, y_1)\): a known point on the line

When to use: When you know one point and the gradient

3. General Form (Standard Form):

\[ ax + by + d = 0 \]

where a, b, d are constants

When to use: For certain algebraic manipulations or when coefficients need to be integers

Converting Between Forms:

• From point-gradient to gradient-intercept: Expand and simplify
• From gradient-intercept to general: Rearrange to get all terms on one side
• From general to gradient-intercept: Solve for y

⚠️ Common Pitfalls & Tips:

• In \(y = mx + c\), c is the y-intercept, not the x-intercept
• The gradient m can be positive, negative, or zero
• Horizontal lines have m = 0 (e.g., y = 3)
• Vertical lines cannot be written as y = mx + c (they're x = k)

Gradient Formula:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x} = \frac{\text{rise}}{\text{run}} \]

Given two points \((x_1, y_1)\) and \((x_2, y_2)\)

Interpreting Gradient:

• Positive gradient (m > 0): Line slopes upward from left to right
• Negative gradient (m < 0): Line slopes downward from left to right
• Zero gradient (m = 0): Horizontal line
• Undefined gradient: Vertical line (division by zero)

Practical Meaning:

If m = 2, the line rises 2 units for every 1 unit it moves to the right

If m = -0.5, the line falls 0.5 units for every 1 unit it moves to the right

⚠️ Common Pitfalls & Tips:

• Consistency matters: Use same order for both numerator and denominator
• Gradient is the same between ANY two points on the line
• Don't confuse rise/run: rise is vertical (y), run is horizontal (x)
• Use GDC to verify gradient from graph or equation

y-intercept:

The point where the line crosses the y-axis

Coordinates: \((0, c)\)

To find: Set x = 0 in the equation

In \(y = mx + c\), the y-intercept is simply c

x-intercept:

The point where the line crosses the x-axis

Coordinates: \((x, 0)\)

To find: Set y = 0 in the equation and solve for x

\[ 0 = mx + c \implies x = -\frac{c}{m} \]

Special Cases:

• Horizontal line \(y = k\): y-intercept is k, no x-intercept (unless k = 0)
• Vertical line \(x = k\): x-intercept is k, no y-intercept (unless k = 0)
• Line through origin: Both intercepts are (0, 0)

⚠️ Common Pitfalls & Tips:

• y-intercept: set x = 0; x-intercept: set y = 0 (don't confuse these)
• Intercepts are POINTS, not just numbers: write as (x, y)
• Not all lines have both intercepts
• Use GDC to find intercepts graphically

Solution:

Step 1: Find the gradient

Using \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

\[ m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2 \]

Step 2: Use point-gradient form

Using point A(2, 5) and m = 2:

\[ y - 5 = 2(x - 2) \]

Step 3: Expand and simplify

\[ y - 5 = 2x - 4 \]

\[ y = 2x + 1 \]

Verification:

Check point A(2, 5): \(y = 2(2) + 1 = 5\) ✓

Check point B(6, 13): \(y = 2(6) + 1 = 13\) ✓

Answer: y = 2x + 1

Parallel Lines:

Two lines are parallel if they have the SAME gradient and never intersect

\[ \text{If } L_1 \parallel L_2, \text{ then } m_1 = m_2 \]

Example: \(y = 3x + 2\) and \(y = 3x - 5\) are parallel (both have m = 3)

Perpendicular Lines:

Two lines are perpendicular if they meet at a 90° angle

Their gradients multiply to give -1:

\[ \text{If } L_1 \perp L_2, \text{ then } m_1 \times m_2 = -1 \]

\[ \text{Or: } m_2 = -\frac{1}{m_1} \]

Example: \(y = 2x + 1\) and \(y = -\frac{1}{2}x + 3\) are perpendicular

Check: \(2 \times (-\frac{1}{2}) = -1\) ✓

Finding Perpendicular Gradient:

1. Take the negative reciprocal of the original gradient
2. Flip the fraction and change the sign

Examples:

• If \(m_1 = 3\), then \(m_2 = -\frac{1}{3}\)
• If \(m_1 = -\frac{2}{5}\), then \(m_2 = \frac{5}{2}\)
• If \(m_1 = \frac{1}{4}\), then \(m_2 = -4\)

⚠️ Common Pitfalls & Tips:

• Parallel: Same gradient, different intercepts
• Perpendicular: Negative reciprocal gradient
• Don't forget to change BOTH the sign AND flip the fraction
• Horizontal and vertical lines are always perpendicular

Solution:

Step 1: Find gradient of original line

From \(y = 4x - 3\), we have \(m_1 = 4\)

Step 2: Find perpendicular gradient

Perpendicular gradient is negative reciprocal:

\[ m_2 = -\frac{1}{m_1} = -\frac{1}{4} \]

Step 3: Use point-gradient form

Using point P(2, 5) and \(m_2 = -\frac{1}{4}\):

\[ y - 5 = -\frac{1}{4}(x - 2) \]

Step 4: Simplify

\[ y - 5 = -\frac{1}{4}x + \frac{1}{2} \]

\[ y = -\frac{1}{4}x + \frac{1}{2} + 5 \]

\[ y = -\frac{1}{4}x + \frac{11}{2} \]

Verification:

Check gradients: \(4 \times (-\frac{1}{4}) = -1\) ✓

Check point: \(y = -\frac{1}{4}(2) + \frac{11}{2} = -\frac{1}{2} + \frac{11}{2} = 5\) ✓

Answer: \(y = -\frac{1}{4}x + \frac{11}{2}\) or \(y = -0.25x + 5.5\)

Steps to Find Perpendicular Bisector:

Step 1: Find the midpoint

For points \((x_1, y_1)\) and \((x_2, y_2)\):

\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]

Step 2: Find gradient of original segment

\[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} \]

Step 3: Find perpendicular gradient

\[ m_{\perp} = -\frac{1}{m_1} \]

Step 4: Write equation

Use point-gradient form with midpoint M and gradient \(m_{\perp}\):

\[ y - y_M = m_{\perp}(x - x_M) \]

Properties:

• Every point on the perpendicular bisector is equidistant from both endpoints
• The perpendicular bisector divides the segment into two equal parts
• Used in geometry, construction, and locus problems

⚠️ Common Pitfalls & Tips:

• Don't forget to find the MIDPOINT first
• Must be PERPENDICULAR (negative reciprocal gradient)
• Use midpoint coordinates in point-gradient form
• Check your answer passes through the midpoint

Solution:

Step 1: Find midpoint M

\[ M = \left(\frac{1 + 7}{2}, \frac{2 + 8}{2}\right) = \left(\frac{8}{2}, \frac{10}{2}\right) = (4, 5) \]

Step 2: Find gradient of AB

\[ m_{AB} = \frac{8 - 2}{7 - 1} = \frac{6}{6} = 1 \]

Step 3: Find perpendicular gradient

\[ m_{\perp} = -\frac{1}{1} = -1 \]

Step 4: Write equation using M(4, 5)

\[ y - 5 = -1(x - 4) \]

\[ y - 5 = -x + 4 \]

\[ y = -x + 9 \]

Verification:

Passes through M(4, 5): \(y = -4 + 9 = 5\) ✓

Perpendicular: \(1 \times (-1) = -1\) ✓

Answer: y = -x + 9

Forms of Lines

• \(y = mx + c\) (gradient-intercept)
• \(y - y_1 = m(x - x_1)\) (point-gradient)
• \(ax + by + d = 0\) (general)

Gradient

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
• Positive: slopes up
• Negative: slopes down

Parallel & Perpendicular

• Parallel: \(m_1 = m_2\)
• Perpendicular: \(m_1 \times m_2 = -1\)
• \(m_{\perp} = -\frac{1}{m}\)

Perpendicular Bisector

• Find midpoint
• Find perpendicular gradient
• Use point-gradient form