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IB Mathematics AI SL Equations of perpendicular bisectors MAI Study Notes - New Syllabus

IB Mathematics AI SL Equations of perpendicular bisectors MAI Study Notes

LEARNING OBJECTIVE

  • Given either two points, or the equation of a line segment and its midpoint.

Key Concepts: 

  • Equations of perpendicular bisectors.

MAI HL and SL Notes – All topics

EQUATIONS OF PERPENDICULAR BISECTORS

A perpendicular bisector is a line that:

  • Passes through the midpoint of a given line segment.

  • Is perpendicular to that segment.

♦ Midpoint

Given two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$, the midpoint formula is:

$\text { Midpoint }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

 Example

If we have a line segment with endpoints $(2,3)$ and $(6,7)$, the midpoint would be:

$\left(\frac{2+6}{2}, \frac{3+7}{2}\right)=(4,5)$

♦ PERPENDICULAR LINES

Consider two lines: \( L_1: y = m_1 x + c_1 \) and \( L_2: y = m_2 x + c_2 \)

Parallel lines:

\( L_1 // L_2 \) if \( m_1 = m_2 \)

Perpendicular lines:

\( L_1 \perp L_2 \) if \( m_2 = -1 / m_1 \)

For example,

The lines \( y = 3x + 5 \) and \( y = 3x + 8 \) are parallel
The lines \( y = 3x + 5 \) and \( y = -\frac{1}{3}x + 8 \) are perpendicular

♦A POINT AND A SLOPE

The line which passes through point \( P(x_0, y_0) \)  has slope \( m \) is given by

\( y – y_0 = m(x – x_0) \)

Example

Find the equation of the perpendicular bisector of the line segment joining

\( A(2, 4) \) and \( B(6, 8) \).

in form of $y=mx+c$

▶️Answer/Explanation

Solution:

$M = \left( \frac{2+6}{2}, \frac{4+8}{2} \right) = (4, 6)$

slope of \( AB \)
$m_{AB} = \frac{8-4}{6-2} = \frac{4}{4} = 1$

Slope of the perpendicular bisector
$m_{\perp} = -\frac{1}{m_{AB}} = -1$

Equation using point-slope form:
$y – 6 = -1(x – 4)$

$y = -x + 10$

♦ TWO POINTS

The line which passes through the points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) has slope

\( m = \frac{\Delta y}{\Delta x} = \frac{y_2 – y_1}{x_2 – x_1} \)

and its equation is again given by the formula

\( y – y_1 = m(x – x_1) \)

Example

Find the intersection point of the perpendicular bisectors of A & B and of A & C:

 

▶️Answer/Explanation

Solution:

Slope between A & B =

$
\frac{10 – 6}{2 – 8} = \frac{4}{-6} = -\frac{2}{3}
$

Perpendicular to that = $\frac{3}{2}$

Midpoint =

$
\left( \frac{2 + 8}{2}, \frac{10 + 6}{2} \right) = (5, 8)
$

Using point-slope form:

$
y – y_1 = m(x – x_1) \Rightarrow y – 8 = \frac{3}{2}(x – 5)
$

$
y = \frac{3}{2}x + \frac{1}{2}
$

Same for A & C gives:

$
y = -\frac{2}{3}x + 7
$

Intersection point:

$
\frac{3}{2}x + \frac{1}{2} = -\frac{2}{3}x + 7
$

$
\frac{9}{6}x + \frac{3}{6} = -\frac{4}{6}x + \frac{42}{6}
\Rightarrow 9x + 3 = -4x + 42
\Rightarrow 13x = 39
\Rightarrow x = 3
$

$
y = -\frac{2}{3}(3) + 7 = -2 + 7 = 5
$

Hence: (3, 5)

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