Equation of a Perpendicular Bisector

Line a has the equation \(y = 3x + 6\), is perpendicularly bisected by the line l. What is the gradient of line a?

1. Identify the original gradient: In the equation y = mx + c, m is the gradient. Therefore, the gradient of the original line is 3.

2. Find the gradient of the slope of the perpendicular bisector: Substitute the original gradient, 3, into the formula \(-\frac{1}{m}\) to find the inverse reciprocal because it is perpendicular. Therefore, the gradient of the line is \(-\frac{1}{3}\).

Line 1 stems from (3, 3) to (9, -21) and is perpendicularly bisected by Line 2. What is the gradient of the slope of Line 2?

1. Identify the original gradient: As we don't have the equation for line 1 we'll need to calculate the gradient of its slope. To find the gradient of Line 1, you'll need to substitute the coordinates into the gradient formula: \( gradient = \frac{change \, in \, y}{change \, in \, x} \). Therefore, \(\frac{-21 - 3}{9 - 3} = \frac{-24}{6} = -4\).
2. Find the gradient of the perpendicular bisector: Substitute -4 into the formula \(-\frac{1}{m}\), because the lines are perpendicular. Therefore, the gradient is \(\frac{-1}{-4}\), which is equal to \(\frac{1}{4}\).

A segment of a line has the endpoints (-1, 8) and (15, 10). Find the coordinates of the midpoint.

• Using \(\left(\frac{a+c}{2},\frac{b+d}{2}\right)\), substitute in the endpoints (-1, 8) and (15, 10) to get \(\left(-\frac{1+15}{2}, \frac{8+10}{2}\right)\)= (7, 9)

AB is a segment of a line with a midpoint of (6, 6). Find B when A is (10, 0).

• You can partition \(\left(\frac{a+c}{2}, \frac{b+d}{2}\right)\)into parts relating to the x- and y- coordinate where the centre is (m, n)
  • X coordinate: \(\frac{a + c}{2}\)= m
  • Y coordinates: \(\frac{b+d}{2} = n\)

• Then, you can substitute the known coordinates into these new Equations

  • X coordinates: \(\frac{10+c}{2} = 6\)

  • Y coordinates:\(\frac{0+d}{2}=6\)

• Rearranging these Equations would give you c = 2 and d = 12. Therefore, B = (2, 12)

A segment of a line from (4,10) to (10, 20) is perpendicularly bisected by line 1. What is the equation of the perpendicular bisector?

1. Find the gradient of the slope of the original line: \(\frac{20 - 10}{10 - 4} = \frac{10}{6} = \frac{5}{3}\)
2. Find the gradient of the slope of line 1: \(-\frac{1}{m} = -\frac{1}{\frac{5}{3}} = -\frac{3}{5}\)
3. Find the midpoint of the line segment: \(\left(\frac{4+10}{2}, \frac{10+20}{2}\right) = (7, 15)\)
4. Substitute into a formula: \(y - 15 = -\frac{3}{5}(x - 7)\)

A segment of a line from (-3, 7) to (6, 14) is perpendicularly bisected by line 1. What is the equation of the perpendicular bisector?

1. Find the gradient of the slope of the original line: \(\frac{14-7}{6 - (-3)} = \frac{7}{9}\)
2. Find the gradient of the slope of line 1: \(-\frac{1}{m} = -\frac{1}{\frac{7}{9}} = -\frac{9}{7}\)
3. Find the midpoint of the line segment: \(\left(-\frac{3}{2}+6, \frac{7}{2}+14 \right) = \left(\frac{3}{2}, \frac{21}{2}\right)\)
4. Substitute into a formula: \(y - \frac{21}{2} = -\frac{9}{7}(x - \frac{21}{2})\)

Therefore, the equation for the perpendicular bisector of the line segment is

\(y - \frac{21}{2} = - \frac{9}{7}(x - \frac{21}{2})\) \(y - \frac{21}{2} = - \frac{9}{7}x + \frac{189}{14}\frac{9}{7}x + y = \frac{189}{14} + \frac{21}{2}\frac{9}{7}x + y = \frac{189+147}{14}\frac{9}{7}x + y = \frac{189+147}{14}\frac{9}{7}x + y = \frac{336}{14}\frac{9}{7}x + y = 24\frac{9}{7}x + y - 24 = 0\)