Equation of a perpendicular bisector Problem set B One question at a time

A Voronoi diagram has the three sites \(D(1, 7)\), \(E(9, 3)\), \(F(5, 11)\). For each pair of sites, write down the midpoint and find the equation of the perpendicular bisector, in the form \(y=mx+c\).

1. (a) Sites D and E.

   • Midpoint: \((5, 5)\)
   • Gradient of the line through the two points: \(-\frac{1}{2}\)
   • Perpendicular gradient: \(2\)
   • \(y-5=2(x-5)\)
   • \(y=2x-5\)

2. (b) Sites D and F.

   • Midpoint: \((3, 9)\)
   • Gradient of the line through the two points: \(1\)
   • Perpendicular gradient: \(-1\)
   • \(y-9=-(x-3)\)
   • \(y=-x+12\)

3. (c) Sites E and F.

   • Midpoint: \((7, 7)\)
   • Gradient of the line through the two points: \(-2\)
   • Perpendicular gradient: \(\frac{1}{2}\)
   • \(y-7=\frac{1}{2}(x-7)\)
   • \(y=\frac{1}{2}x+\frac{7}{2}\)

Find the equation of the perpendicular bisector of the line segment joining \((-6, -1)\) and \((2, 5)\), in the form \(y=mx+c\).

• Midpoint: \((-2, 2)\)
• Gradient of the line through the two points: \(\frac{3}{4}\)
• Perpendicular gradient: \(-\frac{4}{3}\)
• \(y-2=-\frac{4}{3}(x+2)\)
• \(y=-\frac{4}{3}x-\frac{2}{3}\)

The line segment AB has equation \(x+4y=12\) and midpoint \((4, 2)\). Find the equation of the perpendicular bisector of AB, in the form \(px+qy=r\), where p, q and r are integers.

• Rearrange: \(y=-\frac{1}{4}x+3\), so the gradient of the line is \(-\frac{1}{4}\)
• Perpendicular gradient: \(4\)
• \(y-2=4(x-4)\)
• Multiply out and rearrange: \(4x-y=14\)

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

1. (a) \((2, -3)\) and \((2, 5)\);

   • Midpoint: \((2, 1)\)
   • The sites are in the same vertical line, so the bisector is horizontal: \(y=1\)

2. (b) \((-4, 6)\) and \((2, 6)\).

   • Midpoint: \((-1, 6)\)
   • The sites are at the same height, so the bisector is vertical: \(x=-1\)

Two kiosks are sites in a Voronoi diagram. One is at \(U(-4, 3)\). The edge between the two cells has equation \(y=-1\). Find the coordinates of the other kiosk, V.

• The edge is horizontal, so UV is vertical: V has the same x-coordinate as U, \(-4\)
• The midpoint of UV is \((-4, -1)\), because it is on the edge
• \(V=2M-U=(2\times(-4)-(-4),\ 2\times(-1)-3)=(-4, -5)\)

Two schools are at \(S(1, 2)\) and \(T(7, 10)\), where the units are hundreds of metres. Each pupil goes to the nearer school. The boundary between the two catchment areas is the perpendicular bisector of ST. A pupil lives at \(H(5, 7)\).

1. (a) Find the equation of the boundary, in the form \(y=mx+c\).

   • Midpoint: \((4, 6)\)
   • Gradient of the line through the two points: \(\frac{4}{3}\)
   • Perpendicular gradient: \(-\frac{3}{4}\)
   • \(y-6=-\frac{3}{4}(x-4)\)
   • \(y=-\frac{3}{4}x+9\)

2. (b) Find which school the pupil goes to, by working out the two distances.

   • \(SH=\sqrt{41}=6.40\)
   • \(TH=\sqrt{13}=3.61\)
   • T is nearer, because its distance is smaller.

Two lifeguard towers are at \(T(2, 1)\) and \(U(6, 9)\) on a map. The line through them has equation \(2x-y=3\) and its midpoint is \((4, 5)\). Each swimmer is helped by the nearer tower. Find the equation of the boundary between the two areas, in the form \(px+qy=r\).

• Gradient of the line: \(2\); perpendicular gradient: \(-\frac{1}{2}\)
• \(y-5=-\frac{1}{2}(x-4)\)
• Multiply out and rearrange: \(x+2y=14\)

Two fire stations are at \(R(11, 5)\) and \(T(1, 7)\), where the units are kilometres. A straight cycle path has equation \(x-y=4\). A drinking fountain on the path must be the same distance from the two stations.

1. (a) Find the equation of the perpendicular bisector of RT, in the form \(y=mx+c\).

   • Midpoint: \((6, 6)\)
   • Gradient of the line through the two points: \(-\frac{1}{5}\)
   • Perpendicular gradient: \(5\)
   • \(y-6=5(x-6)\)
   • \(y=5x-24\)

2. (b) Find the coordinates of the fountain.

   • The path: \(x-y=4\), so \(y=x-4\)
   • Solve with the bisector: \(5x-24=x-4\)
   • \(x=5\), \(y=1\)
   • Check: \(RF=7.21\) and \(TF=7.21\)

The sites \(C(1, 6)\) and \(D(7, 2)\) are in a Voronoi diagram, and K is the point \((0, -2)\).

1. (a) Find the equation of the perpendicular bisector of CD, in the form \(y=mx+c\).

   • Midpoint: \((4, 4)\)
   • Gradient of the line through the two points: \(-\frac{2}{3}\)
   • Perpendicular gradient: \(\frac{3}{2}\)
   • \(y-4=\frac{3}{2}(x-4)\)
   • \(y=\frac{3}{2}x-2\)

2. (b) Show that K is on the bisector by showing that K is the same distance from C and from D.

   • Put \(x=0\) into the equation: \(y=-2\), so K is on the line
   • \(KC=\sqrt{65}=8.06\)
   • \(KD=\sqrt{65}=8.06\)
   • The distances are equal.

\(C(3, 2)\) is a site. The perpendicular bisector of CD has equation \(y=-3x+21\). Find the coordinates of D.

• The bisector has gradient \(-3\), so the line CD has gradient \(\frac{1}{3}\)
• Line CD: \(y-2=\frac{1}{3}(x-3)\)
• The midpoint of CD is where this line meets the bisector: \((6, 3)\)
• The midpoint is halfway between C and D, so D = \(2M-C\): \((9, 4)\)