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

A Voronoi diagram has the three sites \(P(0, 2)\), \(Q(6, 8)\), \(R(10, 0)\). 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 P and Q.

   • Midpoint: \((3, 5)\)
   • Gradient of the line through the two points: \(1\)
   • Perpendicular gradient: \(-1\)
   • \(y-5=-(x-3)\)
   • \(y=-x+8\)

2. (b) Sites P and R.

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

3. (c) Sites Q and R.

   • Midpoint: \((8, 4)\)
   • Gradient of the line through the two points: \(-2\)
   • Perpendicular gradient: \(\frac{1}{2}\)
   • \(y-4=\frac{1}{2}(x-8)\)
   • \(y=\frac{1}{2}x\)

Find the equation of the perpendicular bisector of the line segment joining \((-3, 6)\) and \((5, 0)\), in the form \(y=mx+c\).

• Midpoint: \((1, 3)\)
• Gradient of the line through the two points: \(-\frac{3}{4}\)
• Perpendicular gradient: \(\frac{4}{3}\)
• \(y-3=\frac{4}{3}(x-1)\)
• \(y=\frac{4}{3}x+\frac{5}{3}\)

The line segment AB has equation \(3x-2y=-1\) and midpoint \((3, 5)\). Find the equation of the perpendicular bisector of AB, in the form \(px+qy=r\), where p, q and r are integers.

• Rearrange: \(y=\frac{3}{2}x+\frac{1}{2}\), so the gradient of the line is \(\frac{3}{2}\)
• Perpendicular gradient: \(-\frac{2}{3}\)
• \(y-5=-\frac{2}{3}(x-3)\)
• Multiply out and rearrange: \(2x+3y=21\)

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

1. (a) \((4, 1)\) and \((4, 7)\);

   • Midpoint: \((4, 4)\)
   • The sites are in the same vertical line, so the bisector is horizontal: \(y=4\)

2. (b) \((-1, 5)\) and \((9, 5)\).

   • Midpoint: \((4, 5)\)
   • The sites are at the same height, so the bisector is vertical: \(x=4\)

Two cafés are sites in a Voronoi diagram. One is at \(S(3, 1)\). The edge between the two cells has equation \(y=5\). Find the coordinates of the other café, T.

• The edge is horizontal, so ST is vertical: T has the same x-coordinate as S, 3
• The midpoint of ST is \((3, 5)\), because it is on the edge
• \(T=2M-S=(2\times3-3,\ 2\times5-1)=(3, 9)\)

Two phone masts are at \(M(2, 9)\) and \(N(8, 3)\), where the units are kilometres. Each house uses the nearer mast. The boundary between the two areas is the perpendicular bisector of MN. A house is at \(H(3, 7)\).

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

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

2. (b) Find which mast the house at H uses, by working out the two distances.

   • \(MH=\sqrt{5}=2.24\)
   • \(NH=\sqrt{41}=6.40\)
   • M is nearer, because its distance is smaller.

Two wells are at \(W(1, 10)\) and \(V(9, 0)\). The line through them has equation \(5x+4y=45\) and its midpoint is \((5, 5)\). Each household uses the nearer well. Find the equation of the boundary between the two areas, in the form \(px+qy=r\).

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

Two farms are at \(P(9, 5)\) and \(Q(5, 3)\), where the units are kilometres. A straight canal has equation \(y=x+6\). A pump on the canal must be the same distance from the two farms.

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

   • Midpoint: \((7, 4)\)
   • Gradient of the line through the two points: \(\frac{1}{2}\)
   • Perpendicular gradient: \(-2\)
   • \(y-4=-2(x-7)\)
   • \(y=-2x+18\)

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

   • The canal: \(y=x+6\)
   • Solve with the bisector: \(-2x+18=x+6\)
   • \(x=4\), \(y=10\)
   • Check: \(PX=7.07\) and \(QX=7.07\)

The sites \(A(2, 1)\) and \(B(8, 5)\) are in a Voronoi diagram, and K is the point \((3, 6)\).

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

   • Midpoint: \((5, 3)\)
   • Gradient of the line through the two points: \(\frac{2}{3}\)
   • Perpendicular gradient: \(-\frac{3}{2}\)
   • \(y-3=-\frac{3}{2}(x-5)\)
   • \(y=-\frac{3}{2}x+\frac{21}{2}\)

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

   • Put \(x=3\) into the equation: \(y=6\), so K is on the line
   • \(KA=\sqrt{26}=5.10\)
   • \(KB=\sqrt{26}=5.10\)
   • The distances are equal.

\(A(2, 1)\) is a site. The perpendicular bisector of AB has equation \(y=-2x+10\). Find the coordinates of B.

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