AI HL · 12.8
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Equation of a perpendicular bisector
12.8.2
Drag the sites A and B. The working finds the perpendicular bisector line by line. Check that it matches the line on the grid.
Drag A or B.
The line segment AB has equation \(4x+3y=24\) and midpoint \((3, 4)\). Find the equation of the perpendicular bisector of AB, in the form \(px+qy=r\), where p, q and r are integers.
Have a go first. Then open the working one line at a time.
\(3y=-4x+24,\ y=-\frac{4}{3}x+8\)
Rearrange to y = mx + c. The gradient is the coefficient of x.
\(m=\frac{3}{4}\)
Flip the fraction and change the sign.
\(y-4=\frac{3}{4}(x-3)\)
Use the midpoint (3, 4) as the point.
\(4y-16=3x-9,\ 3x-4y=-7\)
Multiply by 4 to clear the fraction, then collect x and y on the left.
Find the equation of the perpendicular bisector of the line segment joining \(A(-4, 3)\) and \(B(2, 6)\), in the form \(y=mx+c\).
Have a go first. Then open the working one line at a time.
\(\left(\frac{-4+2}{2},\frac{3+6}{2}\right)=\left(-1,\frac{9}{2}\right)\)
Add the coordinates and halve.
\(\frac{6-3}{2-(-4)}=\frac{1}{2}\)
Change in y over change in x.
\(-2\)
Flip \(\frac{1}{2}\) and change the sign.
\(y-\frac{9}{2}=-2(x+1)\), \(y=-2x+\frac{5}{2}\)
Check: at x = −1 the line gives y = 4.5, the midpoint.
A Voronoi diagram has the sites \(A(1, 5)\), \(B(7, 9)\) and \(C(9, 1)\). For each pair of sites, write down the midpoint and the equation of the perpendicular bisector, in the form \(y=mx+c\). Give the gradient of the bisector of BC.
Have a go first. Then open the working one line at a time.
Midpoint \((4, 7)\); \(y=-\frac{3}{2}x+13\)
The gradient of AB is \(\frac{2}{3}\), so the bisector has gradient \(-\frac{3}{2}\).
Midpoint \((5, 3)\); \(y=2x-7\)
The gradient of AC is \(-\frac{1}{2}\), so the bisector has gradient \(2\).
Midpoint \((8, 5)\); \(y=\frac{1}{4}x+3\)
The gradient of BC is \(-4\), so the bisector has gradient \(\frac{1}{4}\).
Find the perpendicular bisector of the sites \(P(-2, 4)\) and \(Q(6, 4)\). Then find the perpendicular bisector of \(R(5, -1)\) and \(S(5, 7)\).
Have a go first. Then open the working one line at a time.
P and Q both have y = 4
The line PQ is horizontal, so the bisector is vertical.
Midpoint \((2, 4)\); \(x=2\)
A vertical line is x = the midpoint's x-coordinate.
R and S both have x = 5
The line RS is vertical, so the bisector is horizontal.
Midpoint \((5, 3)\); \(y=3\)
A horizontal line is y = the midpoint's y-coordinate.
Two schools are at \(A(2, 8)\) and \(B(10, 4)\), where the units are kilometres. A straight road has equation \(y=x-2\). A bus stop on the road must be the same distance from the two schools. Find the coordinates of the bus stop.
Have a go first. Then open the working one line at a time.
Midpoint \((6, 6)\); gradient of AB \(-\frac{1}{2}\); bisector gradient \(2\); \(y-6=2(x-6)\), so \(y=2x-6\)
Every point the same distance from A and B is on the perpendicular bisector of AB.
\(2x-6=x-2\), so \(x=4\) and \(y=4-2=2\)
The stop is on the road and on the bisector, so it is where the two lines meet.
\(SA=\sqrt{40}=6.32\) and \(SB=\sqrt{40}=6.32\)
The two distances are equal. The bus stop is at (4, 2).
A is a site at \((2, 7)\). The perpendicular bisector of AB has equation \(2x-3y=-4\). Find the coordinates of B.
Have a go first. Then open the working one line at a time.
Bisector gradient \(\frac{2}{3}\); AB gradient \(-\frac{3}{2}\); \(y-7=-\frac{3}{2}(x-2)\), so \(y=-\frac{3}{2}x+10\)
AB is perpendicular to the bisector and passes through A.
\(2x-3(-\frac{3}{2}x+10)=-4\), so \(\frac{13}{2}x=26\), \(x=4\), \(y=4\); midpoint \((4, 4)\)
The midpoint of AB is on both lines. Solve the two equations together.
\(B=(2\times4-2,\ 2\times4-7)=(6, 1)\)
B is as far past the midpoint as A is before it. Check: the midpoint of (2, 7) and (6, 1) is (4, 4).