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Steps

1. 1

   Find the midpoint of PQ: the line goes through it.

2. 2

   Find the gradient of PQ, then flip it and change the sign.

3. 3

   Put the midpoint into $y=mx+c$ to find c.

Worked example

1. 1

   $P(-2,1),Q(4,5)$ — The two points. The bisector goes through their middle, at a right angle to PQ.

2. 2

   $M=\left(\frac{-2+4}{2},\frac{1+5}{2}\right)$ — Midpoint: add the two x's and halve, then the two y's. Watch the sign: −2 + 4 = 2.

3. 3

   $M=(1,3)$ — 2 ÷ 2 = 1 and 6 ÷ 2 = 3. This is the one point we know is on the bisector.

4. 4

   $m_{PQ}=\frac{5-1}{4-(-2)}=\frac{4}{6}=\frac{2}{3}$ — Gradient of PQ: change in y over change in x. 4 − (−2) = 6, because taking away a negative adds.

5. 5

   $m=-\frac{3}{2}$ — Flip 2/3 to 3/2 and change the sign, because perpendicular gradients multiply to −1: $\frac{2}{3}\times \left(-\frac{3}{2}\right)=-1$.

6. 6

   $3=-\frac{3}{2}(1)+c$ — Put M into $y=mx+c$: x = 1 and y = 3. Use M, not P or Q.

7. 7

   $3=-\frac{3}{2}+c$ — −3/2 × 1 = −3/2.

8. 8

   $c=3+\frac{3}{2}=\frac{9}{2}$ — Add 3/2 to both sides to get c on its own.

9. 9

   $y=-\frac{3}{2}x+\frac{9}{2}$ — Write m and c into $y=mx+c$. Stop here if the question asks for this form.

10. 10

    $2y=-3x+9$ — Times every term by 2 to clear the halves, for questions that want whole numbers.

11. 11

    $3x+2y-9=0$ — Add 3x and take 9 from both sides. This is the form $ax+by+c=0$.

Worked example

1. 1

   $P(-3,2),Q(2,5)$ — The two points. −3 + 2 = −1 is odd, so expect a half.

2. 2

   $M=\left(\frac{-3+2}{2},\frac{2+5}{2}\right)=\left(-\frac{1}{2},\frac{7}{2}\right)$ — Midpoint: add and halve each pair. Keep the halves as fractions, not decimals.

3. 3

   $m_{PQ}=\frac{5-2}{2-(-3)}=\frac{3}{5}$ — Gradient of PQ: change in y over change in x. 2 − (−3) = 5.

4. 4

   $m=-\frac{5}{3}$ — Flip 3/5 and change the sign. Check: $\frac{3}{5}\times \left(-\frac{5}{3}\right)=-1$.

5. 5

   $\frac{7}{2}=-\frac{5}{3}\times \left(-\frac{1}{2}\right)+c$ — Put M into $y=mx+c$. Brackets keep the minus on x.

6. 6

   $\frac{7}{2}=\frac{5}{6}+c$ — Minus times minus is plus: $-\frac{5}{3}\times \left(-\frac{1}{2}\right)=+\frac{5}{6}$.

7. 7

   $c=\frac{21}{6}-\frac{5}{6}=\frac{16}{6}=\frac{8}{3}$ — Take 5/6 from both sides. Write 7/2 as 21/6 so the bottoms match, then simplify.

8. 8

   $y=-\frac{5}{3}x+\frac{8}{3}$ — Write m and c into $y=mx+c$.

9. 9

   $3y=-5x+8$ — Times every term by 3 to clear the thirds.

10. 10

    $5x+3y=8$ — Add 5x to both sides. Whole numbers only, as “px + qy = r” questions ask.

Worked example

1. 1

   $\text{Kite: }AB=AD,CB=CD,B(1,6),D(7,2)$ — Find the line AC. A is the same distance from B and D, and so is C. So AC is the perpendicular bisector of BD.

2. 2

   $M=\left(\frac{1+7}{2},\frac{6+2}{2}\right)=(4,4)$ — Midpoint of BD: add and halve. AC passes through it.

3. 3

   $m_{BD}=\frac{2-6}{7-1}=\frac{-4}{6}=-\frac{2}{3}$ — Gradient of BD: change in y over change in x. y goes down 4, so the gradient is negative.

4. 4

   $m=\frac{3}{2}$ — Flip 2/3 to 3/2 and change the sign: minus becomes plus. Check: $-\frac{2}{3}\times \frac{3}{2}=-1$.

5. 5

   $4=\frac{3}{2}(4)+c$ — Put M(4, 4) into $y=mx+c$.

6. 6

   $4=6+c$ — 3/2 × 4 = 6.

7. 7

   $c=-2$ — Take 6 from both sides: 4 − 6 = −2.

8. 8

   $y=\frac{3}{2}x-2$ — Write m and c into $y=mx+c$.

9. 9

   $2y=3x-4$ — Times every term by 2 to clear the half. This is line AC.

Worked example

1. 1

   $AB=AC,A(5,9),B\text{ and }C\text{ on }2y=x+4$ — Find the line of symmetry of triangle ABC. A is the same distance from B and C, so A is on the perpendicular bisector of BC.

2. 2

   $y=\frac{1}{2}x+2$ — Divide the line BC by 2 so y is on its own. Its gradient is the number in front of x: 1/2.

3. 3

   $m=-2$ — The line of symmetry is perpendicular to BC: flip 1/2 to 2 and change the sign.

4. 4

   $9=-2(5)+c$ — Put A(5, 9) into $y=mx+c$. A takes the place of the midpoint, which we do not know.

5. 5

   $9=-10+c$ — −2 × 5 = −10.

6. 6

   $c=19$ — Add 10 to both sides.

7. 7

   $y=-2x+19$ — Write m and c into $y=mx+c$.

8. 8

   $2x+y=19$ — Add 2x to both sides, for the form px + qy = r.

Worked example

1. 1

   $\text{bisector: }y=2x-1$ — The bisector's gradient is the number in front of x: 2.

2. 2

   $m_{AB}=-\frac{1}{2}$ — AB crosses the bisector at a right angle, so flip 2 to 1/2 and change the sign.

3. 3

   $A(-1,7),B(7,k)$ — The question: find k.

4. 4

   $\frac{k-7}{7-(-1)}=-\frac{1}{2}$ — Write the gradient of AB from the two points, and set it equal to −1/2.

5. 5

   $\frac{k-7}{8}=-\frac{1}{2}$ — 7 − (−1) = 8, because taking away a negative adds.

6. 6

   $k-7=-4$ — Times both sides by 8: 8 × (−1/2) = −4.

7. 7

   $k=3$ — Add 7 to both sides.

8. 8

   $M=\left(\frac{-1+7}{2},\frac{7+3}{2}\right)=(3,5)$ — Now that k is known, find the midpoint of AB: add and halve.

9. 9

   $2(3)-1=5✓$ — Put x = 3 into the bisector: y = 5, which matches M. So M is on the bisector, as it must be.

Steps

1. 1

   Open the compasses to more than half of AB. Less than half, and the arcs will not meet.

2. 2

   Put the point on A. Draw one arc above AB and one below it.

3. 3

   Keep the same width. Put the point on B and draw two arcs that cross the first two.

4. 4

   With a ruler, join the two crossing points. That line is the perpendicular bisector. Where it crosses AB is the midpoint.