Coordinate Geometry of Shapes

Worked example

1. 1

   $A(1,1),B(6,2),C(8,6)$ — The question. The letters go in order round the shape, so D is the corner joined to both A and C.

2. 2

   $A\text{to}B:+5,+1$ — Subtract: 6 − 1 = 5 across and 2 − 1 = 1 up. This is the step along side AB.

3. 3

   $D\text{to}C:+5,+1$ — DC is the side opposite AB. Opposite sides of a parallelogram are parallel and equal, so it is the same step.

4. 4

   $D=(8-5,6-1)$ — D is the start of that step and C is the end, so undo the step: take it away from C.

5. 5

   $D=(3,5)$ — 8 − 5 = 3 and 6 − 1 = 5.

Worked example

1. 1

   $A(-1,1),B(3,3)$ — The question. AB is one side of the square, and you want the next side, BC.

2. 2

   $m_{AB}=\frac{3-1}{3-(-1)}=\frac{2}{4}=\frac{1}{2}$ — Gradient is change in y ÷ change in x. Brackets keep the minus safe: 3 − (−1) = 4.

3. 3

   $m_{BC}=-2$ — BC is perpendicular to AB, so flip ½ to 2 and change the sign. Check: ½ × (−2) = −1.

4. 4

   $3=-2(3)+c$ — B is on line BC, so put x = 3 and y = 3 into y = −2x + c.

5. 5

   $3=-6+c$ — −2 × 3 = −6.

6. 6

   $c=9$ — Add 6 to both sides: 3 + 6 = 9.

7. 7

   $y=-2x+9$ — Put m = −2 and c = 9 into y = mx + c.

8. 8

   $2x+y-9=0$ — Add 2x and take 9 from both sides, so everything is on one side. Some questions ask for this form.

Worked example

1. 1

   $A(2,1),B(5,2)$ — The question: square ABCD, and C has two positive coordinates. Find C and D.

2. 2

   $A\text{to}B:+3,+1$ — Subtract: 5 − 2 = 3 across and 2 − 1 = 1 up. This is the step along side AB.

3. 3

   $-1,+3$ — Turn it 90°: swap the two numbers and change one sign. The side stays the same length and now meets AB at a right angle.

4. 4

   $C=(5-1,2+3)=(4,5)$ — BC is the next side, so add the turned step to B.

5. 5

   $D=(2-1,1+3)=(1,4)$ — AD is parallel to BC and the same length, so add the same turned step to A.

6. 6

   $\text{other turn }(+1,-3):C=(6,-1)$ — Turning the other way gives a second square. Reject it: −1 is not positive, and the question said C is.

Worked example

1. 1

   $A(-3,-1),B(1,7),C(5,a),D(b,c)$ — The question: ABCD is a rectangle. Find a, b and c.

2. 2

   $m_{AB}=\frac{7-(-1)}{1-(-3)}=\frac{8}{4}=2$ — Change in y ÷ change in x. Brackets keep the minus signs safe: 7 − (−1) = 8 and 1 − (−3) = 4.

3. 3

   $m_{BC}=-\frac{1}{2}$ — BC meets AB at a right angle, so flip 2 to ½ and change the sign.

4. 4

   $\frac{a-7}{5-1}=-\frac{1}{2}$ — Write BC's gradient a second way, using B and C, and make the two equal.

5. 5

   $\frac{a-7}{4}=-\frac{1}{2}$ — 5 − 1 = 4.

6. 6

   $a-7=-2$ — Multiply both sides by 4.

7. 7

   $a=5$ — Add 7 to both sides. So C is (5, 5).

8. 8

   $B\text{to}C:+4,-2$ — Subtract: 5 − 1 = 4 across and 5 − 7 = −2 up. This is the step along BC.

9. 9

   $D=(-3+4,-1-2)=(1,-3)$ — AD is opposite BC, so it is the same step. Add it to A: b = 1 and c = −3.

Worked example

1. 1

   $M(2,3),AC:y=2x-1$ — The question: a rhombus whose diagonals cross at M. Find the line BD.

2. 2

   $m_{AC}=2$ — Read it from y = 2x − 1: the gradient is the number in front of x.

3. 3

   $m_{BD}=-\frac{1}{2}$ — A rhombus's diagonals cross at a right angle, so BD is perpendicular: flip 2 to ½ and change the sign.

4. 4

   $3=-\frac{1}{2}(2)+c$ — BD passes through M, so put x = 2 and y = 3 into y = −½x + c.

5. 5

   $3=-1+c$ — −½ × 2 = −1.

6. 6

   $c=4$ — Add 1 to both sides.

7. 7

   $y=-\frac{1}{2}x+4$ — Put m and c together. BD meets the y-axis at (0, 4), because c is where the line crosses it.

Worked example

1. 1

   $A(0,1),B(1,4),C(6,4)$ — The question: ABCD is a kite with AB = AD and CB = CD. Find D.

2. 2

   $m_{AC}=\frac{4-1}{6-0}=\frac{1}{2}$ — Gradient of the diagonal AC: change in y ÷ change in x.

3. 3

   $AC:y=\frac{1}{2}x+1$ — A(0, 1) is on the y-axis, so c = 1.

4. 4

   $m_{BD}=-2$ — A kite's diagonals cross at a right angle, so flip ½ to 2 and change the sign.

5. 5

   $4=-2(1)+c,c=6$ — BD passes through B(1, 4): 4 = −2 + c, so add 2 to both sides.

6. 6

   $\frac{1}{2}x+1=-2x+6$ — M is on both diagonals, so make the two y's equal.

7. 7

   $x+2=-4x+12$ — Multiply every term by 2 to clear the fraction.

8. 8

   $5x=10,x=2$ — Add 4x and take 2 from both sides, then divide by 5.

9. 9

   $M=(2,2)$ — Put x = 2 into y = ½x + 1: y = 1 + 1 = 2. This is where the diagonals cross.

10. 10

    $B\text{to}M:+1,-2$ — Subtract: 2 − 1 = 1 across and 2 − 4 = −2 up.

11. 11

    $D=(2+1,2-2)=(3,0)$ — AC cuts BD in half, so M is halfway. Do the same step again from M.

Worked example

1. 1

   $A(-2,1),B(1,5),C(6,5)$ — The question. You must show your working.

2. 2

   $AB:1-(-2)=3,5-1=4$ — Across and up from A to B. Brackets keep the minus safe: −(−2) = +2.

3. 3

   $AB=\sqrt{3^2+4^2}=\sqrt{25}=5$ — Pythagoras: square each, add, then square root.

4. 4

   $BC:6-1=5,5-5=0,\text{so}BC=5$ — Up is 0, so BC is flat and its length is just the 5 across.

5. 5

   $AC:6-(-2)=8,5-1=4$ — Across and up from A to C.

6. 6

   $AC=\sqrt{8^2+4^2}=\sqrt{80}=8.94$ — Square, add, square root: 64 + 16 = 80. √80 = 8.94 to 3 s.f.

7. 7

   $AB=BC\ne AC$ — Exactly two sides are equal, so the triangle is isosceles.

Worked example

1. 1

   $y=-\frac{1}{2}x+4\text{and the axes}$ — The question: find the area of the triangle made by this line and the two axes.

2. 2

   $x=0:y=4$ — On the y-axis x is 0, so the line crosses at (0, 4). This upright side is the height.

3. 3

   $y=0:0=-\frac{1}{2}x+4$ — On the x-axis y is 0.

4. 4

   $\frac{1}{2}x=4$ — Add ½x to both sides.

5. 5

   $x=8$ — Multiply both sides by 2. The line crosses at (8, 0), so the flat side, the base, is 8.

6. 6

   $\frac{1}{2}\times 8\times 4=16$ — Area of a triangle = ½ × base × height. The axes meet at a right angle, so the height is upright.