Coordinate Geometry Problems

Add the two x-coordinates and halve the result, then do the same with the y-coordinates.

1. x: \;\dfrac{-7+3}{2}=\dfrac{-4}{2}=-2
2. y: \;\dfrac{4+(-10)}{2}=\dfrac{-6}{2}=-3
3. Answer: (-2, -3)

The step from A to M is the same as the step from M to B.

1. From A to M: x goes up by 4-(-3)=7 and y goes down by 5-(-1)=6
2. Do the same step again from M to B
3. x: \;4+7=11
4. y: \;-1-6=-7
5. (Check: \dfrac{-3+11}{2}=4 and \dfrac{5+(-7)}{2}=-1)
6. B=(11, -7)

Find the change in x and the change in y, then use Pythagoras; look for a square number factor at the end.

1. Change in x: \;4-(-2)=6
2. Change in y: \;1-(-3)=4
3. Pythagoras: \;PQ^2=6^2+4^2=36+16=52
4. PQ=\sqrt{52}=\sqrt{4}\times\sqrt{13}
5. PQ=2\sqrt{13}

AB and DC are both horizontal, so ABCD is a trapezium: find the two parallel sides and the distance between them.

1. AB and DC are horizontal, so they are parallel and ABCD is a trapezium
2. AB=6-(-2)=8
3. DC=4-0=4
4. Height (the distance between y=-1 and y=5) =6
5. Area =\dfrac12(8+4)\times6
6. Answer: 36 square units

AC : CB = 3 : 4 means C is \dfrac37 of the way from A to B.

1. From A to B: x changes by 9-(-5)=14 and y changes by -12-9=-21
2. C is \dfrac37 of the way along
3. \dfrac37\times14=6 and \dfrac37\times(-21)=-9
4. C=(-5+6,\ 9-9)
5. Answer: (1, 0)

Write the coordinates of the midpoint in terms of k, then substitute them into y=x+4.

1. Midpoint: \left(\dfrac{k+3k}{2}, \dfrac{5+(-1)}{2}\right)=(2k, 2)
2. It lies on y=x+4, so 2=2k+4
3. 2k=-2
4. Answer: k=-1

In parallelogram ABCD the step from B to A is the same as the step from C to D.

1. The letters go round the shape in order, so the step from C to D is the same as the step from B to A
2. From B to A: x goes down by 6 and y goes down by 2
3. Do the same from C: (7-6,\ 8-2)
4. (Check: the diagonals AC and BD both have midpoint (2.5, 4.5))
5. Answer: D=(1, 6)

Keep the change in x as a fraction, then use Pythagoras' theorem.

1. Change in x: \;\dfrac32-\dfrac14=\dfrac54
2. Change in y: \;1-(-2)=3
3. EF^2=\left(\dfrac54\right)^2+3^2=\dfrac{25}{16}+\dfrac{144}{16}=\dfrac{169}{16}
4. EF=\sqrt{\dfrac{169}{16}}=\dfrac{13}{4}
5. Answer: \dfrac{13}{4}

The gradient of QR must be the same as the gradient of PQ.

1. Gradient of PQ=\dfrac{1-7}{2-(-2)}=\dfrac{-6}{4}=-\dfrac32
2. Gradient of QR=\dfrac{-8-1}{k-2}=\dfrac{-9}{k-2}
3. \dfrac{-9}{k-2}=-\dfrac32, so k-2=6
4. Answer: k=8

Work out the gradients of AB and BC (change in y over change in x) and multiply them.

1. Gradient of AB=\dfrac{5-3}{4-0}=\dfrac12
2. Gradient of BC=\dfrac{1-5}{6-4}=-2
3. \dfrac12\times(-2)=-1, so AB is perpendicular to BC
4. (The gradients are not equal, and AB^2+AC^2=20+40=60, which is not BC^2=20)
5. Answer: gradient of AB\times gradient of BC=-1, so angle ABC=90^\circ

BC is 3 parts of the ratio: find the step for 1 part, then go back 2 parts from B.

1. From B to C: x changes by 6 and y changes by -9
2. BC is 3 parts, so 1 part is x: +2, y: -3
3. AB is 2 parts: x: +4, y: -6 going from A to B
4. So A=(1-4,\ 4-(-6))
5. Answer: A=(-3, 10)

Find the gradient of AB; the gradient of BC is its negative reciprocal.

1. Gradient of AB=\dfrac{4-2}{1-(-3)}=\dfrac24=\dfrac12
2. Perpendicular gradient =-2
3. Gradient of BC=\dfrac{k-4}{4-1}=\dfrac{k-4}{3}
4. \dfrac{k-4}{3}=-2, so k-4=-6
5. Answer: k=-2

Use Pythagoras' theorem with a horizontal distance of 8 to find the vertical distance from A to P, which can be up or down.

1. Horizontal distance from A to P: 0-(-8)=8
2. 8^2+(k-3)^2=10^2, so (k-3)^2=100-64=36
3. k-3=6 or k-3=-6
4. Answer: k=9 or k=-3

Solve the two equations together to find P; its y-coordinate is the height of the triangle on base OQ.

1. At P: x+3x=8, so x=2 and y=6
2. At Q: y=0, so x=8
3. Base OQ=8, height =6 (the y-coordinate of P)
4. Area =\dfrac12\times8\times6
5. Answer: 24 square units

The perpendicular bisector passes through the midpoint of AB and its gradient is the negative reciprocal of the gradient of AB.

1. Midpoint of AB: \left(\dfrac{-1+5}{2}, \dfrac{6+2}{2}\right)=(2, 4)
2. Gradient of AB=\dfrac{2-6}{5-(-1)}=\dfrac{-4}{6}=-\dfrac23
3. Perpendicular gradient =\dfrac32
4. y-4=\dfrac32(x-2)
5. Multiply by 2: 2y-8=3x-6
6. Answer: 3x-2y=-2

Write one equation in p and q from the midpoint and another from the gradient, then solve them simultaneously. Answer part (a) before attempting part (b).

Part (a)

1. Midpoint: \left(\dfrac{p+9}{2}, \dfrac{3+q}{2}\right)
2. It lies on y=x, so 3+q=p+9, which gives q=p+6
3. Gradient: \dfrac{q-3}{9-p}=-\dfrac13
4. So 3(q-3)=-(9-p), which gives 3q=p
5. Substitute q=p+6: \;3(p+6)=p
6. 2p=-18
7. Answer: p=-9

Part (b)

1. q=p+6=-9+6
2. Answer: q=-3

Part (c)

1. P=(-9, 3) and Q=(9, -3)
2. Change in x: 18; change in y: -6
3. PQ^2=18^2+6^2=324+36=360
4. \sqrt{360}=\sqrt{36}\times\sqrt{10}
5. Answer: 6\sqrt{10}

Find A and B by putting x=0 and then y=0 into the equation of L. Answer part (a) before attempting part (b).

Part (a)

1. At A, x=0: 4y=24, so A=(0, 6)
2. At B, y=0: 3x=24, so B=(8, 0)
3. AB^2=8^2+6^2=100
4. Answer: AB=10

Part (b)

1. From A to B: x changes by 8 and y changes by -6
2. BC is \dfrac14 of AB: x: +2, y: -1.5
3. C=(8+2,\ 0-1.5)
4. Answer: C=(10, -1.5)

Part (c)

1. Use OA (on the y-axis) as the base: OA=6
2. The height is the horizontal distance from the y-axis to C, which is 10
3. Area =\dfrac12\times6\times10
4. Answer: 30 square units

Rearrange the equation of L to y=mx+c to read its gradient, then find the equation of the perpendicular line through A. Answer part (a) before attempting part (b).

Part (a)

1. Rearrange L: y=2x-2, so the gradient of L is 2
2. Perpendicular gradient =-\dfrac12
3. Line through A: y-4=-\dfrac12(x+2), so y=-\dfrac12x+3
4. Solve with y=2x-2: \;2x-2=-\dfrac12x+3
5. \dfrac52x=5, so x=2 and y=2\times2-2=2
6. Answer: F=(2, 2)

Part (b)

1. From A(-2, 4) to F(2, 2): change in x is 4, change in y is -2
2. AF^2=4^2+2^2=20
3. \sqrt{20}=\sqrt4\times\sqrt5
4. Answer: 2\sqrt5

Call P the point (x, x+1), use Pythagoras' theorem to write an equation for AP^2, and solve the quadratic. Answer part (a) before attempting part (b).

Part (a)

1. Let P=(x, x+1)
2. AP^2=(x-4)^2+(x+1)^2=25
3. x^2-8x+16+x^2+2x+1=25
4. 2x^2-6x-8=0, so x^2-3x-4=0
5. (x-4)(x+1)=0, so x=4 or x=-1
6. Smaller x: x=-1, y=-1+1=0
7. Answer: (-1, 0)

Part (b)

1. x=4, so y=4+1=5
2. Answer: (4, 5)

Part (c)

1. The vertices are A(4, 0), (-1, 0) and (4, 5)
2. The side from (-1, 0) to A lies on the x-axis: length 5
3. (4, 5) is directly above A, so the height is 5
4. Area =\dfrac12\times5\times5
5. Answer: 12.5 square units

Points that are the same distance from A and B lie on the perpendicular bisector of AB; find its equation first.

Part (a)

1. P is the same distance from A and B, so it lies on the perpendicular bisector of AB
2. Midpoint of AB: \;M=\left(\dfrac{-4+4}{2}, \dfrac{3+(-1)}{2}\right)=(0, 1)
3. Gradient of AB=\dfrac{-1-3}{4-(-4)}=\dfrac{-4}{8}=-\dfrac12
4. Perpendicular gradient =2, so the perpendicular bisector is \;y=2x+1
5. Substitute into 2x+y=17: \;2x+2x+1=17, so 4x=16 and x=4
6. y=2\times4+1=9
7. (Check: PA^2=8^2+6^2=100 and PB^2=0^2+10^2=100)
8. P=(4, 9)

Part (b)

1. Triangle ABP is isosceles (PA=PB), so PM is perpendicular to AB and is the height
2. Base: \;AB=\sqrt{8^2+4^2}=\sqrt{80}=4\sqrt5
3. Height: \;PM=\sqrt{4^2+8^2}=\sqrt{80}=4\sqrt5
4. Area =\dfrac12\times4\sqrt5\times4\sqrt5=\dfrac12\times80
5. Area =40 square units