,
,
Useful formulas: distance is given by: a2 = b2 + c2, or l 2  ( x2  x1 ) 2  ( y 2  y1 ) 2
Midpoint =
y  y1
A B
Vertical change
 x  x2 y1  y 2 
,
or M   1
or m  2
 ; Gradient = m =
2 
x2  x1
2
Horizontal change
 2
where (x1, y1) and (x2, y2) are the two points being measured.
In each of the following exercises, draw a diagram to illustrate the method and/or result.
Distance:
Ex1 Use the distance formula to find the distance between the two points:
a) (2, 2) and (5, 6)
b) (-6, 4) and (6, 9)
c) (4, 3) and (2, 7)
d) (-3, -1) and (-1, 2)
Ex2 By substituting the given coordinates into the distance formula, find and simplify an expression
for the distance between the two points:
a) (0, 0) and (3a, 4a)
b) (-3b, b) and (-6b, 5b)
c) (2c, 3c) and (4c, 6c)
d) (d-1, 5) and (-1, d+5)
Ex3 By substituting all of the given information into the distance formula, form an equation with
the unknown value. Then solve the equation to find the two possible values.
a) (e, 3) and (0, 0), dist2 = 25
b) (3, f) and (1, 4), dist = √13
c) (-2, -1), (4, g + 3), dist = √40
d) (2, 3) and (d + 1, -1), dist2 = 41
Ex4 In each part a triangle is given by three points. Use the distance formula to find the length2 of
each side. Hence decide if it is isosceles and/or right-angled (use Pythagoras’ theorem) or neither.
a) (4, 1), (1, 4), (-2, -5)
b) (3, -10), (3, -2), (7, -6)
c) (1, 16), (-3, 12), (2, 5)
d) (-2, 11), (5, 16), (-9, 16)
Created by petermerrick
Midpoint:
Ex1 Use the midpoint formula to find the midpoints of the lines joining the following points:
a) (2, 2) and (4, 6)
b) (-6, 3) and (4, 9)
c) (4, 4) and (2, 7)
d) (-3, -1) and (-1, 2)
Ex2 Substitute the given coordinates into the midpoint formula, then simplify the result:
a) (0, 0) and (3h, 4h)
b) (-3j, j) and (-6j, 5j)
c) (2k, 3k) and (4l, 6l)
d) (m-1, 5) and (-1, m+5)
Ex3 Use the midpoint formula to form equations for x and y and hence find the unknown values:
a) (n, 2n) and (0, 0), M = (3, 6)
b) (p, q) and (1, 4), M = (3, 6)
c) (r, -1) and (4, s + 3), M = (-3, 2)
d) (t, 5) and (3, -2), M = (4½, u + 2)
Gradient
Ex1 Use the gradient formula to find the gradients of the line segments between the points:
a) (2, 2) and (5, 6)
b) (-6, 4) and (6, 9)
c) (4, 3) and (2, 7)
d) (-3, -1) and (-1, 2)
Ex2 By substituting the given coordinates into the gradient formula, find and simplify an expression
for the gradient between the two points:
a) (0, 0) and (3t, 4t)
b) (-3u, u) and (-6u, 5u)
c) (2v, 3v) and (4v, 6v)
d) (w – 1, 5) and (-1, w+5)
Ex3 Use the gradient formula to form an equation and hence find the unknown value:
a) (x, 3) and (0, 0), grad = 3/4
b) (3, y) and (1, 4), grad = 3
c) (-2, -1), (4, z + 3), grad = 5/6
d) (5, 2) and (a, a), grad = 4
Created by petermerrick
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,
Distance:
Ex1 Use the distance formula to find the distance between the two points:
e) (2, 2) and (5, 6)
f) (-6, 4) and (6, 9)
X = 5 – 2 = 3; Y = 6 – 2 = 4
X = 6 – -6 = 12; Y = 9 – 4 = 5
dist = √(32 + 42) = √25 = 5
dist = √(122 + 52) = √169 = 13
g) (4, 3) and (2, 7)
h) (-3, -1) and (-1, 2)
X = 2 – 4 = -2; Y = 7 – 3 = 4
X = -1 – -3 = 2; Y = 2 – -1 = 3
2
2
dist = √(2 + 4 ) = √20
dist = √(22 + 32) = √13
Ex2 By substituting the given coordinates into the distance formula, find and simplify an expression
for the distance between the two points:
e) (0, 0) and (3a, 4a)
f) (-3b, b) and (-6b, 5b)
X = 3a – 0 = 3a; Y = 4a
X = -6b – -3b = -3b; Y = 5b – b = 4b
2
2
2
2
dist = √((3a) + (4a) ) = √(9a + 16a )
dist = b × √(32 + 42) = 5b
2
= √(25a ) = 5a
Or: dist = a × √(32 + 42) = 5a
g) (2c, 3c) and (4c, 6c)
h) (d-1, 5) and (-1, d+5)
X = 4c – 2c = 2c; Y = 6c – 3c = 3c
X = d + 5 – 5 = d; Y = -1 – (d – 1) = -d
dist = c × √(22 + 32) = √13 c
dist = d × √(12 + 12) = √2 d
Ex3 By substituting all of the given information into the distance formula, form an equation with
the unknown value. Then solve the equation to find the two possible values.
e) (e, 3) and (0, 0), dist2 = 25
X = 0 – e = -e; Y = 0 – 3 = -3
dist2 = e2 + 32 = 25
so e2 = 16 and e = ±4
Check: 42 + 32 = 25, so it’s correct!
g) (-2, -1), (4, g + 3), dist = √40
X = 4 – -2 = 6; Y = g + 3 - -1 = g + 4
dist2 = 62 + (g + 4)2 = 40
so (g + 4)2 = 4 and g + 4 = ±2
g = -4 ± 2 = -2 or -6
f) (3, f) and (1, 4), dist = √13
X = 1 – 3 = -2; Y = 4 – f
dist2 = 22 + (4 – f)2 = 13
so (4 – f)2 = 9 and 4 – f = ±3
f = 4 ± 3 = 7 or 1
h) (2, 3) and (h + 1, -1), dist2 = 41
X = h + 1 – 2 = h – 1; Y = -1 – 3 = -4
dist2 = (h – 1)2 + 42 = 41
so (h – 1)2 = 25 and h – 1 = ±5
h = 1 ± 5 = 6 or -4
Ex4 In each part a triangle is given by three points. Use the distance formula to find the length2 of
each side. Hence decide if it is isosceles and/or right-angled (use Pythagoras’ theorem) or neither.
e) A(4, 1), B(1, 4), C(-2, -5)
f) D(3, -10), E(3, -2), F(7, -6)
AB2 = 32 + 32 = 18
DE2 = 02 + 82 = 64
2
2
2
AC = 6 + 6 = 72
DF2 = 42 + 42 = 32
BC2 = 32 + 92 = 90
EF2 = 42 + 42 = 32
Not isosceles (3 different sides)
Isosceles (2 same sides)
2
2
2
Right-angled as AB + AC = BC
Right-angled as DF2 + EF2 = DE2
g) G(1, 16), H(-3, 12), I(2, 5)
h) J(-2, 11), K(5, 16), L(-9, 16)
2
2
2
GH = 4 + 4 = 32
JK2 = 72 + 52 = 74
2
2
2
GI = 1 + 11 = 122
JL2 = 72 + 52 = 74
HI2 = 52 + 72 = 74
KL2 = 142 + 02 = 196
Not isosceles (3 different sides)
Isosceles (2 same sides)
2
2
2
Not right-angled as GH + HI ≠ GI
Not right-angled as JK2 + JL2 ≠ KL2
Midpoint:
Ex1 Use the midpoint formula to find the midpoints of the lines joining the following points:
e) (2, 2) and (4, 6)
24 26
,

 = (3, 4)
2 
 2
Created by petermerrick
f) (-6, 3) and (4, 9)
 6 4 39
,

 = (-1, 6)
2 
 2
g) (4, 4) and (2, 7)
h) (-3, -1) and (-1, 2)
42 47
  3  1  1  2 
,
,

 = (-2, ½)

 = (3, 5½)
2 
2 
 2
 2
Ex2 Substitute the given coordinates into the midpoint formula, then simplify the result:
e) (0, 0) and (3h, 4h)
f) (-3j, j) and (-6j, 5j)
 0  3h 0  4h   3h

  3i  6i i  5i    9i 
,
,
,3i 

   ,2h 


2   2
2
2   2
 2



g) (2k, 3k) and (4l, 6l)
h) (m-1, 5) and (-1, m+5)
3k
m
 2k  4l 3k  6l  

 m  1  1 5  m  5   m

,
,

   k  2l ,  3l 

    1,  5 
2  
2
2
2
2
 2


 2

Ex3 Use the midpoint formula to form equations for x and y and hence find the unknown values:
e) (n, 2n) and (0, 0), M = (3, 6)
 n  0 2n  0 
,

  (3, 6)
2 
 2
n
X   0 and Y  n  6
2
So n = 6
g) (r, -1) and (4, s + 3), M = (-3, 2)
 r  4 1 s  3 
,

  (-3, 2)
2
 2

r4
s2
 3 and
2
2
2
So r = -10 and s = 2
f) (p, q) and (1, 4), M = (3, 6)
 p 1 q  4 
,

  (3, 6)
2 
 2
p 1
q4
 3 and
6
2
2
So p = 5 and q = 8
h) (t, 5) and (3, -2), M = (4½, u + 2)
 t  3 5  2 
,

  (4½, u + 2)
2 
 2
t 3
3
 4 12 and  u  2
2
2
So t = 6 and u = -½
Gradient
Ex1 Use the gradient formula to find the gradients of the line segments between the points:
e) (2, 2) and (5, 6)
f) (-6, 4) and (6, 9)
62 4
94
5


52 3
6  6 12
g) (4, 3) and (2, 7)
h) (-3, -1) and (-1, 2)
73
4
2  1 3

 2

24 2
 1  3 2
Ex2 By substituting the given coordinates into the gradient formula, find and simplify an expression
for the gradient between the two points:
e) (0, 0) and (3t, 4t)
f) (-3u, u) and (-6u, 5u)
4t  0 4t 4
4t  0 4t 4
 
 
3t  0 3t 3
3t  0 3t 3
g) (2v, 3v) and (4v, 6v)
h) (w – 1, 5) and (-1, w+5)
6v  3v 3v 3
w55
w



 1
4v  2v 2v 2
 1  ( w  1)  w
Ex3 Use the gradient formula to form an equation and hence find the unknown value:
e) (x, 3) and (0, 0), grad = 3/4
03 3 3
 
0 x x 4
so x = 4
g) (-2, -1), (4, z + 3), grad = 5/6
z  3  1 z  4 5


4  2
6
6
z + 4 = 5 so z = 1
Created by petermerrick
f) (3, y) and (1, 4), grad = 3
4 y 4 y

3
1 3
2
4 – y = -6, so y = 2
h) (5, 2) and (a, a), grad = 4
a2
 4 , so a – 2 = 4(a – 5) = 4a – 20
a 5
18 = 3a, so a = 6.
Created by petermerrick