Coordinate Proof Practice:
1. Given: ΔABC with A:(0, 0), B:(2, 3) and C(4, 0)
Prove: ΔABC is isosceles
2. Given: A:(1, 1), B:(2, 5), C:(5, 7) and D:(7, 5)
Prove: ABCD is a trapezoid
3. Prove that quadrilateral A(1,2), B(2,5), C(5,7) and D(4,4) is a
parallelogram by using slopes.
4. Prove that A(1,1), B(4,4), C(6,2) are the vertices of a right triangle.
5. Prove that if A(-3,2), B(-2,6), C(2,7) and D(1,3), ABCD is a rhombus.
6. Prove that if A(4,-1), B(5,6), C(1,3), ΔABC is an isosceles right triangle.
7. Prove that the diagonals of a parallelogram bisect each other.
8. Prove that the diagonals of a rhombus are perpendicular.
9. Prove that all 3 medians of a triangle are concurrent.
10. Prove that all 3 perpendicular bisectors of a triangle are concurrent and
that the point of intersection is equidistant from all 3 vertices.
11. Prove that all 3 altitudes of a triangle are concurrent.
1. Given: ΔABC with A:(0, 0), B:(2, 3) and C(4, 0)
Prove: ΔABC is isosceles
AB  22  32  13 , BC  22  32  13 . Since AB = BC, ΔABC is isosceles.
2. Given: A:(1, 1), B:(2, 5), C:(5, 7) and D:(7, 5)
Prove: ABCD is a trapezoid
5 1 4 2
7 5 2
Since the slopes are equal, AD ║ BC, and
  , slope(BC ) 

7 1 6 3
52 3
this makes ABCD a trapezoid.
slope(AD) 
3. Prove that quadrilateral A(1,2), B(2,5),C(5,7) and D(4,4) is a
parallelogram by using slopes.
slope(AB) 
52 3
74 3
  3 , slope(CD) 
  3 . Since the slopes are equal, AB ║ CD.
2 1 1
54 1
5  7 2 2
2  4 2 2

 , slope(AD) 

 . Since the slopes are equal, BC ║
2  5 3 3
1  4 3 3
AD. A quadrilateral with 2 opposite sides parallel is a parallelogram, so ABCD is a ║gram.
slope(BC ) 
4. Prove that A(1,1), B(4,4), C(6,2) are the vertices of a right triangle.
Way #1: AB  32  32  18  3 2 , BC  22  22  8  2 2 , AC  52  12  26
Since AB2  BC 2  18  8  26  AC 2 , by the converse of the Pythagorean Theorem, ΔABC
is a right triangle.
4 1 3
4 2
2
  1 , slope(BC ) 

 1 . Since AB and BC have
4 1 3
4  6 2
opposite reciprocal slopes, these segments are perpendicular, so ΔABC is a right triangle.
Way #2: slope(AB) 
5. Prove that if A(-3,2), B(-2,6), C(2,7) and D(1,3), ABCD is a rhombus.
AB  12  42  17 , BC  42  12  17 , CD  12  42  17 , and AD  42  12  17 .
Since a quadrilateral with 4 congruent sides is a rhombus, ABCD is a rhombus.
6. Prove that if A(4,-1), B(5,6), C(1,3), ΔABC is an isosceles right triangle.
Way #1: AB  12  72  50  5 2 , BC  42  32  25  5 , AC  42  32  25  5
Since BC 2  AC 2  25  25  50  AB2 , by the converse of the Pythagorean Theorem,
ΔABC is a right triangle. Furthermore, since BC = AC, ΔABC is isosceles.
63 3
3  1
4
. Since BC and AC have opposite
 , slope(AC ) 

5 1 4
14
3
reciprocal slopes, these segments are perpendicular, so ΔABC is a right triangle.
Way #2: slope(BC ) 
Furthermore, AC  42  32  25  5 and BC  42  32  25  5 , ΔABC is isosceles.
7. Prove that the diagonals of a parallelogram bisect each other.
Let ABCD be a parallelogram with vertices A:(0, 0), B:(a, 0), D:(b, c) and C:(a+b, c).
(Note, the set-up of a coordinate proof is important. It can affect the difficulty of the proof.)
a b c 
The diagonals of ABCD are AC and BD. The midpoint of AC is 
,  The midpoint of
 2 2
a b c 
BD is 
,  . Since the 2 segments have a common midpoint, they must bisect each
 2 2
other.
8. Prove that the diagonals of a rhombus are perpendicular.
Let ABCD be a rhombus with vertices A:(a, 0), B:(0, b), C:(-a, 0) and D:(0, -b). (Note: This
set-up insures that all sides of ABCD are congruent, because they all have length
a2  b2 .)
The diagonals of ABCD are AC and BD. The slope of AC is 0/2a = 0 and the slope of BD =
2b/0 = undefined. These are slopes of perpendicular lines, so the diagonals are
perpendicular. OR These diagonals lie on the x and y axes which are perpendicular so the
diagonals are perpendicular.
9. Prove that all 3 medians of a triangle are concurrent.
Let ΔABC have vertices A:(0, 0), B:(2m, 2n) and C:(2k, 0) (Note: I’m using these coordinates
because I know that finding medians will require knowing midpoints and I’d like nicer values for the
midpoints.)
The midpoint of AB is (m, n) and let’s call it P. CP is a median. The slope of CP is
so the equation of the line CP is y 
n
(x  2k )
m  2k
The midpoint of AC is (k, 0) and let’s call it R. BR is a median. The slope of BR is
so the equation of the line BR is y 
n
,
m  2k
2n
,
2m  k
2n
(x  k ) .
2m  k
The midpoint of BC is (m + k, n) and let’s call it Q. AQ is a median. The slope of AQ is
n
n
(x ) .
, so the equation of the line CP is y 
mk
mk
Let’s find the intersection of CP and AQ and show that it lies on BR. The intersection of
CP and AQ can be found by solving
n
n
(x  2k ) 
(x ) . First divide both sides by n:
m  2k
mk
(x  2k )
(x )
. Then cross- multiply: (x  2k)(m  k)  (m 2k)(x) . Distribute:

m  2k
m k
xm xk 2km 2k2  mx  2kx  3kx  2km  2k 2  3x  2m  2k so x 
That means y 
2m  2k
.
3
n
n  2m  2k  2n
. So that point of intersection is:
(x ) 


mk
m k 
3
 3
2m  2k
 2m  2k 2n 
,
,  . Furthermore, this point lies on BR because if we let x 

3
3
3


2n  2m  2k
2n  2m  k  2n
. So the medians are
y
k 



2m  k 
3
 2m  k  3  3
concurrent at  2m  2k 2n  and it’s called the centroid.
, 

3
3

10. Prove that all 3 perpendicular bisectors of a triangle are concurrent and
that the point of intersection is equidistant from all 3 vertices.
Let ΔABC have vertices A:(0, 0), B:(2m, 2n) and C:(2k, 0)
The midpoint of AB is (m, n). The slope of AB is
AB, then its slope is
n
. If l1 is a perpendicular bisector of
m
m
m
and its equation is y  n 
(x  m)
n
n
The midpoint of AC is (k, 0). The slope of AC is 0. If l2 is a perpendicular bisector of AB,
then its slope is undefined and its equation is x = k.
The midpoint of BC is (m + k, n) The slope of AC is
of AB, then its slope is
n
. If l3 is a perpendicular bisector
m k
k m
k m
and its equation is y  n 
(x  (m  k )) .
n
n
Let’s find the intersection of l1 and l2 and show it lies on l3. The intersection of l1 and l2 has
x = k and y  n 
 m2  km  n2 
m
km  m2 n2 m2  km  n2
(k  m)  y 


so it is:  k,
.
n
n
n
n
n


If it lies on l3, then if we let x = k:
y n 
y n 
k m
(k  (m  k )) .
n
Distributing, we get
k m
k m
km  m2
km  m2
m2  km  n2
(k  m  k ) 
( m) 
y 
n 
.
n
n
n
n
n
 m2  km  n2 
Therefore:  k,
 is the point of intersection of the perpendicular bisectors and
n


it is called the circumcenter.
11. Prove that all 3 altitudes of a triangle are concurrent. We did this in
class. ;-)