SME
Grade 10 Math · Problem type
Prove a right triangle with slopes
Use this for “Is the triangle A(−1, 1), B(1, 5), C(5, 3) a right triangle?”
Chart
Slopes: AB 2, BC −1/2, CA 1/3.
B: right angle, the sides there perpendicular (2 × −1/2 = −1).
This is an example. Type your own values to start a new problem.
xA
x of A
example
yA
y of A
example
xB
x of B
example
yB
y of B
example
xC
x of C
example
yC
y of C
example
mAB
Slope of AB
worked out (not typed)
mBC
Slope of BC
worked out (not typed)
mAC
Slope of AC
worked out (not typed)
p
Product at B, mAB × mBC
worked out (not typed)
q
Product at C, mBC × mAC
worked out (not typed)
w
Product at A, mAB × mAC
worked out (not typed)
z
Corner with the right angle
worked out (not typed)
Formulas
mAB
=
yB − yA
xB − xA
2
=
5 − 1
1 − (−1)
mBC
=
yC − yB
xC − xB
−
1
2
=
3 − 5
5 − 1
mAC
=
yC − yA
xC − xA
1
3
=
3 − 1
5 − (−1)
p
=
mAB
×
mBC
−1
=
2
×
(−
1
2
)
q
=
mBC
×
mAC
−
1
6
=
−
1
2
×
1
3
w
=
mAB
×
mAC
2
3
=
2
×
1
3
z
:
the
corner
where
p
,
q
or
w
is
−1
−1 = −1, so B
Assumptions
•
Two sides are perpendicular when their slopes multiply to −1.
•
A vertical side has no slope; it is perpendicular to a level side (slope 0).
•
Check the two slopes at each corner: a product of −1 is a right angle there.
Step-by-step
WE KNOW
xA = −1, yA = 1, xB = 1, yB = 5, xC = 5, yC = 3
FIND
slope of AB (mAB), slope of BC (mBC), slope of AC (mAC), product at B, mAB × mBC (p), product at C, mBC × mAC (q), product at A, mAB × mAC (w), corner with the right angle (z)
Step 1 · Find slope of AB (mAB)
Use mAB = (yB − yA) ÷ (xB − xA)
Rise over run from A to B.
mAB
=
5 − 1
1 − (−1)
mAB
=
4
2
mAB = 2
Step 2 · Find slope of BC (mBC)
Use mBC = (yC − yB) ÷ (xC − xB)
Rise over run from B to C.
mBC
=
3 − 5
5 − 1
mBC
=
−
2
4
mBC
=
−
1
2
Step 3 · Find slope of AC (mAC)
Use mAC = (yC − yA) ÷ (xC − xA)
Rise over run from A to C.
mAC
=
3 − 1
5 − (−1)
mAC
=
2
6
mAC
=
1
3
Step 4 · Find product at B, mAB × mBC (p)
Use p = mAB × mBC
Multiply the two slopes at B: −1 means a right angle.
p
=
2
×
(−
1
2
)
p
=
−1
Step 5 · Find product at C, mBC × mAC (q)
Use q = mBC × mAC
Multiply the two slopes at C: −1 means a right angle.
q
=
−
1
2
×
1
3
q
=
−
1
6
Step 6 · Find product at A, mAB × mAC (w)
Use w = mAB × mAC
Multiply the two slopes at A: −1 means a right angle.
w
=
2
×
1
3
w
=
2
3
Step 7 · Find corner with the right angle (z)
Use z: the corner where p, q or w is −1
p
=
−1:
AB
⊥
BC,
so
the
right
angle
is
at
B.
−1 = −1: B
z
=
the
corner
where
−1,
−
1
6
or
2
3
is
−1
z
=
B
Check
Put the numbers back in. Both sides should match.
2
=
5 − 1
1 − (−1)
✓
−
1
2
=
3 − 5
5 − 1
✓
1
3
=
3 − 1
5 − (−1)
✓
−1
=
2
×
(−
1
2
)
✓
−
1
6
=
−
1
2
×
1
3
✓
2
3
=
2
×
1
3
✓
−1 = −1, so B ✓
Topic check
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