Geometry: Coordinates and Gradients
The distance between
A
(
5,
–1
)
and
B
(
2,
0
)
is
to 3 dps
Worked Solution
By the Pythagoras Theorem, the distance
AB
is
√((B
x
-A
x
)
2
+ (B
y
-A
y
)
2
)
= √( (2 – 5)
2
+ (0 – (–1))
2
)
= √((–3)
2
+ 1
2
)
= √10
= 3.162
to 3 dps.
©
MEI Produced by Dr Ron Knott, 28 April 2004
Test reference:
GeomCoordsGrads.html?ref=18695,qu=dist
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