Show that the points (4,4), (5,-1), (-6,2) are the vertices of a right angled triangle
Answers
Answer:
let , A(4,4) ;B(5,-1) ;C(-6,2)
by distance formula
AB= root ( (5-4)^2+(-1-4)^2)
=root(1+25)
=root(26)units
BC=root( ( -6-5)^2+(2+1)^2)
=root(121+9)
=root(130)units
AC=root( ( -6-4)^2+(2-4)^2)
=root(121+4)
=root(125)
=5root(5)unit
if abc is a right angled triagle then,
BC^2=AB^2+AC^2
130=26+125
130=151
SO, abc isnot a right sngled triangle
Answer:
let, A=(4,4 )
B=(5,-1)
C=(-6,2)
Now,
Distance AB=√(5-4)²+(-1-4)²
=√1²+(-5)²
=√1+25
=√26
Distance BC=√(-6-5)²+(2+1)²
=√(-11)²+3²
=√121+9
=√140
Distance CA=√(4+6)²+(4-2)²
=√10²+2²
=√100+4
=√104
By Pythagoras theorem,
BC²=AB²+CA²
=>√130²=√26²+√104²
=> 130 = 26+ 104
=> 130=130
Hence Proof
•°•(4,4),(5,-1)&(-6,2) are the vertices of a right angeled traingle
Step-by-step explanation:
1.In distance CA,4-(-6) is written as 4+6 as minus×minus is plus.
2.In Pythagoras Theorem part , we cancel the root with the squares , i.e. ².
Hope you are satisfied with the answers .
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