Without using Pythagoras theorem, show that the points (1,-4),(2,-3) and (4,-7) form a right angled triangle.
Answers
Answer:
†D=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Solution :
Let Point be,
A ( 1 , - 4 )
B ( 2 , - 3 )
C ( 4 , - 7 )
\tt\mapsto AB = \sqrt{ {(2 - 1)}^{2} + {( - 4 + 3)}^{2} }↦AB=
(2−1)
2
+(−4+3)
2
\tt\mapsto AB = \sqrt{ {(1)}^{2} + {( - 1)}^{2} }↦AB=
(1)
2
+(−1)
2
\tt\mapsto AB = \sqrt{ 2} \: units.↦AB=
2
units.
\rule{40}1
\tt \mapsto BC = \sqrt{ {(4 - 2)}^{2} + {( - 7 + 3)}^{2} }↦BC=
(4−2)
2
+(−7+3)
2
\tt \mapsto BC = \sqrt{ {(2)}^{2} + {( - 4)}^{2} }↦BC=
(2)
2
+(−4)
2
\tt \mapsto BC = \sqrt{ 4 + 16 }↦BC=
4+16
\tt \mapsto BC = \sqrt{20 } \: units↦BC=
20
units
\rule{40}1
\tt \mapsto AC = \sqrt{ {(4 - 1)}^{2} + {(7 - 4)}^{2} }↦AC=
(4−1)
2
+(7−4)
2
\tt \mapsto AC = \sqrt{9 + 9 }↦AC=
9+9
\tt \mapsto AC = \sqrt{18 } \: units.↦AC=
18
units.
\rule{100}1
Yes,
AB + AC = BC.
And given point ABC lies with 90° angle.
Verification :
\tt \mapsto H^{2} = P^{2} + B^{2}↦H
2
=P
2
+B
2
\tt \mapsto BC^{2} = AB^{2} + AC^{2}↦BC
2
=AB
2
+AC
2
\tt \mapsto( \sqrt{20})^{2} =( \sqrt{ 2})^{2} + (\sqrt{18 } )^{2}↦(
20
)
2
=(
2
)
2
+(
18
)
2
\tt \mapsto 20 = 2+18↦20=2+18
\tt \mapsto 20= 20.↦20=20.