Verify the vertices (2, 3), (7,-1) and (-1.3) form a right angled triangle.
Answers
Step-by-step explanation:
Given Question:-
Verify the vertices (2, 3), (7,-1) and (-1.3) form a right angled triangle.
Correct question:-
Verify the vertices (2, 3), (7,-1) and (-1,3) form a right angled triangle.
Given:-
The vertices (2, 3), (7,-1) and (-1,3)
To show:-
Verify the vertices (2, 3), (7,-1) and (-1,3) form a right angled triangle.
Solution:-
Given vertices are (2,3) ; (7,-1);(-1,3)
Let A=(2,3)
B=(7,-1)
C=(-1,3)
To show that the given vertices form a right angled triangle then we have to show that
"The square of any side is equal to the sum of the squares of other two sides"
And we know that
The distance between the two points(x1,y1) and (x2,y2) is √{(x2-x1)²+(y2-y1)²} units
Now,
Distance between A and B:-
Let (x1, y1)=A(2,3) =>x1=2 and y1=3
(x2,y2)=B(7,-1)=>x2=7 and y2=-1
AB=√{(x2-x1)²+(y2-y1)²} units
=>AB=√{(7-2)²+(-1-3)³}
=>AB=√{5²+(-4)²}
=>AB=√(25+16)
=>AB=√41
On squaring both sides
AB²=41 units----------(1)
Distance between B and C:-
Let (x1, y1)=B(7,-1)=>x1=7 and y1=-1
(x2,y2)=C(-1,3)=>x2=-1 and y2=3
BC=√{(x2-x1)²+(y2-y1)²} units
=>BC=√{(-1-7)²+(3-(-1))²}
=>BC=√{(-8)²+(4)²}
=>BC=√(64+16)
=>BC=√80
on squaring both sides
BC²=80 units ----------(2)
Distance between C and A:-
Let (x1,y1)=C(-1,3)=>x1=-1 and y1=3
(x2, y2)=A(2,3)=>x2=2 and y2=3
CA=√{(x2-x1)²+(y2-y1)²} units
=>CA=√[{(2-(-1)}²+(3-3)²]
=>CA=√{3²+0²}
=>CA=√9
on squaring both sides
=>CA²=9 units----------(3)
From (1),(2)&(3)
It is clearly that
The sum of the squares of any two sides is not equal to the square of the third side .
Answer:-
Given vertices (2, 3), (7,-1) and (-1.3) does not form a right angled triangle.
Using Concept:-
- In a right angled triangle,"The square of the Hypotenuse is equal to sum of the squares of other two sides".This is Pythagoras theorem.
- The distance between the two points (x1, y1) and (x2, y2) is √{(x2-x1)²+(y2-y1)²} units
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