Prove that A(0,7), B(4, 3), C(6,5) form the vertices of a right-angled triangle.
Answers
Answer:
Consider the problem
Given points are
A(7,5),B(2,3),C(6,−7)
Let the given lines of triangle be AB,BC,AC
Let , By distance formula
Distance of AB
AB=
(3−5)
2
+(2−7)
2
=
(−2)
2
+(−5)
2
=
4+25
=
29
Similarly,
BC=
(−7−3)
2
+(6−2)
2
=
100+16
=
116
AC=
(−7−5)
2
+(6−7)
2
=
145
Now, Pythagoras theorem
AC
2
=BC
2
+AB
2
(
145
)
2
=(
116
)
2
+(
29
)
2
145=116+29
145=145
L.H.S.=R.H.S
Hence, prove that they are at right angles.
Concept
A right angled triangle is a triangle that has one of its interior angle as 90.
Given
A(0,7), B(4, 3), C(6,5)
Find
Prove these three points are vertices of right angled triangle
Solution
Points
A(0,7), B(4, 3), C(6,5)
Let the given sides of triangle be AB, BC, AC
Using Distance Formula
Distance = √((x₂-x₁)² + (y₂-y₁)²)
Distance² = (x₂-x₁)² + (y₂-y₁)²)
We will be calculating Sides square
AB² = (4-0)² + (3-7)²
AB² = 16 + 16
AB² = 32
Similarily Calculating BC² and AC²
BC² = (6-4)² + (5-3)²
BC² = 4 + 4
BC² = 8
AC² = (6-0)² + (5-7)²
AC² = 36 +4
AC² = 40
So according to property of right angled triangle AC² = AB² + BC²
Hence, A(0,7), B(4, 3), C(6,5) form the vertices of a right-angled triangle.
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