Prove that the points (3, 0), (6, 4) and (- 1, 3) are vertices of a right-angled isosceles triangle.
Answers
Given : Points (3, 0), (6, 4) and (- 1, 3) are vertices of a triangle.
To prove : Vertices of a right-angled isosceles triangle
Solution :
Let A(3, 0), B(6, 4) and C (- 1, 3)
By using distance formula : √(x2 - x1)² + (y2 - y1)²
Vertices : A(3, 0), B(6, 4)
Length of side AB = √(6 - 3)² + (4 - 0)²
AB = √3² + (4)²
AB = √9 + 16
AB = √25 units
Vertices : B(6, 4) and C (- 1, 3)
Length of side BC = √(- 1 - 6)² + ( 3 - 4)²
BC = √(-7)² + (-1)²
BC = √49 + 1
BC = √50 units
Vertices : A(3, 0), C (- 1, 3)
Length of side AC = √(- 1 - 3)² + (3 - 0)²
AC = √(-4)² + (3)²
AC = √16 + 9
AC = √25 units
Since the 2 sides AB = AC = √25 .
Therefore ∆ is an isosceles.
Now, in ∆ABC, by using Pythagoras theorem
BC² = AB² + AC²
(√50)² = (√25)². + (√25)²
50 = 25 + 25
50 = 50
Since BC² = AB² + AC²
Hence, the given triangle is a right angled triangle.
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Answer:
Prove that the points (3,0) (6,4) (-1,3) are the vertices of a right angle ... CA =root {((3+1)^2+(-3)^2}=5. which is related to Pythagoras theorem . hence , ABC is the vertices of right angle isoceles triangle . in which A is right angle .