prove that the points A(1,-3) B(-3,0)C(4,1) are the vertices of a right isosceles triangle (hint prove that in fig AB=BC AND AC^2=AB^2+BC^2)
Answers
Answered by
7
Given:
Vertices A (1,-3) , B(-3,0) and C (4,1)
Answer:
Using distance formula
√[(x2-x1)²+(y2-y1)²]
AB=√ [ (−3−1)²+(0−(−3))² ]
=√ [ (−4)²+(0+3)²]
=√ (16+9)
=√25
AB = 5
BC = √ [ (4-(-3))²+(1-0)² ]
= √ [ (4+3)²+ 1 ]
= √ [ (7)²+1 ]
= √ (49+1)
= √ (50)
BC= 5√2
AC = √ [ (4-1)² + (1-(-3))² ]
= √ [(3)²+ (1+3)² ]
=√ [ (9)+ (4)² ]
=√ (9+ 16)
=√25
AC = 5
AB = AC = 5
As two sides of the given triangle are equal, it is an isosceles trianlge.
Now consider,
AB² + AC² = BC²
5² + 5² = (5√2)²
25 +25 = (25×2) 《(5)² × (√2)²》
50 = 50
Since it satisfied pythagoras theorem, it is a right angled isosceles trianlge.
If it helps you then Mark as brainliest.
Similar questions