prove that points (0,0) (5,5) and (-5,5) are the vertices of a right angled isoceles triangle.
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Answered by
52
Hi friend,
Let A(0,0) , B(5,5) and C(-5,5) be the given points.
AB = ✓(5-0)² + (5-0)² = ✓25+25 = ✓50
BC = ✓(-5-5)² + (5-5)² = ✓100 = 10
And,
AC = ✓(-5-0)² + (5-0)² = ✓25+25 = ✓50.
Thus,
AB = AC = ✓50
THEREFORE, ∆ ABC is a isosceles triangle.
Also, (AB² + AC² ) = (✓50)² + (✓50)² = 50+50= 100.
And,
BC² = (10)² = 100.
Thus,
AB² + AC² = BC².
This shows that ∆ ABC is right - angled at C.
Let A(0,0) , B(5,5) and C(-5,5) be the given points.
AB = ✓(5-0)² + (5-0)² = ✓25+25 = ✓50
BC = ✓(-5-5)² + (5-5)² = ✓100 = 10
And,
AC = ✓(-5-0)² + (5-0)² = ✓25+25 = ✓50.
Thus,
AB = AC = ✓50
THEREFORE, ∆ ABC is a isosceles triangle.
Also, (AB² + AC² ) = (✓50)² + (✓50)² = 50+50= 100.
And,
BC² = (10)² = 100.
Thus,
AB² + AC² = BC².
This shows that ∆ ABC is right - angled at C.
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