Are the points (1,1),(13,1), (13,6) the points of the vertices of a triangle?
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Answered by
4
Let the points be A(1,1) , B(13,1) and C(13,6).
By Distance Formula,
AB² = (1 - 13)² + (1 - 1)²
= (-12)² + 0²
= 144 + 0
∴AB² = 144
BC² = (13 - 13)² + (1 - 6)²
= 0² + (-5)²
= 0 + 25
∴BC² = 25
AC² = (1 - 13)² + (1 - 6)²
= (-12)² + (-5)²
= 144 + 25
∴AC² = 169
Now, AB² + BC² = 144 + 25
= 169
= AC²
∴AB² + BC² = AC²
Thus, (1,1) , (13,1) and (13,6) are the vertices of a triangle, and that too a right angled triangle.
By Distance Formula,
AB² = (1 - 13)² + (1 - 1)²
= (-12)² + 0²
= 144 + 0
∴AB² = 144
BC² = (13 - 13)² + (1 - 6)²
= 0² + (-5)²
= 0 + 25
∴BC² = 25
AC² = (1 - 13)² + (1 - 6)²
= (-12)² + (-5)²
= 144 + 25
∴AC² = 169
Now, AB² + BC² = 144 + 25
= 169
= AC²
∴AB² + BC² = AC²
Thus, (1,1) , (13,1) and (13,6) are the vertices of a triangle, and that too a right angled triangle.
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Answered by
0
yes these are points are of a right angled triangle triangle.
we can check it using distance formula
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