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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.

yes these are points are of a right angled triangle triangle.

we can check it using distance formula