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AB = 6, BC = 8 and AC = 10. Since 6² + 8² = 10², the triangle is right-angled.
AB = 5, BC = 5 and AC = 10. Since AB + BC = AC, the points are collinear.
Equating squared distances gives (x−2)² = (x−6)², hence x = 4.
Equating the squared distances and simplifying gives x + y = 9.
Let the point be (0,y). Equating distances gives 9+(y−4)² = 25+(y−2)², hence y = 5.
Equating squared distances and simplifying gives x − y = 2.
Midpoint = ((−4+8)/2, (6−2)/2) = (2,2).
Midpoint = ((3−7)/2, (−5+9)/2) = (−2,2).
P = ((3×11+2×1)/5, (3×7+2×2)/5) = (7,5).
P = ((1×5+3×(−3))/4, (1×(−2)+3×4)/4) = (−1,5/2).
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