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Answer:

Answer:

by distance formula,

AB= √(-2-7)^2+(5-10)^2

= √81+25

= √106

BC= √(3+2)^2+(-4-5)^2

= √25+81

= √106

AC= √(3-7)^2+(-4-7)^2

= √16+196

= √ 212

By Pythagoras theorem,

AB^2 + BC^2= 106+106= 212 => AC^2

Therefore, ΔABC is a right-angle triangle.

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