Answer:
a) Acute triangle
b) Right triangle
Step-by-step explanation:
∵ A triangle having sides a, b and c is called,
Acute : If a² + b² > c² or a² + c² > b² or b² + c² > a²,
Obtuse : if a² + b² < c² or a² + c² < b² or b² + c² < a²
Right : a² + b² = c² or a² + c² = b² or b² + c² = a²,
a) Let A≡(-1, 1), B≡(7,-2) and C≡(1,-5),
By the distance formula,

Similarly,


∵ The sum of any two sides is greater than third side,
So, ABC is a triangle,
Now,

⇒ ABC is an acute triangle.
b) Let P≡(0,6), Q≡(1,2) and R≡(5,3),
By the distance formula,
PQ = √17 unit,
QR = √17 unit,
RP = √34 unit,
∵ The sum of any two sides of PQR is greater than third side,
⇒ PQR is a triangle ,
Also,

Hence, PQR is a right triangle.