(1) Given points are L(6,4) , M(-5,-3) , N(-6,8).

By distance formula LM = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

= √[(-5-6)^{2}+(-3-4)^{2}]

= √[(-11)^{2}+(-7)^{2}]

= √[121+49]

= √170

LM = √170 …………(i)

By distance formula MN = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

= √[(-6-(-5))^{2}+(8-(-3))^{2}]

= √[(-1)^{2}+(11)^{2}]

= √(1+121)

= √(122)

MN = √122 ……………(ii)

By distance formula LN = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

= √[(-6-6)^{2}+(8-4)^{2}]

= √[(-12^{2}+(4)^{2}]

= √(144+16)

= √(160)

LN = √160 …………..(iii)

(MN +LN ) LM

These points are not collinear.

We can construct a triangle through 3 non collinear points.

LM ≠ MN ≠ LN

Triangle formed is a scalene triangle.

(2) Given points are P(-2,-6) , Q(-4,-2), R(-5,0)

By distance formula PQ = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

= √[(-4-(-2)^{2}+(-2-(-6))^{2}]

= √[(-2)^{2}+(4)^{2}]

= √[4+16]

= √20

= 2√5

PQ = 2√5 …………(i)

By distance formula QR = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

= √[(-5-(-4))^{2}+(0-(-2))^{2}]

= √[(-1)^{2}+(2)^{2}]

= √(1+4)

= √(5)

QR = √5 ……………(ii)

By distance formula PR = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

= √[(-5-(-2))^{2}+(0-(-6))^{2}]

= √[(-3^{2}+(6)^{2}]

= √(9+36)

= √(45)

= 3√5

PR = 3√5 …………..(iii)

Add (i) and (ii)

2√5+√5 = 3√5

d(P,Q)+d(Q,R) = d(P,R)

P,Q,R are collinear points.

So we cannot construct a triangle with these collinear points.