In the following examples, can the segment joining the given points form a triangle ? If triangle is formed, state the type of the triangle considering sides of the triangle.

(1) Given points are L(6,4) , M(-5,-3) , N(-6,8).
By distance formula LM = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(-5-6)²+(-3-4)²]
= √[(-11)²+(-7)²]
= √[121+49]
= √170
LM = √170 …………(i)
By distance formula MN = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(-6-(-5))²+(8-(-3))²]
= √[(-1)²+(11)²]
= √(1+121)
= √(122)
MN = √122 ……………(ii)
By distance formula LN = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(-6-6)²+(8-4)²]
= √[(-12²+(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 = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(-4-(-2)²+(-2-(-6))²]
= √[(-2)²+(4)²]
= √[4+16]
= √20
= 2√5
PQ = 2√5 …………(i)
By distance formula QR = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(-5-(-4))²+(0-(-2))²]
= √[(-1)²+(2)²]
= √(1+4)
= √(5)
QR = √5 ……………(ii)
By distance formula PR = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(-5-(-2))²+(0-(-6))²]
= √[(-3²+(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.