Question

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) L(6,4) , M(-5,-3) , N(-6,8)
(2) P(-2,-6) , Q(-4,-2), R(-5,0)
(3) A( √2,√2 ), B( -√2,-√2 ), C( -√6,√6)

Answer 1

(1) L ≡ (6, 4) , M ≡ (-5, -3) and N ≡ (-5, 0)
find side length LM , MN and NL
LM = $\bf{\sqrt{(-5-6)^2+(-3-4)^2}=\sqrt{(-11)^2+(-7)^2}=\sqrt{170}}$
MN = = $\bf{\sqrt{(-5+5)^2+(0+3)^2}=3}$
NL = $\bf{\sqrt{(-5-6)^2+(0-4)^2}=\sqrt{121+16} = \sqrt{137}}$

Here you see, LM ≠ MN ≠ NL
so, LMN is scalene triangle.

(2) P ≡ (-2,-6) , Q ≡ ( -4, - 2) and R ≡ (-5,0)
PQ = $\bf{\sqrt{(-4+2)^2+(-2+6)^2}=\sqrt{4+16}=\sqrt{20}}$
QR = $\bf{\sqrt{(-5+4)^2+(0+2)^2}=\sqrt{5}}$
RS = $\bf{\sqrt{(-5+2)^2+(0+6)^2}=\sqrt{9+36}=\sqrt{45}}$
Here, you see PQ , QR and RS follow Pythagoras triplet
So, ∆PQR is right angle triangle.

(3) A ≡(√2, √2) , B ≡ (-√2, -√2) and C ≡ (-√6 , √6)
AB = $\bf{\sqrt{(-\sqrt{2}-\sqrt{2})^2+(-\sqrt{2}-\sqrt{2})^2}=4}$
BC = $\bf{\sqrt{(-\sqrt{6}+\sqrt{2})^2+(\sqrt{6}+\sqrt{2})^2}=4}$
CA = $\bf{\sqrt{(-\sqrt{6}-\sqrt{2})^2+(\sqrt{6}-\sqrt{2})^2}=4}$
Here, you see AB = BC = CA
So, ABC is equilateral triangle

Answer 2

Answer:

In the following example, can the segment joining the given point form a triangle? If a triangle is formed, state the type of the triangle considering the side of the triangle.L(6, 4), M(–5, –3In the following example, can the segment joining the given point form a triangle? If a triangle is formed, state the type of the triangle considering the side of the triangle.L(6, 4), M(–5, –3

Step-by-step explanation: