Question

detemine if the points
$(1.5)(2.3)and( - 2. - 11)are coliner$
​

Answer 1

Correct Question:-

Determine if the points (1,5), (2,3), (-2,-11) are collinear.

GiveN:-

• (1,5)
• (2,3)
• (-2,-11)

To FinD:-

To determine are the points collinear or not.

SolutioN:-

Let the three points be A, B and C.

Collinear points are the points which form on the same line.

There are three cases possible :

✝ Case 1 :

A, B, C are collinear if,

AB + BC = AC

✝ Case 2 :

A, B, C are collinear if,

BA + AC = BC

✝ Case 3 :

A, B, C are collinear if,

BC + CA = BA

If neither case 1, case 2 or case 3 is not satisfied then the points are not collinear.

Let the points be :

• A(1,5)
• B(2,3)
• C(-2,-11)

To FinD the AB :

$\normalsize{\underline{\boxed{\sf{Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}}}$

where,

• x₁ = 1
• x₂ = 2
• y₁ = 5
• y₂ = 3

Substituting the values in the required formula,

$\\ : \normalsize\implies{\sf{Distance=\sqrt{(2-1)^2+(3-5)^2}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{(1)^2+(-2)^2}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{1+4}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{5}}}$

∴ Distance of AB = √5

_________________________________

To FinD the BC :

$\normalsize{\underline{\boxed{\sf{Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}}}$

where,

• x₁ = 2
• x₂ = -2
• y₁ = 3
• y₂ = -11

Substituting the values in the required formula,

$\\ :\normalsize\implies{\sf{Distance=\sqrt{(-2-2)^2+(-11-3)^2}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{(-4)^2+(-14)^2}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{16+196}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{212}}}$

∴ Distance of BC = √212

_________________________________

To FinD the AC :

$\normalsize{\underline{\boxed{\sf{Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}}}$

where,

• x₁ = 1
• x₂ = -2
• y₁ = 5
• y₂ = -11

Substituting the values in the required formula,

$\\ :\normalsize\implies{\sf{Distance=\sqrt{(-2-1)^2+(-11-5)^2}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{(-3)^2+(-16)^2}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{9+256}}}$

$\\ :\normalsize\implies{\sf{Distance=\sqrt{265}}}$

∴ Distance of AC = √265

_________________________________

Now we have to check the three cases:-

We have AB = √5, BC = √212 and AC = √265

✝ Case 1 :

AB + BC = AC

where,

• AB = √5
• BC = √212
• AC = √265

Substituting the values,

⇒ √5 + √212 = √265

⇒ √217 = √265

⇒ √217 ≠ √265

∴ LHS ≠ RHS

• Hence case 1 is not true.

✝ Case 2 :

BA + AC = BC

where,

• BA = √5
• BC = √212
• AC = √265

Substituting the values,

⇒ √5 + √265 = √212

⇒ √270 = √212

⇒ √270 ≠ √212

∴ LHS ≠ RHS

• Hence case 2 is not true.

✝ Case 3 :

BC + CA = BA

where,

• BA = √5
• BC = √212
• CA = √265

Substituting the values,

⇒ √212 + √265 = √5

⇒ √477 = √5

⇒ √477 ≠ √5

∴ LHS ≠ RHS

• Hence case 3 is not true.

Since all three cases are not satisfied the three points are not collinear.

A(1,5), B(2,3) and C(-2,-11) are not collinear.