Question

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

Answer 1

Let the points$(1,5),(2,3),and(-2,-11)$ be represented as $A,B,andC.$

They must lie on the same line for A, B, and C to be collinear.

Hence, we will have to check$\mathrm{if}\mathrm{AB}+\mathrm{BC}=\mathrm{AC}\mathrm{or}\mathrm{BC}+\mathrm{AC}=\mathrm{AB}\mathrm{or}\mathrm{AB}+\mathrm{AC}=\mathrm{BC}.$

We know that the distance between any two points is given by,

Distance Formula = $\sqrt{{(x₂-x₁)}^2+{(y₂-y₁)}^2}....\left(1\right)$

To find AB, the Distance between the Points$A(1,5)andB(2,3)$,

Let, $x₁=1,y₁=5,x₂=2,y₂=3$

$\therefore AB=\sqrt{{(2-1)}^2+{(3-5)}^2}$ (By Substituting in (1))

$AB=\sqrt{5}$

Now, Distance between Points $B(2,3)andC(-2,-11),letx₁=2,y₁=3,x₂=-2,y₂=-11$

Therefore, $BC=\sqrt{{(-2-2)}^2+{(-11-3)}^2}$

$=\sqrt{{(-4)}^2+{(-14)}^2}$ (By Substituting in the Equation (1))

$=\sqrt{16+196}$

$=\sqrt{212}$

Now, Distance between Points $A(1,5)andC(-2,-11),letx₁=1,y₁=5,x₂=-2,y₂=-11$

Therefore,$CA=\sqrt{(-2-1)²+(-11-5)²}$

$=\sqrt{(-3)²+(-16)²}$ (By Substituting in the Equation (1))

$=\sqrt{9+256}$

$=\sqrt{265}$

$AB=\sqrt{5},BC=\sqrt{212},CA=\sqrt{265}$

Since $\mathrm{AB}+\mathrm{AC}\ne \mathrm{BC}\mathrm{and}\mathrm{BC}+\mathrm{AC}\ne \mathrm{AB}\mathrm{and}\mathrm{AB}+\mathrm{BC}\ne \mathrm{AC}$,

Hence, the points $(1,5),(2,3),and(-2,-11)$ are not collinear.