Question

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

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

★Assumption :-

P(1 , 5)

Q(2 , 3)

R(-2 , -11)

★They are the given points

★Hence we have :-

$\large{\sf\:{PQ=\sqrt{(2-1)^2+(3-5)^2}}}$

$\large{\sf\:{PQ=\sqrt{(1)^2+(-2)^2}}}$

$\large{\sf\:{PQ=\sqrt{1+4}}}$

$\large{\sf\:{PQ=\sqrt{5}}}$

★Now :-

$\large{\sf\:{QR=\sqrt{(-2-2)^2+(-11-3)^2}}}$

$\large{\sf\:{QR=\sqrt{(-4)^2+(-14)^2}}}$

$\large{\sf\:{QR=\sqrt{16+196}}}$

$\large{\sf\:{QR=\sqrt{212}}}$

$\large{\sf\:{QR=\sqrt{4\times 53}}}$

$\large{\sf\:{QR=2\sqrt{53}}}$

★Now :-

$\large{\sf\:{PR=\sqrt{(-2-1)^2+(-11-5)^2}}}$

$\large{\sf\:{PR=\sqrt{(-3)^2+(-16)^2}}}$

$\large{\sf\:{PR=\sqrt{9+256}}}$

$\large{\sf\:{PR=\sqrt{265}}}$

In above cases we get :-

QR ≠ PQ + PR

PQ ≠ QR + PR

PR ≠ QR

$\fbox{Therefore\;we\;get - P, Q , R\;are\;not\;collinear}$

Answer 2

Answer:

Step-by-step explanation:

Let A=(1,5),B=(2,3) and C=(-2,-11)

By applying distance formula we get,

Distance formula=$\sqrt(x_{2}-x_{1})+(y_{2} -y_{1})$

AB=√(2-1)²+(3-5)²=√(1)²+(-2)²=√1+4=√5

BC=√(-2-2)²+(-11-3)²=√(-4)²+(-14)²=√16+196=√212=2√53

CA=√(-2-1)²+(-11-5)²=√(-3)²+(-16)²=√9+256=√265=

Since AB+AC≠BC,BC+AC≠AB and AC≠BC.

Therefore,the points A,B and C are not collinear.

Therefore,please mark it as brainlist answer