Question

show that the following point are collinear:(0,1),(1,2) and (-2,-1)​

Answer 1

$\huge\bf\underline \green{Solution:-}$

If these three points are Collinear then the area of triangle formed by these three points is 0.

Area of triangle = $\bf\pink{\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}$

Let the points be A (0,1) ,B(1,2) , C(-2-1)

• x1 = 0 , x2 = 1 , x3 = -2
• y1 = 1 , y2 = 2 , y3 = -1

$: \implies \rm\: \small\frac{1}{2}[0(2 - ( - 1)) + 1( - 1 - 1) - 2(1 - 2)] \\ \\ : \implies \rm\: \frac{1}{2}[0(3) + 1( - 2) - 2( - 1)] \\ \\: \implies \rm\: \frac{1}{2}[0 - \cancel2 + \cancel2] \\ \\ : \implies \rm\: \frac{1}{2}[0] \\ \\: \implies \rm\: 0$

Since,

Area becomes zeroes then,

(0,1),(1,2) and (-2,-1) are collinear points.

Other method :-

________________________

B⠀⠀⠀⠀⠀⠀A⠀⠀⠀⠀⠀⠀⠀C⠀⠀

A (0,1) ,B(1,2) , C(-2-1)

AB = √(1-0)²+(2-1)²

➝ AB = √1 +1

➝ AB = √2 units

AC = √(-2-0)²+(-1-1)²

➝ AC = √4 +4

➝ AC = √8

➝ AC = 2√2

BC = √(-2-1)²+(-1-2)²

➝ BC = √9+9

➝ BC = √18

➝ BC = 3√2 units

for collinear points AB +AC = BC

➝ √2 + 2√2 = 3√2

➝ 3√2 = 3√2

Thus, AB + AC = BC.

Hence ,the given points are Collinear.

Answer 2

$\red{\underline{\underline{Answer:}}}$

$\orange{Given:}$

$\sf{Co-ordinates \ of \ points \ are:-}$

$\sf{\implies{(01);}}$

$\sf{\implies{(1,2);}}$

$\sf{\implies{(-2-1)}}$

$\sf\pink{To \ prove:}$

$\sf{Points \ are \ collinear.}$

$\sf\green{Proof:}$

$\sf{Let \ given \ points \ be}$

$\sf{A(0,1); \ B(1,2) \ and \ C(-2,-1)}$

$\sf{Method \ (I)}$

$\sf{If \ points \ are \ collinear,}$

$\sf{Then \ slope \ of \ line \ formed \ by \ joining}$

$\sf{the \ line \ will \ be \ same.}$

$\sf\boxed{Slope \ of \ line(m)=\frac{y2-y1}{x2-x1}}$

$\sf{\therefore{Slope \ of \ AB=\frac{1-0}{2-1}}}$

$\sf{\therefore{Slope \ of \ AB=1...(1)}}$

$\sf{Slope \ of \ BC=\frac{-1-2}{-2-1}}$

$\sf{\therefore{Slope \ of \ BC=1...(2)}}$

$\sf{Slope \ of \ AC=\frac{-1-1}{-2-0}}$

$\sf{\therefore{Slope \ of \ AC=1...(3)}}$

$\sf{...from \ (1), \ (2) \ and \ (3)}$

$\sf{Slopes \ of \ AB, \ BC \ and \ AC}$

$\sf{are \ equal.}$

$\sf\purple{\tt{\therefore{Points \ (0,1); \ (1,2) \ and \ (-2,-1)}}}$

$\sf\purple{\tt{are \ collinear.}}$

$\sf{Method \ (II)}$

$\sf\boxed{Distance \ formula=\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}}$

$\sf{\therefore{d(A,B)=\sqrt(1-0)^{2}+(2-1)^{2}}}}$

$\sf{\therefore{d(A,B=\sqrt2...(1)}}$

$\sf{d(B,C)=\sqrt{(-2-1)^{2}+(-1-2)^{2}}}$

$\sf{d(B,C)=\sqrt{9\times2}}$

$\sf{d(B,C)=3\sqrt2...(2)}$

$\sf{d(A,C)=\sqrt{(0-2)^{2}+(1+1)^{2}}}$

$\sf{d(A,C)=\sqrt{4\times2}}$

$\sf{d(A,C)=2\sqrt2...(3)}$

$\sf{If \ maximum \ distance=sum \ of other \ distances}$

$\sf{Then \ points \ are collinear}$

$\sf{3\sqrt2 \ is \ maximum \ distance.}$

$\sf{2\sqrt2+\sqrt2=3\sqrt2}$

$\sf{...from \ (1), \ (2) \ and \ (3)}$

$\sf{d(B,C)=d(A,B)+d(A,C)}$

$\sf\purple{\tt{\therefore{Points \ (0,1); \ (1,2) \ and \ (-2,-1)}}}$

$\sf\purple{\tt{are \ collinear.}}$