State True/ False:The graph represents the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3,1),B(6,4)   and C(8,6)   respectively. And A, B and C points are in line.

State True/ False:The graph represents the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at   and   respectively. And A, B and C points are in line.

1. A
True
2. B
False

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Solution:

Consider the given question,
Here,
We know that,
If AB + BC = AC, then the points are collinear.
So, let’s calculate AB, BC and AC.
Using the distance formula;
AB = $\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
= $\sqrt{{\left(6-3\right)}^{2}+{\left(4-1\right)}^{2}}$
= $\sqrt{{3}^{2}+{3}^{2}}$
= $\sqrt{{2\left(3\right)}^{2}}$
= $\sqrt{2}\left(3\right)$
= $3\sqrt{2}$
Thus, AB is $3\sqrt{2}$.
Similarly, calculating BC,
BC =
= $\sqrt{{\left(8-\left(6\right)\right)}^{2}+{\left(6-4\right)}^{2}}$
= $\sqrt{{\left(2\right)}^{2}+{\left(2\right)}^{2}}$
Solving further,
= $2\sqrt{2}$
Thus, BC is $2\sqrt{2}$.
Calculating AC;
AC =
= $\sqrt{{\left(8-\left(3\right)}^{2}+{\left(6-1\right)}^{2}}$
= $\sqrt{{\left(5\right)}^{2}+{\left(5\right)}^{2}}$
= $\sqrt{2{\left(5\right)}^{2}}$
= $5\sqrt{2}$
Now, finding if AB + BC = AC            ……(1)
= $3\sqrt{2}$ + $2\sqrt{2}$  = $5\sqrt{2}$
Thus, AB + BC = AC
Therefore, the points A, B and C are collinear.
Hence, option 1 is true.

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