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Hint: Take the points (1, 0), (0, 1), (1, 1) as A, B, C respectively, obtains the values of AB, BC and CA using formula \[\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}\], if \[\left( {{x}_{1}},{{x}_{2}} \right)\] and \[\left( {{y}_{1}},{{y}_{2}} \right)\] are points. Then try to find similarities and relations to get desired results.

Let us take A as (1,0), B as (0,1) and C as (1,1).

Then we will find length of AB, BC and CA using following formula:

If points are $\left( {{x}_{1}},{{y}_{2}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then its distance is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .

So, if A is (1,0) and B is (0,1), then

\[AB=\sqrt{{{\left( 0-1 \right)}^{2}}+{{\left( 1-0 \right)}^{2}}}=\sqrt{1+1}\]

Hence, $AB=\sqrt{2}$

Now if B is (0,1) and C is (1,1), then

\[BC=\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 1-1 \right)}^{2}}}=\sqrt{{{1}^{2}}+{{0}^{2}}}\]

Hence, BC = 1.

Now, if C is (1,1) and A is (1,0), then

\[CA=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 0-1 \right)}^{2}}}=\sqrt{{{0}^{2}}+{{1}^{2}}}\]

Hence, CA = 1.

So, we can say that $BC=CA\ne AB$.

Therefore, we can say that it is an isosceles triangle and not an equilateral triangle because in isosceles triangle two sides are equal.

Now we will apply Pythagoras theorem,

$\begin{align}

& B{{C}^{2}}+C{{A}^{2}}=A{{B}^{2}} \\

& {{1}^{2}}+{{1}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\

\end{align}$

2 = 2.

Here the left hand side is equal to the right hand side. So, the triangle ABC is a right angled triangle.

Therefore, triangle ABC is an acute angled isosceles triangle and not an obtuse angled triangle.

Hence, the correct answer is option (a) and (c).

Note: Another approach is by plotting the points (1, 0), (0, 1) and (1,1) on the graph and joining the points one will just compare the lengths of triangles to understand what will be the answer. Students often stop their solution once they get to know that the triangle is an isosceles triangle. We should check for acute and obtuse angled conditions too.

__Complete step-by-step answer:__Let us take A as (1,0), B as (0,1) and C as (1,1).

Then we will find length of AB, BC and CA using following formula:

If points are $\left( {{x}_{1}},{{y}_{2}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then its distance is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .

So, if A is (1,0) and B is (0,1), then

\[AB=\sqrt{{{\left( 0-1 \right)}^{2}}+{{\left( 1-0 \right)}^{2}}}=\sqrt{1+1}\]

Hence, $AB=\sqrt{2}$

Now if B is (0,1) and C is (1,1), then

\[BC=\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 1-1 \right)}^{2}}}=\sqrt{{{1}^{2}}+{{0}^{2}}}\]

Hence, BC = 1.

Now, if C is (1,1) and A is (1,0), then

\[CA=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 0-1 \right)}^{2}}}=\sqrt{{{0}^{2}}+{{1}^{2}}}\]

Hence, CA = 1.

So, we can say that $BC=CA\ne AB$.

Therefore, we can say that it is an isosceles triangle and not an equilateral triangle because in isosceles triangle two sides are equal.

Now we will apply Pythagoras theorem,

$\begin{align}

& B{{C}^{2}}+C{{A}^{2}}=A{{B}^{2}} \\

& {{1}^{2}}+{{1}^{2}}={{\left( \sqrt{2} \right)}^{2}} \\

\end{align}$

2 = 2.

Here the left hand side is equal to the right hand side. So, the triangle ABC is a right angled triangle.

Therefore, triangle ABC is an acute angled isosceles triangle and not an obtuse angled triangle.

Hence, the correct answer is option (a) and (c).

Note: Another approach is by plotting the points (1, 0), (0, 1) and (1,1) on the graph and joining the points one will just compare the lengths of triangles to understand what will be the answer. Students often stop their solution once they get to know that the triangle is an isosceles triangle. We should check for acute and obtuse angled conditions too.

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