
If (1, 0), (0, 1), (1, 1) are the coordinates of vertices of a triangle. The triangle is _______ triangle.
(a) Isosceles
(b) Obtuse angled
(c) Acute angled
(d) Equilateral
Answer
600.6k+ views
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.
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.
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.
Recently Updated Pages
In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

In cricket, what is a "tail-ender"?

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

The draft of the Preamble of the Indian Constitution class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

How many members did the Constituent Assembly of India class 10 social science CBSE

Write an application to the principal requesting five class 10 english CBSE

The Constitution of India was adopted on A 26 November class 10 social science CBSE

