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Q. Let A (a ,0), B (3a ,0) and C (x ,y) be the vertices of a triangle ABC. If a, x, & y are integers then 

   (A) ABC can be an equilateral triangle                                  (B) ABC cannot be equilateral triangle

   (C) ABC cannot be right angled triangle                                (D) none of these

Dear Student,

Please find below the solution to the asked query:

We have : Let A ( a ,0 ), B ( 3a ,0) and C ( x , y ) be the vertices of a triangle ABC. 

We know distance formula :  d  = $\sqrt{{\left(x_2 - x_1\right)}^2 + {\left(y_2 - y_1\right)}^2}$

For AB we get : x₁ = a , y₁ = 0 and  x₂ = 3 a , y₁ = 0 , So 

AB  = $\sqrt{{\left( 3 \mathrm{a} - \mathrm{a}\right)}^2 + {\left(0 - 0\right)}^2} = \sqrt{{\left(2 \mathrm{a}\right)}^2}= \mathbf{2}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{ }\mathbf{unit}$

For AC we get : x₁ = a , y₁ = 0 and  x₂ = x , y₁ = y , So 

AC  = $\sqrt{{\left(x - a\right)}^2 + {\left(y - 0\right)}^2} = \sqrt{{\left(\mathbf{x}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{a}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{y}}^{\mathbf{2}}}\mathbf{ }\mathbf{ }\mathbf{unit}$

And for BC we get : x₁ = 3 a , y₁ = 0 and  x₂ = x , y₁ = y , So 

BC  = $\sqrt{{\left(x - 3a\right)}^2 + {\left(y - 0\right)}^2} = \sqrt{{\left(\mathbf{x}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{3}\mathbf{a}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{y}}^{\mathbf{2}}}\mathbf{ }\mathbf{ }\mathbf{unit}$

If we assume given triangle is a equilateral , Then we can say :  AB = AC  =  BC , So

Here we take AC =  BC and substitute values and get :

$$\begin{gathered} \sqrt{{\left(\mathrm{x} - \mathrm{a}\right)}^2 + {\mathrm{y}}^2} = \sqrt{{\left(\mathrm{x} - 3\mathrm{a}\right)}^2 + {\mathrm{y}}^2} \\ \mathrm{Taking} \mathrm{whole} \mathrm{square} \mathrm{on} \mathrm{both} \mathrm{hand} \mathrm{side} \mathrm{we} \mathrm{get} : \\ \Rightarrow {\left(\mathrm{x} - \mathrm{a}\right)}^2 + {\mathrm{y}}^2 = {\left(\mathrm{x} - 3\mathrm{a}\right)}^2 + {\mathrm{y}}^2 \\ \Rightarrow {\left(\mathrm{x} - \mathrm{a}\right)}^2 = {\left(\mathrm{x} - 3\mathrm{a}\right)}^2 \\ \mathrm{We} \mathrm{know} : {\left(\mathrm{m} - \mathrm{n}\right)}^2 = {\mathrm{m}}^2 + {\mathrm{n}}^2 - 2 \mathrm{m} \mathrm{n} , \mathrm{So} \\ \Rightarrow {\mathrm{x}}^2 + {\mathrm{a}}^2 - 2 \mathrm{a} \mathrm{x} ={\mathrm{x}}^2 + 9{\mathrm{a}}^2 - 6 \mathrm{a} \mathrm{x} \\ \Rightarrow 4 \mathrm{a} \mathrm{x} = 8 {\mathrm{a}}^2 \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{a} \end{gathered}$$
Here we can have value of x in integer as we take ' a '  also as an integer , Here we can say that given triangle can be equilateral when a and x are integers

At x  =  2 a we get our length of AC , As :

AC  = $\sqrt{{\left(2 \mathrm{a}- \mathrm{a}\right)}^2 + {\mathrm{y}}^2} = \sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{y}}^{\mathbf{2}}}\mathbf{ }\mathbf{ }\mathbf{unit}$

And we take AB  =  AC and substitute values and get :

$$\begin{gathered} \sqrt{{\mathrm{a}}^2 + {\mathrm{y}}^2} = 2 \mathrm{a} \\ \mathrm{Taking} \mathrm{whole} \mathrm{square} \mathrm{on} \mathrm{both} \mathrm{hand} \mathrm{side} \mathrm{we} \mathrm{get} : \\ \Rightarrow {\left(\sqrt{{\mathrm{a}}^2 + {\mathrm{y}}^2}\right)}^2= {\left(2 \mathrm{a}\right)}^2 \\ \Rightarrow {\mathrm{a}}^2 + {\mathrm{y}}^2=4 {\mathrm{a}}^2 \\ \Rightarrow {\mathrm{y}}^2=3 {\mathrm{a}}^2 \\ \Rightarrow \mathrm{y} =\sqrt{3} \mathrm{a} \end{gathered}$$
From above equation we can say that value of ' y '  always be an ' Irrational number ' ( As we know $\sqrt{3}$ is a Irrational number and Irrational number $\times$ Integer = Irrational number )

So,

Here we can say that given triangle ABC can not be equilateral when a , x and y are integers .

Therefore,

Option ( B )                                                                ( Ans )


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