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Question

The straight lines x+y=0,3x+y-4=0,x+3y-4=0form a triangle which is


A

isosceles

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B

equilateral

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C

right angled

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D

none of these

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Solution

The correct option is A

isosceles


Explanation for the correct option:

Compute the required term:

Given:

x+y=0...(i)3x+y-4=0...(ii)x+3y-4=0...(iii)

From (i) we get x=-y

Put x=-y in (ii)

-3y+y-4=0-2y=4

So y=-2

Then x=2

So line (i) and (ii) intersect at A=(2,-2)

Similarly solving (ii) and (iii), we get x=1,y=1.

So line (ii) and (iii) intersect at B=(1,1)

Similarly solving (i) and (iii), we get x=-2,y=2

So line (i) and (iii) intersect at C=(-2,2)

AB=x2-x12+y2-y12 (distance formula)

=1-22+1+22=1+9=10

Similarly

BC=-2-12+2-12=10

AC=-2-22+2+22=32

Here AB=BC

Here the two sides of the triangle are equal. So the triangle is isosceles.

Hence, option A is the correct answer.


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