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The straight lines x + y = 0, 3x + y = 4...

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

A

isoscles

B

equilateral

C

right angled

D

None of the above

Text Solution

Verified by Experts

The points of intersection of three lines are `A(1,1)`, `B(2,-2)`, `C(-2,2)`.
Now, `|AB|=sqrt(1+9)=sqrt(10)`,
`|BC|=sqrt(16+16)=4sqrt(2)`,
and `|CA|=sqrt(9+1)=sqrt(10)`
`:.` Triangle is in isosceles.
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