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If the straight lines $x+3 y=4,3 x+y=4$ and $x+y$ $=0$ form a triangle, then the triangle is
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Verified Answer
The correct answer is:
isosceles
isosceles
Let equation of $A B: x+3 y=4$
Let equation of $B C: 3 x+y=4$
Let equation of $C A: x+y=0$
Now, By solving these equations we get $A=(-2,2), B=(1,1)$ and $C=(2,-2)$
Now, $A B=\sqrt{9+1}=\sqrt{10}$,
$$
B C=\sqrt{1+9}=\sqrt{10}
$$
and $C A=\sqrt{16+16}=\sqrt{32}$
Since, length of $A B$ and $B C$ are same
therefore triangle is isosceles.
Let equation of $B C: 3 x+y=4$
Let equation of $C A: x+y=0$
Now, By solving these equations we get $A=(-2,2), B=(1,1)$ and $C=(2,-2)$
Now, $A B=\sqrt{9+1}=\sqrt{10}$,
$$
B C=\sqrt{1+9}=\sqrt{10}
$$
and $C A=\sqrt{16+16}=\sqrt{32}$
Since, length of $A B$ and $B C$ are same
therefore triangle is isosceles.
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