Question

# The straight lines x + y = 0, 3x + y- 4 = 0 and x + 3y - 4 = 0 form 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 B isosceles Solving equation 3x + y - 4 = 0 & x + y = 0 Substituting in (ii) y = - 2 Coordinate & B = (2, -2) Solving equation x + y = 0 & x + 3y - 4 = 0 Substituting in equation (i) we get x = - 2 Coordinate of A≡(−2,2) Solving equation 3x + y - 4 = 0 and x + 3y - 4 = 0 (3x+y=4)×3 x+3y=4 Substituting in (vi) we get y = 1 coordinate of C≡(1,1) So, we get the vertices of triangle as A(−2,2), B(2,−2), C(1,1). The distance between two points (x1,y1)and(x2,y2) is given by √(x2−x1)2+(y2−y1)2 AB=√(2−(−2))2+(−2−2)2 AB=√(4)2+(−4)2 AB=√16+16=4√2 units BC=√(2−1)2+(−2−1)2 BC=√12+(−3)2=√10 units CA=√(1−(−2))2+(1−2)2 CA=√(3)2+(−1)2 CA=√9+1=√10 units Here the two sides are equal hence the triangle is isosceles.

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