Question:
The intersection of three lines \(x-y=0\), \(x+2y=3\) and \(2x+y=6\) is a :
The intersection of three lines \(x-y=0\), \(x+2y=3\) and \(2x+y=6\) is a :
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Find the three vertices from pairwise intersections and compare the side lengths.
Updated On: Nov 4, 2026
- Right angled triangle
- Equilateral triangle
- None of the above
- Isosceles triangle
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The Correct Option is D
Solution and Explanation
Vertices: \(x-y=0\) with \(x+2y=3\) gives \(A(1,1)\); \(x-y=0\) with \(2x+y=6\) gives \(B(2,2)\); \(x+2y=3\) with \(2x+y=6\) gives \(C(3,0)\).
\(AB=\sqrt{1+1}=\sqrt2\), \(BC=\sqrt{1+4}=\sqrt5\), \(AC=\sqrt{4+1}=\sqrt5\).
Two sides are equal (\(BC=AC\)), and \(AB^2+AC^2=7\ne5=BC^2\) (similarly no other pair satisfies Pythagoras), so the triangle is isosceles but not right-angled or equilateral.
Answer: (4) Isosceles triangle
\(AB=\sqrt{1+1}=\sqrt2\), \(BC=\sqrt{1+4}=\sqrt5\), \(AC=\sqrt{4+1}=\sqrt5\).
Two sides are equal (\(BC=AC\)), and \(AB^2+AC^2=7\ne5=BC^2\) (similarly no other pair satisfies Pythagoras), so the triangle is isosceles but not right-angled or equilateral.
Answer: (4) Isosceles triangle
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