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

The straight lines `x+y=0, 3x+y-4=0 and x+3y-4=0` form a triangle which is (A) isosceles (B) right angled (C) equilateral (D) scalene

A

isoscles

B

equilateral

C

right angled

D

None of the above

Text Solution

AI Generated Solution

To determine the type of triangle formed by the lines \(x+y=0\), \(3x+y-4=0\), and \(x+3y-4=0\), we will follow these steps: ### Step 1: Find the intersection points of the lines 1. **Intersection of \(x+y=0\) and \(3x+y-4=0\)**: - From \(x+y=0\), we have \(y = -x\). - Substitute \(y = -x\) into \(3x + y - 4 = 0\): \[ 3x - x - 4 = 0 \implies 2x - 4 = 0 \implies x = 2 ...
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