The vertices of a triangle are A(-1, 3), B(-2, 2) and C(3,-1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :
- -x+y-(2- 2)=0
- x+y-(2- 2)=0
- x+y+(2- 2)=0
- x-y-(2+ 2)=0
Answer
(2) x+y-(2- 2)=0
Solution
AB: x-y+4=0, AC: x+y-2=0
BC: 3x+5y-4=0
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Shift each side 1 unit towards the opposite vertex (inwards):
AB (towards C, where x-y+4>0): x-y+4= 2 distance from O = 4- 2 2 1.8
AC (towards B, where x+y-2<0): x+y-2=- 2 distance from O = 2- 2 2 0.41
BC (towards A, away from O): 3x+5y-4= 34 distance from O = 4+ 34 34 1.7
nearest side: x+y-(2- 2)=0
Answer: (2)