if A(-3,2),B(a,b) and C(-1,4) are the vertices of an isosceles triangle,show that a+b =1, if AB=BC
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A(-3,2) (x1 y1)
B(a,b) (x2 y2)
C(-1,4)(x3 y3)
AB = BC
by distance formula,
AB = √[(x1-x2)² + (y1-y2)²]
= √[(-3-a)² + (2-b)²]
= √[9+a²+6a+4+b²-4b]
BC = √[(x3-x2)² + (y3-y2)²]
= √[(-1-a)² + (4-b)²]
= √[1+a²+2a+16+b²-8b]
as AB = BC
therefore
√[13+a²+b²+6a-4b] = √[17+a²+b²+2a-8b]
13+6a-4b = 17+2a-8b
4a+4b = 4
a+b=1
B(a,b) (x2 y2)
C(-1,4)(x3 y3)
AB = BC
by distance formula,
AB = √[(x1-x2)² + (y1-y2)²]
= √[(-3-a)² + (2-b)²]
= √[9+a²+6a+4+b²-4b]
BC = √[(x3-x2)² + (y3-y2)²]
= √[(-1-a)² + (4-b)²]
= √[1+a²+2a+16+b²-8b]
as AB = BC
therefore
√[13+a²+b²+6a-4b] = √[17+a²+b²+2a-8b]
13+6a-4b = 17+2a-8b
4a+4b = 4
a+b=1
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