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In ∆BDC,
BC = BD√2
BC = 5√2
THEREFORE, C(5√2,0), A(0,5√2)
EQUATION,
y = mx + C
y = (-1)x + 5√2
x + y = 5√2
BC = BD√2
BC = 5√2
THEREFORE, C(5√2,0), A(0,5√2)
EQUATION,
y = mx + C
y = (-1)x + 5√2
x + y = 5√2
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AsmitPhuyal:
This is wrong
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