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If `|(a,b-c,b+c),(a+c,b,c-a),(a-b,a+b,c)|=0` then the line `ax+by+c=0` passes through the fixed point which isA. `(1,2)`B. `(1,1)`C. `(-2,1)`D. `(1,0)` |
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Answer» Correct Answer - B Applying `C_(1) to aC_(1)` and then `C_(1) to C_(1) +bC_(2) +cC_(3)` and taking `(a^(2)+b^(2)+c^(2))` common from `C_(1) ` we get ` Delta =((a^(2)+b^(2)+c^(2)))/(a) |{:(1,,b-c,,c+b),(1,,b,,c-a),(1,,b+a,,c):}|` `=((a^(2)+b^(2)+c^(2)))/(a) |{:(1,,b-c,,c+b),(0,,c,,-a-b),(0,,a+c,,-b):}|` `(R_(2) to R_(2)-R_(1),R_(3) to R_(3)-R_(1))` `=((a^(2)+b^(2)+c^(2)))/(a) (-bc+a^(2)+ab+ac+bc)` (expanding along `C_(1)`) `=(a^(2)+b^(2)+c^(2))(a+b+c)` hence `Delta =0 rArr a+b+c=0` Therefore line ax+by+c=0 passes through the fixed point (1,1) |
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