A) \[y=\frac{1}{2}(x-1)\]
B) \[y=\frac{x}{2}\]
C) \[y=x-1\]
D) \[2y=x+3\]
Correct Answer: B
Solution :
\[\angle ABC=\tan \theta =\frac{\frac{1}{2}-1}{1+\frac{1}{2}}=-\frac{1}{3}\] (Here \[{{m}_{1}}=\frac{1}{2},\,{{m}_{2}}=1)\] \[\because \] \[AB=AC\]; \ \[\angle ABC=\angle ACB\] Hence \[-\frac{1}{3}=\frac{m-1}{1+m}\] Þ \[m=\frac{1}{2}\] (Here m is the gradient of line AC) \ Equation of line AC is\[y-1=\frac{1}{2}(x-2)\] Þ \[y=\frac{x}{2}\].You need to login to perform this action.
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