question_answer

In an isosceles triangle ABC, the coordinates of the points B and C on the base BC are respectively (1, 2) and (2, 1). If the equation of the line AB is \[y=2x\], then the  equation of the line AC is                                                                [Roorkee 2000]

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 \[y=\frac{x}{2}\].