question_answer

A line \[4x+y=1\]passes through the point \[A(2,\ -\ 7)\] meets the line BC whose equation is \[3x-4y+1=0\] at the point B. The equation to the line AC so that AB = AC, is                                [IIT 1971]

A)            \[52x+89y+519=0\]                   

B)            \[\beta \]

C)            \[89x+52y+519=0\]                   

D)            \[89x+52y-519=0\]

Correct Answer: A

Solution :

           Slopes of \[AB\] and BC are ? 4 and \[\frac{3}{4}\] respectively. If \[\alpha \] be the angle between \[AB\] and \[BC\], then \[\tan \alpha =\frac{-4-\frac{3}{4}}{1-4\left( \frac{3}{4} \right)}=\frac{19}{8}\]                 .....(i)                    Since \[AB=AC\]                    \[\Rightarrow \angle ABC=\angle ACB=\alpha \]                                Thus the line AC also makes an angle \[\alpha \]with BC. If m be the slope of the line AC, then its equation is \[y+7=m(x-2)\]                                                  .....(ii)                    Now \[\tan \alpha =\pm \left[ \frac{m-\frac{3}{4}}{1+m.\frac{3}{4}} \right]\Rightarrow \frac{19}{8}=\pm \frac{4m-3}{4+3m}\]                    Þ \[m=-4\]or ?\[\frac{52}{89}\].                     But slope of AB is ? 4, so slope of AC is \[-\frac{52}{89}\].                    Therefore the equation of line AC given by (ii) is \[52x+89y+519=0\].