question_answer

A line \[4x+y=1\] passes through the point \[A(2,-7)\] meets the line 5C whose equation is \[3x-4y+1=0\] at the point5. The equation to the line AC so that \[AB=AC,\] is

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

B) \[52x+59y-519=0\]

C) \[89x+52y+519=0\] 

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

Correct Answer: A

Solution :

Slopes of the lines \[AB\equiv 4x+y=1\]and \[BC\equiv 3x-4y+1=0\]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)
Given,  \[AB=\text{ }AC\]
\[\Rightarrow \,\,\,\angle ABC=\angle ACB=\alpha \]
Thus the line A C 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 \[\tan \alpha =\pm \left[ \frac{m-\frac{3}{4}}{1+m.\frac{3}{4}} \right]\Rightarrow \frac{19}{8}=\pm \frac{4m-3}{4+3m}\]
\[\Rightarrow \,\,m=-4\] or \[-\frac{52}{89}\].
But slope of AB is \[-4,\] therefore slope of AC is \[-\frac{52}{89}\].
Hence, the equation of line AC given by (ii) is \[52x+89y+519=0\]