question_answer

The line \[x+3y-2=0\]bisects the angle between a pair of straight lines of which one has equation\[x-7y+5=0\]. The equation of the other line is

A) \[3x+3y-1=0\]

B) \[x-3y+2=0\]

C) \[5x+5y-3=0\]

D) None of these

Correct Answer: C

Solution :

[c] The family of line through the given lines is \[L\equiv x-7y+5+\lambda (x+3y-2)=0\]                    ?. (i) For line L =0 in the diagram, the distance of any point say (2, 0) on the line \[x+3y-2=0\] from the line \[x-7y+5=0\] and the line \[L=0\] must be the same. Therefore, \[\left| \frac{2+5}{\sqrt{50}} \right|=\left| \frac{2+2\lambda +5-2\lambda }{\sqrt{{{(1+\lambda )}^{2}}+{{(3\lambda -7)}^{2}}}} \right|\] or \[10{{\lambda }^{2}}-40\lambda =0\] i.e., \[\lambda =4\,\,or\,\,0\] Hence, \[L=0,\,\,\lambda =4.\] Therefore, the required line is \[5x+5y-3=0.\]