The bisector of the acute angle formed between the lines 4x- 3y + 7 = 0 and 3x - 4y + 14 = 0 has the equation

The equations of given straight lines, by making constant terms positive, are4x−3y+7=0 and 3x−4y+14=0∵ 4×3+(−3)(−4)=24>0 i.e. a1a2+b1b2>0So, the bisector of the acute angle is given by4x−3y+742+(−3)2=−3x−4y+1432(−4)2⇒x−y+3=0