The equation of straight line bisecting the acute angle between the lines represented by the equation 7x2+6xy−y2−5x+3y−2=0

7x2+6xy−y2−5x+3y−2=(7x−y+n1)(x+y+n2)coefficient of x is −5=n1+7n2coefficient of y is 3=n1−n2solving we get n1=2,n2=−1 lines are7x−y+2=0,x+y−1=0⇒−x−y+1=0a1=7,b1=−1,c1=2,a2=−1,b2=−1,c2=1c1c2>0,a1a2+b1b2<0acute angle bisector is a1x+b1y+c1a12+b12=+a2x+b2y+c2a22+b22⇒12x+4y−3=0