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

$7x^2+6xy-y^2-5x+3y-2=(7x-y+n_1)(x+y+n_2)$

coefficient of x is $-5=n_1+7n_2$

coefficient of y is $3=n_1-n_2$

solving we get $n_1=2,n_2=-1$ lines are

$7x-y+2=0,x+y-1=0\Rightarrow -x-y+1=0$

$a_1=7,b_1=-1,c_1=2,a_2=-1,b_2=-1,c_2=1$

$c_1{c}_2>0,a_1{a}_2+b_1{b}_2<0$

acute angle bisector is $\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=+\frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$

$\Rightarrow 12x+4y-3=0$