Blurred Lines

Debtanu Bhuiya's method is better. But just to give another method.
With vertex at (0, 0), form an isosceles triangle with l and m.
n will pass through (0, 0) and midpoint of the base of the triangle.
Take any point on l, say (1, 3). Then (x, 5x) will be the vertex on m.
The sides are of same length,
$1^{2} + 3^{2} = x^{2} + 25x^{2} ............. x = \sqrt{5/13}$
The point on m is $(\sqrt{5/13}, 5\sqrt{5/13} )$
The midpoint is $( (1 + \sqrt{5/13})/2, (3 + 5\sqrt{5/13})/2 )$

Slope of n is $(3 + 5\sqrt{5/13})/2 \div (1 + \sqrt{5/13})/2 = 3.76556$
$\boxed {3.76556}$

Hint : I did it in two ways: 1) using the definition of slope in terms of the tangent of the angle of inclination of the line,or 2) getting the distance of each line from the angle bisector.

It is (arctan5+arctan3)÷2=74.89° Hence tan74.89°= 3.70359