Find the equation of a straight line passing through the point(4,5) and equally inclined to the lines 3x = 4y +7 and 5y= 12x +6

First we have to determine the slope of each lines by transforming to the slope-intercept form:

y=(3x-7/)4; m2= ¾y=(12x+6)/5, m3 = 12/5

The formula to be used in the proceeding steps is    a=tan^-1(m1-m2)/1+m1m2=tan^-1(m1-m2)/1+m1m2
substituting,      a=tan^-1(m1-3/4)/1+3m1/4=tan^-1(m1-12/5)1+12m1/5)     =>(4m1-3)/(4+3m1)=(5m1-12)/(5+12m1)m1 = -1applying this slope 

y -y1 = m(x-x1)
when y1 = 5 and x1 = 4 then,
y - 5 = -1(x-4)
y = -x +4+ 5 ; y = -x +9

I'm sure the answer is a