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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

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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
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