"""Let us consider the situation when the axes are inclined at an angle `omega` . If the coordinates of a point P are `(x_1,y_1)`,  then  `PN = x_1, PM =y_1`  where PM is parallel to the y-axis and PN is parallel to the Xaxis. The straight line through P that makes an angle `theta`  with the x-axisis `RQ = y-y_1, PQ = x - x_1 """" From """" DeltaPQR`, we have `(PQ)/(sin(omega-theta))=(sqrtQR)/(sin theta) or y-y_1 = (sin theta)/(sin(omega-theta))(x-x_1)` writen in the form of `y-y_1=m(x-x_1) """" where """" m=(sin theta)/(sin(omega-theta))`   Therefore, if the slope of the line is m, then the angle of inclination of the line with the x-axis is given by an `tan theta=((m sin omega)/(1+m cos omega))` <br> <img src=""""https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_MAT_XI_VIB_P01_C04_E01_044_Q01.png"""" width=""""80%"""">  <br> The axes being inclined at an angle of `60^(@)`, the angle between the two  straight line y=2x+5 and 2y+x+7=0"""
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