thee distance of the point (2,3) from the lne 2x-3y+9=0 measured along a line x-y+1=0 is Share with your friends Share 5 Rajat Sethi answered this Slope of line 2x - 3y +9 =0 is m1 = 23Slope of line x - y +1 =0 is m2 = 1Now,angle between the two lines is tanθ = m2 - m11 +m1m2 = 1 - 231+1×23= 15sinθ = 126Now,perpendicular distance of(2,3) from 2x - 3y +9=0 isp =2×2 -3×3 + 922 + 32=413 Now distance along the given line = psinθ = 413×26 = 42 units 15 View Full Answer