find the equations of the lines passing through the point (3,-2) and inclined at an angle of 60° to the line root3x + y =1?
Slope of √3x + y = 1 is -√3 As, y= -√3x + 1 which is in the form of y=mx +c where m= -√3. Now,equation of a line passing through (3,-2) is given by y+2=m(x-3)....(1). Since unknown lines and given line are inclined at 60°, tan60=|m1-m2/1+m1m2| √3= |m+√3/1+m(-√3)|...where m1=m=slope of unknown lines and m2=slope of given line=-√3. Now, + or - √3 =m+√3/1-√3m (removing modulus). There are 2 cases where either, √3 =m+√3/1-√3m. √3(1-√3m)=m+√3. √3-3m=m+√3. -3m-m=0 i.e [m=0.] Or -√3=m+√3/1-√3m. -√3(1-√3m)=m+√3. -√3 + 3m=m+√3. 3m-m=√3+√3. 2m=2√3 i.e[ m=√3.]. Replacing both values in (1), we get either y+2=0 (m=0). Or y+2=√3(x-3). √3x-y-3√3-2=0. Thus equations are y+2=0 or √3x-y-3√3-2=0. Hope it helped :)