Login

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?

Dear Student,
Please find below the solution to the asked query:

Let slope of required line L be m.Given line is L1:3x+y=1Slope of linem1=-Coefficient of xCoefficient of y=-31=-3Angle between L and L1 is 60°.⇒tan60°=m-m11+mm1⇒m+31-3m=tan60°⇒m+31-3m=3⇒m+31-3m=±3⇒m+31-3m=3 or m+31-3m=-3⇒m+3=3-3m or m+3=-3+3m⇒m=-3m or 3+3=3m-m⇒4m=0 or 2m=23⇒m=0 or m=3As line passes through 3,-2, hence equation of L is given by:y--2=mx-3⇒y+2=mx-3m⇒mx-y-3m-2=0When m=0, we get-y-2=0⇒y+2=0When m=33x-y-33-2=0

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards

  • 61
View Full Answer
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 :)
  • 5
What are you looking for?