A straight line $$L=0$$ passes through the point $$5$$,−$$4$$ is inclined at an angle of $$60$$° to the line $$3x+y$$−$$3=0$$,then the equation of the $$lineL=0$$ is

Question

A straight line $$L=0$$  passes through  the point $$5$$,−$$4$$ is inclined at an angle of $$60$$° to the line $$3x+y$$−$$3=0$$,then the equation of the $$lineL=0$$ is

Infinity Learn · Accepted Answer

The equation of any line which makes an angle α with the line $$ax+by+c=0$$ and passing  through the point $$x1$$,$$y1$$ is y−$$y1=$$$$tan$$⁡$$($$θ±α$$)x$$−$$x1$$ here $$tan$$⁡θ$$=$$−ab Here $$tan$$⁡θ$$=$$−$$31$$⇒θ$$=120$$∘ and α$$=60$$∘ Equations of the required lines are $$y+4=$$$$tan$$⁡$$120$$∘±$$60$$∘$$(x$$−$$5)It$$ implies that $$y+4=tan180$$°x−$$5y+4=0Ory+4=tan60$$°x−$$5y+4=3x$$−$$5$$ Therefore, the equations of the required lines are $$y+4=0$$,$$3x$$−y−$$53$$−$$4=0$$

A straight line $L = 0$ passes through the point $(5, - 4)$ is inclined at an angle of $60 °$ to the line $\sqrt{3} x + y - 3 = 0$ ,then the equation of the line $L = 0$ is

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A straight line L=0 passes through the point 5,−4 is inclined at an angle of 60° to the line 3x+y−3=0,then the equation of the lineL=0 is

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A straight line $L = 0$ passes through the point $(5, - 4)$ is inclined at an angle of $60 °$ to the line $\sqrt{3} x + y - 3 = 0$ ,then the equation of the line $L = 0$ is