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

Question

A straight line $$L=0$$ passes through  the point $$5$$,−$$4$$  is inclined at an angle of $$60o$$ to the line $$3x+y$$−$$3=0$$ ,then the equation of the line $$L=0$$ 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 $$x1$$,$$y1is$$ y−$$y1=$$$$tan$$θ±αx−$$x1$$ here  $$tan$$θ$$=$$−abHere $$tan$$θ$$=$$−$$31$$⇒θ$$=120$$°  and  α$$=60$$°Equations of the required lines are $$y+4=tan120$$°±$$60$$°x−$$5It$$ implies that $$y+4=tan180$$°x−$$5y+4=0ory+4=tan60$$°x−$$5y+4=3x$$−$$5Therefore$$, 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^o 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 60o to the line 3x+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^o to the line $\sqrt{3} x + y - 3 = 0$ ,then the equation of the line $L = 0$ is