Question

The equation of the straight line whose perpendicular distance from origin is $3\sqrt{2}$ units and this perpendicular makes an angle of ${75}^{\circ }$ with the positive direction of x-axis, is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let AB be the required line and OL be perpendicular to it.

$Given,OL=3\sqrt{2}and\angle LOA={75}^{\circ }$

$\therefore$Equation of line AB is

x cos ${75}^{\circ }$ + y sin ${75}^{\circ }$=$3\sqrt{2}$ …… (1)(Normal form)

Now $cos{75}^{\circ }=cos({30}^{\circ }+{45}^{\circ })$

$=\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}-\frac{1}{2}.\frac{1}{\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$andsin{75}^{\circ }=sin({30}^{\circ }+{45}^{\circ })=\frac{1}{2}.\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$\therefore$ From (1), equation of line AB is

$x\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)+y\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)=3\sqrt{2}or(\sqrt{3}-1)x+(\sqrt{3}+1)y-12=0$