Question

Find the equation of a line whose perpendicular distance from the origin is $\sqrt{8}$ units and the angle between the positive direction of the x-axis and the perpendicular is ${135}^{\circ }$.

Answer 1

Here p = $\sqrt{8}$ units and $\alpha ={135}^{\circ }$.

So, the equation of the given line in normal form is $xcos\alpha +ysin\alpha =p$, where $\alpha ={135}^{\circ }$ and p = $\sqrt{8}$ units

$\iff xcos{135}^{\circ }+ysin{135}^{\circ }=\sqrt{8}$

$\iff x\left(\frac{-1}{\sqrt{2}}\right)+y\left(\frac{1}{\sqrt{2}}\right)=\sqrt{8}$

$[\because \text{cos}{135}^{\circ }=\text{cos}({180}^{\circ }-{45}^{\circ })=-\text{cos}{45}^{\circ }]$

$[\text{sin}{135}^{\circ }=\text{sin}({180}^{\circ }-{45}^{\circ })=\text{sin}{45}^{\circ }]$

$\iff -x+y=4\iff x-y+4=0,$