A straight line is drawn through the point A( | vedclass

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(A) Let the angle of inclination of the line be $	heta$. Since it passes through point $A(1,2)$,the equation of the line in parametric form is $\frac

Vedclass · Accepted Answer

$	heta=15^{\circ}$ and $75^{\circ}$

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(A) Let the angle of inclination of the line be $	heta$. Since it passes through point $A(1,2)$,the equation of the line in parametric form is $\frac{x-1}{\cos 	heta} = \frac{y-2}{\sin 	heta} = r$,where $r = \pm \frac{\sqrt{6}}{3}$.Any point $P$ on the line is given by $(1 + r \cos 	heta, 2 + r \sin 	heta)$.Since $P$ lies on the line $x+y=4$,we have $(1 + r \cos 	heta) + (2 + r \sin 	heta) = 4$.$3 + r(\cos 	heta + \sin 	heta) = 4 \Rightarrow r(\cos 	heta + \sin 	heta) = 1$.Substituting $r = \pm \frac{\sqrt{6}}{3}$,we get $\pm \frac{\sqrt{6}}{3}(\cos 	heta + \sin 	heta) = 1$.Squaring both sides: $\frac{6}{9}(\cos 	heta + \sin 	heta)^2 = 1 \Rightarrow \frac{2}{3}(1 + \sin 2	heta) = 1$.$1 + \sin 2	heta = \frac{3}{2} \Rightarrow \sin 2	heta = \frac{1}{2}$.Thus,$2	heta = 30^{\circ}$ or $150^{\circ}$,which gives $	heta = 15^{\circ}$ or $75^{\circ}$.

$A$ straight line is drawn through the point $A(1,2)$ such that its point of intersection with the straight line $x+y=4$ is at a distance $\frac{\sqrt{6}}{3}$ from the given point $A$. Find the angle which the line makes with the positive direction of $X$-axis.

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