A straight line is drawn through the point \(A(1,2)\) such that its point of intersection with the straight…
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.
\(\theta=15^{\circ}\) and \(75^{\circ}\)
\(\theta=75^{\circ}\) and \(45^{\circ}\)
\(\theta=45^{\circ}\) and \(60^{\circ}\)
\(\theta=60^{\circ}\) and \(30^{\circ}\)
Solution
Let the angle of inclination of line \(\theta\), and as passed through point \(A(1,2)\), so equation of line is
\(\frac{x-1}{\cos \theta}=\frac{y-2}{\sin \theta}= \pm \frac{\sqrt{6}}{3}\)
\(\therefore\) General point on the line is
\(P\left(1 \pm \frac{\sqrt{6}}{3} \cos \theta, 2 \pm \frac{\sqrt{6}}{3} \sin \theta\right)\)
Let the point \(P\) on the straight line \(x+y=4\), so
\(\begin{aligned}
& 3 \pm \frac{\sqrt{6}}{3}(\sin \theta+\cos \theta)=4 \\
& \Rightarrow \quad \pm \frac{\sqrt{6}}{3}(\sin \theta+\cos \theta)=1
\end{aligned}\)
On squaring both sides, we get
\(\begin{gathered}
2(1+\sin 2 \theta)=3 \\
\Rightarrow \sin 2 \theta=\frac{1}{2} \Rightarrow 2 \theta=30^{\circ} \text { and } 150^{\circ} \Rightarrow \theta=15^{\circ} \text { and } 75^{\circ}
\end{gathered}\)