Question

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 $\sqrt{6}/3$ from the given point ' $A$ '. Find the angle which the lines makes with the positive direction of $x$-axis.

• A. $θ=15^{∘}\&75^{∘}$
• B. $θ=75^{∘}\&45^{∘}$
• C. $θ=45^{∘}\&60^{∘}$
• D. $θ=60^{∘}\&30^{∘}$

Answer 1

Answer: (A)

Let the angle of inclination of line $θ$, and as passed through point $A(1,2)$, so equation of line is
$\frac{x−1}{\cos θ}=\frac{y−2}{\sin θ}=Â\pm \frac{\sqrt{6}}{3}$
$∴$ General point on the line is
$P\left(1Â\pm \frac{\sqrt{6}}{3}\cos θ,2Â\pm \frac{\sqrt{6}}{3}\sin θ\right)$
Let the point $P$ on the straight line $x+y=4$, so
$3Â\pm \frac{\sqrt{6}}{3}(\sin θ+\cos θ)=4$
$⇒Â\pm \frac{\sqrt{6}}{3}(\sin θ+\cos θ)=1$
On squaring both sides, we get
$2(1+\sin 2θ)=3$
$⇒\sin 2θ=\frac{1}{2}$
$⇒2θ=30^{∘}$ and $150^{∘}$
$⇒θ=15^{∘}\text{ and }75^{∘}$