Question

If two lines drawn from a point $P(2,3)$ intersect the line $x+y=6$ at a distance $\sqrt{\dfrac{2}{3}}$, then the angle between the lines is

Hint:

Write a general line through $P$ with slope $m$, find where it meets the given line in terms of $m$, then use the given distance to form an equation in $m$ alone. Solving that equation gives both slopes at once, and the angle between them follows from the usual two-line slope formula.

Answer 1

Correct Answer: 1

Step 1: Use formula for angle between pair of lines from a point to a line.
Angle $\theta$ between two lines drawn from point $P$ to a line at perpendicular distance $d$ is given by \[ \sin\frac{\theta}{2}=\frac{d}{\text{distance from point to line}} \] Step 2: Find distance of point from the line.
Distance of $P(2,3)$ from $x+y-6=0$ is \[ D=\frac{|2+3-6|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt2} \] Step 3: Substitute values.
\[ \sin\frac{\theta}{2} =\frac{\sqrt{\frac{2}{3}}}{\frac{1}{\sqrt2}} =\sqrt{\frac{4}{3}} =\frac{2}{\sqrt3} \] \[ \Rightarrow \sin\frac{\theta}{2}=\frac{\sqrt3}{2} \] Step 4: Find $\theta$.
\[ \frac{\theta}{2}=\frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{3} \] Final conclusion.
The angle between the two lines is $\dfrac{\pi}{3}$.

Answer 2

Concept:

• Any line through a fixed point $P$ with slope $m$ can be intersected with a given line to get the coordinates of the intersection point in terms of $m$.
• Setting the distance from $P$ to that intersection point equal to the given length turns the condition into a quadratic equation in $m$, whose two roots are the slopes of the two required lines.
• Once both slopes are known, the angle between the lines follows directly from the standard slope-angle formula.

Step 1: Set up a line through $P(2,3)$ with slope $m$.
$y - 3 = m(x-2)$

Step 2: Find where this line meets $x+y=6$.
Substituting $y = 6-x$:
$6 - x - 3 = m(x-2)$
$3 - x = m(x-2)$
$x(1+m) = 3+2m$
$x = \dfrac{3+2m}{1+m}$, and correspondingly $y = 6-x = \dfrac{3+4m}{1+m}$

Step 3: Write the displacement from $P$ to this intersection point.
$x - 2 = \dfrac{3+2m}{1+m} - 2 = \dfrac{1}{1+m}$
$y - 3 = \dfrac{3+4m}{1+m} - 3 = \dfrac{m}{1+m}$

Step 4: Apply the given distance condition.
$(x-2)^2 + (y-3)^2 = \dfrac{2}{3}$
$\dfrac{1+m^2}{(1+m)^2} = \dfrac{2}{3}$
$3(1+m^2) = 2(1+m)^2$
$3+3m^2 = 2+4m+2m^2$
$m^2 - 4m + 1 = 0$

Step 5: Solve for the two slopes.
$m = \dfrac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt3$
So $m_1 = 2+\sqrt3$ and $m_2 = 2-\sqrt3$.

Step 6: Find the angle between the two lines.
$\tan\theta = \dfrac{m_1-m_2}{1+m_1 m_2}$ (taking the magnitude)
$m_1 - m_2 = 2\sqrt3$, and $m_1 m_2 = (2+\sqrt3)(2-\sqrt3) = 4-3 = 1$
$\tan\theta = \dfrac{2\sqrt3}{1+1} = \sqrt3$
$\theta = \dfrac{\pi}{3}$

Final Answer: $\dfrac{\pi}{3}$