Q.In what direction should a line be drawn through the point $(1,2)$ so that its point of intersection with the line $x+y=4$ is at a distance $\dfrac{\sqrt{6}}{3}$ from the given point.

The key idea is to use the parametric form of a line to represent points at a specific distance from a given point, then find its intersection with the other line.

1. Let the line through $P(1,2)$ have a direction angle $\theta$ with the positive x-axis. Any point $Q$ on this line at a distance $r$ from $P$ can be represented as $(1 + r \cos \theta, 2 + r \sin \theta)$.
2. The point $Q$ lies on the line $x+y=4$. Substituting its coordinates: $(1 + r \cos \theta) + (2 + r \sin \theta) = 4$ $3 + r(\cos \theta + \sin \theta) = 4$ $r(\cos \theta + \sin \theta) = 1$
3. The distance $r$ from $P$ to $Q$ is given as $\frac{\sqrt{6}}{3}$. Substitute this value: $\frac{\sqrt{6}}{3}(\cos \theta + \sin \theta) = 1$ $\cos \theta + \sin \theta = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$
4. To solve for $\theta$, we can square both sides: $(\cos \theta + \sin \theta)^2 = \left(\frac{\sqrt{6}}{2}\right)^2$ $\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta = \frac{6}{4}$ $1 + \sin(2\theta) = \frac{3}{2}$ $\sin(2\theta) = \frac{1}{2}$ …

The problem asks for the "direction" in which a line should be drawn through a given point $P(1,2)$. This "direction" is typically specified by the slope of the line. We are also given a second line, $x+y=4$, and a condition: the point of intersection of our new line with $x+y=4$ must be at a specific distance from $P(1,2)$.

Our strategy will be to:

1. Represent the unknown line passing through $P(1,2)$ using a variable for its slope.
2. Find the coordinates of the intersection point $Q$ between this unknown line and the line $x+y=4$. These coordinates will be expressed in terms of the unknown slope.
3. Use the distance formula between $P(1,2)$ and $Q$ and equate it to the given distance $\frac{\sqrt{6}}{3}$.
4. Solve the resulting equation for the slope, which will give us the required directions.

Let's proceed step-by-step.

1. Represent the line through $P(1,2)$

   Let the slope of the line drawn through $P(1,2)$ be $m$. Using the point-slope form, the equation of this line is:

   $y - 2 = m(x - 1)$

2. Find the intersection point $Q$ with $x+y=4$

   To find the point of intersection $Q(x_Q, y_Q)$, we need to solve the system of equations:

From equation (2), we can express $y$ as $y = 4 - x$. Substitute this into equation (1):
$(4 - x) - 2 = m(x - 1)$
$2 - x = mx - m$
Rearrange to solve for $x$:
$2 + m = mx + x$
$2 + m = x(m + 1)$
So, $x_Q = \frac{2+m}{m+1}$.

> [!WARNING]
> The denominator $m+1$ cannot be zero, so $m \neq -1$. If $m=-1$, the line through $(1,2)$ would be $y-2 = -1(x-1) \implies y-2 = -x+1 \implies x+y=3$. This line is parallel to $x+y=4$ and distinct, meaning they would never intersect. Thus, $m=-1$ is not a valid slope for an intersection to exist.

Now substitute $x_Q$ back into $y = 4 - x$:
$y_Q = 4 - \frac{2+m}{m+1}$
$y_Q = \frac{4(m+1) - (2+m)}{m+1}$
$y_Q = \frac{4m + 4 - 2 - m}{m+1}$
$y_Q = \frac{3m + 2}{m+1}$
So, the intersection point is $Q\left(\frac{2+m}{m+1}, \frac{3m+2}{m+1}\right)$.

3. Apply the distance formula

We are given that the distance between $P(1,2)$ and $Q(x_Q, y_Q)$ is $\frac{\sqrt{6}}{3}$.

> [!FORMULA]

> The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

It's often easier to work with the square of the distance: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.

Here, $P(x_P, y_P) = (1,2)$ and $Q(x_Q, y_Q) = \left(\frac{2+m}{m+1}, \frac{3m+2}{m+1}\right)$.

The squared distance $PQ^2$ is $\left(\frac{\sqrt{6}}{3}\right)^2 = \frac{6}{9} = \frac{2}{3}$.

Let's calculate the differences in coordinates:
$x_Q - x_P = \frac{2+m}{m+1} - 1 = \frac{2+m - (m+1)}{m+1} = \frac{1}{m+1}$ …