Q.Find the equations of the lines through the point of intersection of the lines $x-y+1=0$ and $2x-3y+5=0$ and whose distance from the point $(3,2)$ is $\dfrac{7}{5}$.

The key idea is to represent the family of lines passing through the intersection of two given lines using the $L_1 + \lambda L_2 = 0$ form, then apply the distance formula from a point to a line.

1. The equation of any line passing through the intersection of $x-y+1=0$ and $2x-3y+5=0$ is given by $(x-y+1) + \lambda(2x-3y+5) = 0$.
2. Rearranging this, we get $(1+2\lambda)x + (-1-3\lambda)y + (1+5\lambda) = 0$.
3. The distance from the point $(3,2)$ to this line is $\dfrac{|(1+2\lambda)(3) + (-1-3\lambda)(2) + (1+5\lambda)|}{\sqrt{(1+2\lambda)^2 + (-1-3\lambda)^2}}$. Given this distance is $\dfrac{7}{5}$, we have: $\dfrac{|3+6\lambda - 2-6\lambda + 1+5\lambda|}{\sqrt{1+4\lambda+4\lambda^2 + 1+6\lambda+9\lambda^2}} = \dfrac{7}{5}$ $\dfrac{|2+5\lambda|}{\sqrt{13\lambda^2+10\lambda+2}} = \dfrac{7}{5}$
4. Squaring both sides and simplifying: $25(2+5\lambda)^2 = 49(13\lambda^2+10\lambda+2)$ $25(4+20\lambda+25\lambda^2) = 637\lambda^2+490\lambda+98$ $100+500\lambda+625\lambda^2 = 637\lambda^2+490\lambda+98$ $12\lambda^2 - 10\lambda - 2 = 0$ $6\lambda^2 - 5\lambda - 1 = 0$ Factoring, $(6\lambda+1)(\lambda-1) = 0$, which gives $\lambda = 1$ or $\lambda = -\dfrac{1}{6}$. …

To find the equations of lines that pass through the intersection of two given lines and satisfy another condition, the most elegant and efficient approach is to use the concept of a "family of lines" or "concurrent lines".

Concept: Family of Lines

Consider two distinct lines given by the equations $L_1: A_1x + B_1y + C_1 = 0$ and $L_2: A_2x + B_2y + C_2 = 0$.

Any line passing through the point of intersection of $L_1$ and $L_2$ can be represented by the equation $L_1 + \lambda L_2 = 0$, where $\lambda$ is a real constant.

The intuition behind this is straightforward:

If a point $(x_0, y_0)$ is the intersection of $L_1$ and $L_2$, then it satisfies both $A_1x_0 + B_1y_0 + C_1 = 0$ and $A_2x_0 + B_2y_0 + C_2 = 0$.

Substituting these into $L_1 + \lambda L_2 = 0$, we get $0 + \lambda(0) = 0$, which is always true. This means that for any value of $\lambda$, the point of intersection $(x_0, y_0)$ will always lie on the line represented by $L_1 + \lambda L_2 = 0$. This equation thus represents the entire family of lines passing through that common intersection point.

We will use this concept to set up a general equation for the lines we are looking for, and then use the given distance condition to find the specific values of $\lambda$.

1. Find the point of intersection of the given lines.

   The given lines are:

   $L_1: x - y + 1 = 0 \quad \text{(Eq 1)}$

   $L_2: 2x - 3y + 5 = 0 \quad \text{(Eq 2)}$

   From (Eq 1), we can express $y$ in terms of $x$: $y = x+1$.

   Substitute this into (Eq 2):

   $2x - 3(x+1) + 5 = 0$

   $2x - 3x - 3 + 5 = 0$

   $-x + 2 = 0$

   $x = 2$

   Now, substitute $x=2$ back into $y=x+1$:

   $y = 2+1 = 3$

   The point of intersection is $(2,3)$. While we don't strictly need this point for the family of lines method, it's good to know it.

2. Formulate the equation of the family of lines passing through the intersection.

   Using the concept $L_1 + \lambda L_2 = 0$, the equation of any line passing through the intersection of $x-y+1=0$ and $2x-3y+5=0$ is:

   $(x - y + 1) + \lambda(2x - 3y + 5) = 0$

   Rearrange this into the standard form $Ax+By+C=0$:

   $x - y + 1 + 2\lambda x - 3\lambda y + 5\lambda = 0$

   $(1 + 2\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda) = 0 \quad \text{(Eq 3)}$

   This is the general equation of the lines we are looking for. We need to find the value(s) of $\lambda$ that satisfy the given distance condition.

3. Apply the distance condition.

   The distance from a point $(x_1, y_1)$ to a line $Ax+By+C=0$ is given by the formula:

   $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

   In our case, the point is $(x_1, y_1) = (3,2)$, and the distance $d = \frac{7}{5}$.

   From (Eq 3), we have $A = (1+2\lambda)$, $B = (-1-3\lambda)$, and $C = (1+5\lambda)$.

   Substitute these values into the distance formula:

   $\frac{7}{5} = \frac{|(1+2\lambda)(3) + (-1-3\lambda)(2) + (1+5\lambda)|}{\sqrt{(1+2\lambda)^2 + (-1-3\lambda)^2}}$

   Let's simplify the numerator:

   $|(3 + 6\lambda) + (-2 - 6\lambda) + (1 + 5\lambda)|$

   $|3 + 6\lambda - 2 - 6\lambda + 1 + 5\lambda|$

   $|2 + 5\lambda|$

   Now, simplify the denominator:

   $\sqrt{(1+2\lambda)^2 + (-1-3\lambda)^2}$

   $\sqrt{(1 + 4\lambda + 4\lambda^2) + (1 + 6\lambda + 9\lambda^2)}$

   $\sqrt{13\lambda^2 + 10\lambda + 2}$

   So, the equation becomes:

   $\frac{7}{5} = \frac{|2 + 5\lambda|}{\sqrt{13\lambda^2 + 10\lambda + 2}}$

4. Solve for $\lambda$.

   To eliminate the absolute value and square root, square both sides of the equation:

   $\left(\frac{7}{5}\right)^2 = \frac{(2 + 5\lambda)^2}{13\lambda^2 + 10\lambda + 2}$

   $\frac{49}{25} = \frac{4 + 20\lambda + 25\lambda^2}{13\lambda^2 + 10\lambda + 2}$

   Cross-multiply:

   $49(13\lambda^2 + 10\lambda + 2) = 25(4 + 20\lambda + 25\lambda^2)$ …