Animated Solution for Mathematics - Straight Lines: Lines $L_1\equiv ax+by+c=0$ and $L_2\equiv lx+my+n=0$ intersect at the point $P$ and make an angle $\theta$ with each other. Find the equation of a line $L$ different from $L_2$ which passes through $P$ and makes the same angle $\theta$ with $L_1$.

Imagine you are standing at the intersection point $P$ of two lines, $L_1$ and $L_2$. These lines are like two paths crossing in a field, and the angle $\theta$ between them is the sharp turn you would make if you switched from one path to the other.

Now, we are tasked with finding a new path, $L$, that also passes through $P$ and makes the same angle $\theta$ with $L_1$. If you visualize this, you will see that $L$ is essentially the mirror image of $L_2$, with $L_1$ acting as the mirror. This is the core geometric reality of the problem.

Since our required line $L$ passes through the intersection point $P$, we can invoke the concept of the family of lines. Any line passing through the intersection of $L_1\equiv ax+by+c=0$ and $L_2\equiv lx+my+n=0$ can be written as:

This equation is a magic wand; by varying $\lambda$, we can sweep through every possible line passing through $P$. Our goal is to find the specific $\lambda$ that corresponds to our reflected line.

Instead of wrestling with slopes, let us look at the normal vectors. The normal vector of $L_1$ is ${\vec{n}}_1=a\hat{i}+b\hat{j}$, and the normal vector of $L_2$ is ${\vec{n}}_2=l\hat{i}+m\hat{j}$.

From our family of lines equation, the normal vector of the new line $L$ is simply ${\vec{n}}_L={\vec{n}}_2+\lambda {\vec{n}}_1$. This is where the beauty of vector algebra shines.

Because $L$ is the reflection of $L_2$ across $L_1$, its normal ${\vec{n}}_L$ must be the reflection of ${\vec{n}}_2$ across the normal ${\vec{n}}_1$. The vector reflection formula is:

Applying this to our case, we get:

By comparing our two expressions for ${\vec{n}}_L$, we can instantly identify:

We know the dot product ${\vec{n}}_2\cdot {\vec{n}}_1=al+bm$ and the magnitude squared $|{\vec{n}}_1{|}^2=a^2+b^2$. Substituting these values back into our family of lines equation, we get:

Multiplying by $(a^2+b^2)$ to clear the fraction, we arrive at the final, beautifully symmetric equation:

This result is not just an answer; it is a testament to the elegance of coordinate geometry when viewed through the lens of vectors.