Animated Solution for Mathematics - Straight Lines: Two sides of a rhombus are along the lines, $x-y+1=0$ and $7x-y-5=0$. If its diagonals intersect at $(1,2)$, then which one of the following is a vertex of this rhombus?

Welcome, future engineers! Today, we are going to dive into the elegant world of coordinate geometry. We are looking at a rhombus—a shape of perfect symmetry and balance.

In the JEE Advanced landscape, a rhombus is not just a quadrilateral; it is a playground of parallel lines, equidistant points, and beautiful algebraic cancellations. Let us embark on this journey to find the hidden vertices of our diamond.

Imagine you are standing on a coordinate plane. You have two lines, $L_1:x-y+1=0$ and $L_2:7x-y-5=0$. These are two sides of our rhombus.

The problem statement provides a center at $(1,2)$, but as we analyze the geometry, we realize there is a typo. For these specific lines to form a valid, symmetric rhombus, the center must be at $O(-1,-2)$.

This is a crucial lesson in JEE preparation: always verify the geometric consistency of the problem. If the math feels like it is fighting you, pause and check your premises. With our center $O(-1,-2)$ firmly established, we can proceed.

Since a rhombus is a special type of parallelogram, its opposite sides must be parallel. This is our golden key.

If $L_1$ is $x-y+1=0$, then the side opposite to it, let us call it $L_3$, must have the same slope. Thus, it must take the form $x-y+\lambda =0$.

Similarly, the side opposite to $L_2$ must take the form $7x-y+\mu =0$. We have introduced two unknowns, $\lambda$ and $\mu$, but we have a powerful tool to find them: the distance formula.

The center of a rhombus is the midpoint of its diagonals, which means it is perfectly equidistant from all four sides. This is the 'magic' of the rhombus.

The perpendicular distance from the center $O(-1,-2)$ to $L_1$ must be exactly equal to the distance from $O$ to $L_3$. We use the perpendicular distance formula:

For $L_1$, the distance is $\frac{|-1-(-2)+1|}{\sqrt{1^2+(-1{)}^2}}$. For $L_3$, it is $\frac{|-1-(-2)+\lambda |}{\sqrt{1^2+(-1{)}^2}}$.

Notice how the denominators, $\sqrt{2}$, appear on both sides? They cancel out instantly! We are left with $|-1+2+1|=|-1+2+\lambda |$, which simplifies to $2=|1+\lambda |$.

This gives us two cases: $1+\lambda =2$ (which gives us our original line $L_1$) or $1+\lambda =-2$ (which gives us our new side $L_3$). Thus, $\lambda =-3$, and our line $L_3$ is $x-y-3=0$.

We repeat this exact logic for $L_2$ and $L_4$, finding $\mu =15$, giving us $L_4:7x-y+15=0$.

Now, we have the equations for all four sides of our rhombus. The vertices are simply the points where these lines intersect. To find a vertex, we look at the intersection of adjacent sides, such as $L_2$ and $L_3$:

Subtracting the second equation from the first, the $y$ terms vanish—a moment of pure algebraic satisfaction! We get $6x-2=0$, which leads us to $x=\frac{1}{3}$.

Substituting this back into $L_3$, we find $y=-\frac{8}{3}$. Our vertex is $(\frac{1}{3},-\frac{8}{3})$.

Look at what we have achieved. We navigated a typo, utilized the symmetry of the rhombus, leveraged the distance formula, and solved a system of linear equations.

Geometry is not about memorizing formulas; it is about visualizing the relationships between lines and points. You have successfully dissected this problem. Keep this mindset—stay curious, stay critical, and keep solving. You are doing great!