Animated Solution for Mathematics - Straight Lines: For $a>b>c>0$, the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and $bx+ay+c=0$ is less than $2\sqrt{2}$. Then

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, might seem like a tedious exercise in algebraic manipulation.

But as we peel back the layers, you will see that it is actually a beautiful dance of symmetry and logic. Let us begin.

We are given two lines: with the constraint $a>b>c>0$.

When you see coefficients like $a$ and $b$ swapped between $x$ and $y$, your intuition should immediately scream 'symmetry!' This is not a coincidence; it is a structural feature of the problem.

We want to find the intersection point $P$. Instead of jumping straight into complex substitution, let us look for a way to simplify the system. If we subtract $L_2$ from $L_1$, we get:

Notice how the constant $c$ vanishes instantly? This is the 'spark' of the solution. We are left with:

Or, more cleanly:

Since $a>b$, we know $a - b eq 0$. We can safely divide by $(a-b)$ to find that $x=y$. This is a powerful revelation! Our intersection point $P$ lies on the line $y=x$.

Now that we know $x=y$, finding the coordinates of $P$ becomes trivial. Let us substitute $y=x$ into $L_1$:

Since $x=y$, our intersection point $P$ is simply:

Because $a,b,c>0$, the coordinates are negative, placing $P$ firmly in the third quadrant. We have successfully reduced a system of two lines to a single, elegant point.

We are asked to find the distance between $P$ and the point $Q(1,1)$. Using the distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, we have:

This simplifies to:

Since $a,b,c$ are all positive, the term inside the modulus is clearly positive, so we can drop the bars:

The problem states that $d<2\sqrt{2}$. Let us plug in our expression for $d$:

Dividing both sides by $\sqrt{2}$ gives us:

Since $a+b>0$, we can cross-multiply without flipping the inequality:

This matches our first correct option! But we are not done yet. We must check the other options using our initial constraint $a>b>c>0$.

Since $a>b$, we know $a-b>0$. Adding a positive $c$ to this positive quantity keeps it positive:

This matches our second correct option.

By leveraging the symmetry of the lines and carefully applying the distance formula, we have navigated through the algebra to find that both $a+b-c>0$ and $a-b+c>0$ are true.

This problem is a perfect example of how coordinate geometry rewards those who look for patterns before diving into calculations. Keep practicing, stay curious, and remember: the math is always on your side!