Animated Solution for Mathematics - Straight Lines: If the perpendicular bisector of the line segment joining the points $P(1,4)$ and $Q(k,3)$ has $y$-intercept equal to $-4$, then a value of $k$ is:

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a coordinate geometry problem; we are choreographing a dance between two points, $P(1,4)$ and $Q(k,3)$, and their perpendicular bisector.

This problem is a classic, a beautiful test of your ability to translate geometric intuition into the language of algebra. Let us break it down, step by step, and find the hidden value of $k$.

Imagine you are standing on a coordinate plane. You have a fixed point $P(1,4)$ and a mysterious point $Q(k,3)$. We want to find the perpendicular bisector of the segment $PQ$.

The most important feature of this line is that it must pass through the center of the segment, which we call the midpoint $M$. Using the midpoint formula, which is the average of the coordinates, we find:

This point $M$ is our anchor. Any line that bisects $PQ$ must pass through this coordinate. It is the heart of our geometric construction.

Now, let us determine the slope. The slope of the segment $PQ$ tells us how it is tilted. Using the slope formula $m=\frac{y_2-y_1}{x_2-x_1}$, we calculate the slope of $PQ$ (denoted as $m_1$):

Here is where the magic happens. The perpendicular bisector is, by definition, perpendicular to $PQ$. The product of the slopes of two perpendicular lines is always $-1$.

If the slope of the bisector is $m_2$, then $m_1\cdot m_2=-1$. This implies $m_2=-\frac{1}{m_1}$. Substituting our $m_1$, we get:

The slope of our bisector is simply $k-1$. We now have both the slope and a point on the line, allowing us to build the equation.

Using the point-slope form, $y-y_1=m(x-x_1)$, where $(x_1,y_1)$ is our midpoint $M$ and $m$ is our slope $m_2$, we write:

This expression represents the relationship between $x$ and $y$ that holds true for every point on the perpendicular bisector.

The problem provides one final, crucial piece of information: the $y$-intercept is $-4$. This means when $x=0$, $y=-4$.

Substituting these values into our equation, we obtain:

Simplifying the left side gives $-\frac{15}{2}$. The right side becomes $(k-1)\left(-\frac{k+1}{2}\right)$. Utilizing the algebraic identity $(k-1)(k+1)=k^2-1$, the equation becomes:

Canceling the negative signs and the denominators of $2$ on both sides, we are left with:

Taking the square root, we find $k=\pm 4$. Therefore, the possible values for $k$ are $4$ or $-4$.