Animated Solution for Mathematics - Straight Lines: A line cuts the x-axis at $A(7,0)$ and the y-axis at $B(0,-5)$. A variable line $PQ$ is drawn perpendicular to $AB$ cutting the x-axis in $P$ and the y-axis in $Q$. If $AQ$ and $BP$ intersect at $R$, find the locus of $R$.

Imagine you are standing on a Cartesian plane. You see two fixed anchors: point $A(7,0)$ on the x-axis and point $B(0,-5)$ on the y-axis. These two points define a fixed, unmoving line, $AB$.

Now, imagine a second line, $PQ$, dancing across the axes. It is not just any line; it is a line that maintains a strict, perpendicular relationship with our fixed line $AB$.

As $PQ$ slides, its intercepts $P(p,0)$ and $Q(0,q)$ change. Consequently, the intersection point $R$ of the lines $AQ$ and $BP$ traces a path. Our mission is to uncover the secret shape of this path—the locus of $R$.

First, let us respect the fixed line $AB$. Its slope, $m_1$, is the gatekeeper of our perpendicularity condition. Using the slope formula:

This value is our constant. It tells us exactly how the line $AB$ is tilted.

Any line perpendicular to it must have a slope $m_2$ such that $m_1\cdot m_2=-1$. This means $m_2=-\frac{7}{5}$. This is the fundamental constraint governing our variable line $PQ$.

Now, consider the line $PQ$. It cuts the x-axis at $P(p,0)$ and the y-axis at $Q(0,q)$. Its slope is:

Applying our perpendicularity condition, we set $-\frac{q}{p}=-\frac{7}{5}$, which simplifies beautifully to $5q=7p$.

This is the heartbeat of our problem—a simple, elegant relationship between the intercepts $p$ and $q$ that must hold true at every moment of the dance.

We are interested in the intersection point $R(h,k)$ of lines $AQ$ and $BP$. Using the intercept form, the equation of line $AQ$ is:

Since $R(h,k)$ lies on this line, we have $\frac{h}{7}+\frac{k}{q}=1$. Rearranging this to isolate $q$, we find:

Similarly, the equation of line $BP$ is $\frac{x}{p}-\frac{y}{5}=1$. Substituting $R(h,k)$ gives $\frac{h}{p}-\frac{k}{5}=1$, which leads us to:

We now have $p$ and $q$ expressed in terms of the coordinates of $R(h,k)$. We return to our heartbeat equation: $5q=7p$.

Substituting our expressions for $p$ and $q$, we get:

Notice the symmetry! We can divide both sides by $35$, leaving us with:

Cross-multiplying yields $k(5+k)=h(7-h)$, or $5k+k^2=7h-h^2$. Rearranging everything to one side, we arrive at $h^2+k^2-7h+5k=0$.

Replacing $(h,k)$ with $(x,y)$, we find the locus:

The Circle Revealed

What we have discovered is that the point $R$ does not wander aimlessly. It is bound to the path of a circle.

The perpendicularity of the lines $AB$ and $PQ$ forces the intersection point $R$ to trace this perfect, closed curve. It is a testament to the hidden order in coordinate geometry—where seemingly independent variables $p$ and $q$ are locked in a dance that results in a beautiful, symmetric shape.