## Task B-4.3.

The lines $p_{1} \ldots y=\frac{1}{4} x, p_{2} \ldots y=9 x$ and the point $T(6,-1)$ are given. Determine the equation of the line that passes through the point $T$, and intersects the x-axis, lines $p_{1}, p_{2}$, and the y-axis at points $A, B, C, D$ respectively, such that $|A B|=|C D|$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1020492bf6e1d2c949fdg-17.jpg?height=922&width=736&top_left_y=767&top_left_x=660)

The line passing through point $T$ has the equation:

$$
\begin{aligned}
& y+1=k(x-6) \\
& y=k x-(6 k+1)
\end{aligned}
$$

To intersect the $x$-axis, $p_{1}, p_{2}$, and the $y$-axis, the line must not be parallel to the given lines and coordinate axes, nor pass through the origin, i.e., it must satisfy:

$$
k \neq \frac{1}{4}, k \neq 9, k \neq 0, k \neq-\frac{1}{6}
$$

The coordinates of points $A, B, C, D$ are determined by solving the system of equations of the respective lines:

$$
\begin{aligned}
& A\left(\frac{6 k+1}{k}, 0\right) \\
& B\left(\frac{4(6 k+1)}{4 k-1}, \frac{6 k+1}{4 k-1}\right) \\
& C\left(\frac{6 k+1}{k-9}, \frac{9(6 k+1)}{k-9}\right) \\
& D(0,-6 k-1)
\end{aligned}
$$

Since $|A B|=|C D|$, we have:

$$
\begin{aligned}
& \left(\frac{4(6 k+1)}{4 k-1}-\frac{6 k+1}{k}\right)^{2}+\left(\frac{6 k+1}{4 k-1}\right)^{2}=\left(\frac{6 k+1}{k-9}\right)^{2}+\left(\frac{9(6 k+1)}{k-9}+(6 k+1)\right)^{2} /:(6 k+1)^{2} \\
& \left(\frac{4}{4 k-1}-\frac{1}{k}\right)^{2}+\left(\frac{1}{4 k-1}\right)^{2}=\left(\frac{1}{k-9}\right)^{2}+\left(\frac{9}{k-9}+1\right)^{2} \\
& \frac{1}{(4 k-1)^{2} k^{2}}+\frac{1}{(4 k-1)^{2}}=\frac{1}{(k-9)^{2}}+\frac{k^{2}}{(k-9)^{2}} \\
& \frac{1+k^{2}}{(4 k-1)^{2} k^{2}}=\frac{1+k^{2}}{(k-9)^{2}} \\
& (4 k-1)^{2} k^{2}=(k-9)^{2} \\
& |(4 k-1) k|=|k-9|
\end{aligned}
$$

We have the following possibilities:

1. 

$$
4 k^{2}-k=k-9 \Rightarrow 4 k^{2}-2 k+9=0 \Rightarrow \text { no real solutions. }
$$

2. 

$$
4 k^{2}-k=9-k \Rightarrow 4 k^{2}=9 \Rightarrow k_{1,2}= \pm \frac{3}{2}
$$

There are two lines with the desired property:

$$
p^{\prime} \ldots y=\frac{3}{2} x-10, p^{\prime \prime} \ldots y=-\frac{3}{2} x+8
$$