4. In the coordinate system in the plane, a point $T(2,4)$ is given. Through point $T$ pass lines $p$ and $t$. Line $p$ intersects the $x$-axis at point $A$, and the $y$-axis at point $D$. Line $t$ intersects the $x$-axis at point $B$, and the $y$-axis at point $C$. Determine the equations of lines $p$ and $t$ if triangles $\triangle A B T$ and $\triangle C D T$ are isosceles with bases $\overline{A B}$, respectively $\overline{C D}$, and the ratio of their areas is $P(\triangle A B T): P(\triangle C D T)=9: 4$.
4. Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_c165b3842bf96502feaeg-23.jpg?height=668&width=551&top_left_y=314&top_left_x=387)

$\triangle A B T$ is isosceles, and the height from vertex $T$ divides the base $\overline{A B}$ into two equal parts.

Then $a=|A B|, \frac{a}{2}=|N B|,|T N|=4$ and $P(\triangle A B T)=\frac{a \cdot 4}{2}=2 a$.

$\triangle C D T$ is isosceles, and the height from vertex $T$ divides the base $\overline{C D}$ into two equal parts.

Then $b=|C D|, \frac{b}{2}=|M C|,|T M|=2$ and $P(\Delta C D T)=\frac{b \cdot 2}{2}=b$.

From the given ratio of areas $P(\triangle A B T): P(\triangle C D T)=9: 4$ we get: $2 a: b=9: 4$

$$
\begin{aligned}
9 b & =8 a \\
b & =\frac{8}{9} a
\end{aligned}
$$

In triangle $\triangle A B T$, $\overline{T N} \perp \overline{A B}$, and in triangle $\triangle C D T$, $\overline{T M} \perp \overline{C D}$.

It follows that $\overline{M C} \| \overline{T N}$ and $\overline{T M} \| \overline{N B}$, so triangles $\triangle N B T$ and $\triangle M T C$ are similar by the AA similarity criterion.

From the similarity, the proportion holds: $|N B|:|T N|=|T M|:|M C|$

$$
\begin{gathered}
\frac{a}{2}: 4=2: \frac{b}{2} \\
\frac{a}{2} \cdot \frac{b}{2}=8 \\
a b=32
\end{gathered}
$$

Substituting $b=\frac{8}{9} a$ into the equation $a b=32$ yields: $a \cdot \frac{8}{9} a=32$

$$
\begin{aligned}
& 8 a^{2}=288 \\
& a^{2}=36 \\
& a=6 .
\end{aligned}
$$

The coordinates of point $N$ are $N(2,0)$, and since point $N$ is the midpoint of segment $\overline{A B}$, the coordinates of points $A$ and $B$ are $A(-1,0), B(5,0)$.

The line $p$ is defined by points $A(-1,0)$ and $T(2,4)$, and let its equation be $y=a_{1} x+b_{1}$.

Then:

$$
\begin{aligned}
& 0=a_{1} \cdot(-1)+b_{1} \\
& \underline{4=a_{1} \cdot 2+b_{1}}
\end{aligned}
$$

Solving this system of equations yields $a_{1}=\frac{4}{3}, b_{1}=\frac{4}{3}$.

Thus, $p \ldots y=\frac{4}{3} x+\frac{4}{3}$.

The line $t$ is defined by points $B(5,0)$ and $T(2,4)$, and let its equation be $y=a_{2} x+b_{2}$.

Then:

$$
\begin{aligned}
& 0=a_{2} \cdot 5+b_{2} \\
& 4=a_{2} \cdot 2+b_{2}
\end{aligned}
$$

Solving this system of equations yields $a_{2}=-\frac{4}{3}, b_{2}=\frac{20}{3}$.

Thus, $t \ldots y=-\frac{4}{3} x+\frac{20}{3}$.

Note: On the sketch, the positions of lines $p$ and $t$, points $A$ and $B$, and points $C$ and $D$ could be swapped. In that case, the solutions would also have swapped values.