To determine if two lines are perpendicular, only the slope needs to be considered. The slopes of perpendicular lines are the negative reciprocals of each other. Knowing a single point on the line is not sufficient, as an infinite number of lines can pass through and individual point.

Statement 1 alone establishes nothing about the angle $\displaystyle n$ makes with $\displaystyle m$, as it is not part of the triangle. Statement 2 alone only establishes that $\displaystyle AY = YB$.

Assume both statements are true. Then $\displaystyle \overleftrightarrow{XY}$ is an altitude of an equilateral triangle, making it - and $\displaystyle n$ - perpendicular with the base $\displaystyle \overline {AB}$ - and $\displaystyle m$.

Assume Statement 1 alone. The measure of one of the angles formed is

$\displaystyle x^{2}+ 9 = 9^{2}+ 9= 81 + 9 = 90$ degrees.

Assume Statement 2 alone.

By substituting $\displaystyle x-1$ for $\displaystyle y$, one angle measure becomes

$\displaystyle 11y + 2 = 11 (x-1) +2= 11x-11+2= 11x-9$

The marked angles are a linear pair and thus their angle measures add up to 180 degrees; therefore, we can set up an equation:

$\displaystyle \left ( 11x-9 \right )+ \left ( x^{2}+9 \right ) = 180$

$\displaystyle x^{2}+11x = 180$

$\displaystyle x^{2}+11x - 180 = 0$

$\displaystyle (x+20)(x-9) = 0$

$\displaystyle x= -20$ or $\displaystyle x= 9$

$\displaystyle x= -20$ yields illegal angle measures - for example, 

$\displaystyle x^{2}+ 9 = (-20)^{2}+ 9= 409$

$\displaystyle x= 9$ yields angle measures $\displaystyle 90 ^{\circ }$ for both angles; the angles are right and the lines are perpendicular.

Statement 1 alone establishes by definition that $\displaystyle l \perp \! \! \! \! \! / \; t$, but does not establish any relationship between $\displaystyle m$ and $\displaystyle t$.

By Statement 2 alone, since alternating interior angles are congruent, $\displaystyle l || m$, but no conclusion can be drawn about the relationship of $\displaystyle t$, since the actual measures of the angles are not given.

Assume both statements are true. By Statement 2, $\displaystyle l || m$. $\displaystyle \angle 2$ and $\displaystyle \angle 6$ are corresponding angles formed by a transversal across parallel lines, so $\displaystyle m \angle 6 = m \angle 2 = 89^{\circ }$. $\displaystyle \angle 6$ is not a right angle, so $\displaystyle m\perp \! \! \! \! \! / \; t$.

Assume Statement 1 alone. Then, as a consequence of congruence, $\displaystyle \angle XYA$ and $\displaystyle \angle XYB$ are congruent. They form a linear pair of angles, so they are also supplementary. Two angles that are both congruent and supplementary must be right angles, so $\displaystyle m \perp n$.

Assume Statement 2 alone. Then $\displaystyle \angle AXY \cong \angle BXY$, but without any other information about the angles that $\displaystyle \overleftrightarrow{AX}, \overleftrightarrow{BX},$ or $\displaystyle n$ make with $\displaystyle m$, it cannot be determined whether $\displaystyle m \perp n$ or not.

The lines of the two equations must have slopes that are the opposites of each others reciprocals.

Write each equation in slope-intercept form:

$\displaystyle 4x+5y = 20$

$\displaystyle 5y = -4x+20$

$\displaystyle y = -\frac{4}{5} x+4$

$\displaystyle m_{1}= -\frac{4}{5}$

$\displaystyle Ax+8y=B$

$\displaystyle 8y=-Ax+B$

$\displaystyle y=-\frac{A}{8}x+\frac{B}{8}$

$\displaystyle m_{2}= -\frac{A}{8}$

As can be seen, knowing the value of $\displaystyle A$ is necessary and sufficient to answer the question. The value of $\displaystyle B$ is irrelevant.

The answer is that Statement 1 alone is sufficient to answer the question, but Statement 2 alone is not sufficient to answer the question.

Even with both statements, $\displaystyle AE$ cannot be determined because the length of $\displaystyle BD$ is missing.

For example, we can have $\displaystyle AB = DE = 8$ and $\displaystyle BD=16$, making $\displaystyle AE = 32$; or, we can have  $\displaystyle AB = DE = 10$ and $\displaystyle BD=14$, making $\displaystyle AE = 34$. Neither scenario violates the conditions given.

Both statements are equivalent, as both are equivalent to stating that $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are collinear. Therefore, it suffices to determine whether the fact that the points are collinear is sufficient to answer the question. 

In both of the above figures,  $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are collinear, so the conditions of both statements are met. But in the top figure, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray, since $\displaystyle C$ is on $\displaystyle \overrightarrow{AB}$; in the bottom figure, since $\displaystyle B$ and $\displaystyle C$ are on opposite sides of $\displaystyle A$, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays.

We show Statement 1 alone is insufficient to determine whether the two rays are the same by looking at the figures below. In the first figure, $\displaystyle B$ is the midpoint of $\displaystyle \overline{AC}$.

In both figures, $\displaystyle AC = 2 \cdot AB$. But only in the second figure, $\displaystyle B$ and $\displaystyle C$ are on the opposite side of the line from $\displaystyle A$, so only in the second figure, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays.

Assume Statement 2 alone. If $\displaystyle B$ is the midpoint of $\displaystyle \overline{AC}$, then, as seen in the top figure, $\displaystyle B$ is on $\displaystyle \overrightarrow{AC}$. Therefore, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are the same ray, not opposite rays.

Statement 1 alone does not answer the question.

Case 1: Examine the figure below.

$\displaystyle AB+BC= AB + AB + AC > AC$,

thereby meeting the condition of Statement 1.

Also, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays, since $\displaystyle B$ and $\displaystyle C$ are on opposite sides of the same line from $\displaystyle A$.

Case 2: Suppose $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ are noncollinear. 

The three points are vertices of a triangle, and by the Triangle Inequality Theorem, 

$\displaystyle AB+BC > AC$.

Furthermore, $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are not part of the same line and are not opposite rays.

Now assume Statement 2 alone. As can be seen in the diagram above, if $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are opposite rays, then by segment addition, $\displaystyle AB+BC = AC$, making Statement 2 false. Contrapositively, if Statement 2 holds, and $\displaystyle AB+ AC \ne BC$, then $\displaystyle \overrightarrow{AB}$ and $\displaystyle \overrightarrow{AC}$ are not opposite rays.