To find the distance, first remember the distance formula: $\displaystyle \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$. Plug in so that you have: $\displaystyle \sqrt{(1+2)^2+(4-3)^2}$. Simplify so that you get $\displaystyle \sqrt{3^2+1^2}$. This yields $\displaystyle \sqrt{10}$.

The two marked angles are same-side exterior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for $\displaystyle y$ in this equation:

$\displaystyle (y-17) + (x + 32) = 180$

$\displaystyle y + x-17 + 32= 180$

$\displaystyle y + x+15= 180$

$\displaystyle y + x+15- 15 - x = 180- 15 - x$

$\displaystyle y = 165- x$

The two marked angles are same-side interior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for $\displaystyle y$ in this equation:

$\displaystyle (2y + 15) + (x - 19) = 180$

$\displaystyle 2y + x+ 15 - 19 = 180$

$\displaystyle 2y + x-4 = 180$

$\displaystyle 2y + x-4+4 -x = 180+4 -x$

$\displaystyle 2y = 184 -x$

$\displaystyle 2y \div 2 = \left ( 184 -x \right )\div 2$

$\displaystyle y= 92 - \frac{1}{2}x$

By angle addition, 

$\displaystyle \left ( x - 13\right )+ 100 + (y + 25) = 180$

$\displaystyle x - 13\right+ 100 + y + 25 = 180$

$\displaystyle x + y - 13\right+ 100+ 25 = 180$

$\displaystyle x + y +112 = 180$

$\displaystyle x + y +112 - 112= 180- 112$

$\displaystyle x + y = 68$

Four statements can be eliminated by the various parallel theorems and postulates. Congruence of alternate interior angles or corresponding angles forces the lines to be parallel, so

 $\displaystyle \angle 5 \cong \angle 4 \Rightarrow l \parallel m$  and

$\displaystyle \angle 2 \cong \angle 6 \Rightarrow l \parallel m$ .

Also, if same-side interior angles or same-side exterior angles are supplementary, the lines are parallel, so 

$\displaystyle m \angle 4 + m \angle 7 = 180 ^{\circ } \Rightarrow l \parallel m$ and 

$\displaystyle m \angle 1 + m \angle 6 = 180 ^{\circ } \Rightarrow l \parallel m$.

However, $\displaystyle \angle 5 \cong \angle 8$ whether or not $\displaystyle l \parallel m$ since they are vertical angles, which are always congruent.

Supplementary angles and complementary angles have measures totaling $\displaystyle 180^{\circ }$ and $\displaystyle 90^{\circ }$, respectively.

$\displaystyle m \angle 2 = 100 ^{\circ }$, so its supplement $\displaystyle \angle 1$ has measure 

$\displaystyle m \angle 1 = 180^{\circ } - m \angle 2 = 180^{\circ } - 100 ^{\circ } = 80^{\circ }$

$\displaystyle \angle 3$, the complement of $\displaystyle \angle 1$, has measure 

$\displaystyle m \angle 3 = 90^{\circ } - m \angle 1 = 90^{\circ } - 80 ^{\circ } = 10^{\circ }$

$\displaystyle \angle 1$ and $\displaystyle \angle 2$ form a linear pair, so their angle measures total $\displaystyle 180^{\circ}$. Set up and solve the following equation:

$\displaystyle m \angle 1+ m \angle 2 = 180^{\circ}$

$\displaystyle \left ( x+17 \right ) + \left (7x+11 \right ) = 180$

$\displaystyle 8x+28 = 180$

$\displaystyle 8x=152$

$\displaystyle x = 19$

$\displaystyle m \angle 1=\left ( x+17 \right )^{\circ} = \left ( 19+17 \right )^{\circ} = 36^{\circ}$

Two angles that form a linear pair are supplementary - that is, they have measures that total $\displaystyle 180^{\circ}$. Therefore, we set and solve for $\displaystyle x$ in this equation:

$\displaystyle (2x+72)+(5x-125)= 180$

$\displaystyle 7x - 53 = 180$

$\displaystyle 7x = 233$

$\displaystyle x = \frac{233}{7}= 33 \frac{2}{7}$

The two angles have measure

$\displaystyle 2 \cdot 33 \frac{2}{7} +72= 138 \frac{4}{7} ^{\circ}$

and 

$\displaystyle 5 \cdot 33 \frac{2}{7} -125 = 41\frac{3}{7}^{\circ}$

$\displaystyle 41\frac{3}{7}^{\circ}$ is the lesser of the two measures and is the correct choice.

Two vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure. Therefore, we set up and solve the equation

$\displaystyle 3x+14= 5x-17$

$\displaystyle 14= 2x-17$

$\displaystyle 2x= 31$

$\displaystyle x= 15\frac{1}{2}$

$\displaystyle 3x+14 = 3 \cdot 15\frac{1}{2} + 14 = 46\frac{1}{2} + 14= 60\frac{1}{2} ^{\circ }$

When a transversal such as $\displaystyle t$ crosses two parallel lines, two corresponding angles - angles in the same relative position to their respective lines - are congruent. Therefore, 

$\displaystyle 3x+2y = 4x+21$

$\displaystyle 3x+2y- 3x - 21 = 4x+21 - 3x - 21$

$\displaystyle x = 2y - 21$

Two same-side interior angles are supplementary - that is, their angle measures total 180 - so

$\displaystyle 3x+2y + 2y-27 = 180$

$\displaystyle 3x+4y-27 = 180$

$\displaystyle 3x+4y= 207$

We can solve this system by the substitution method as follows:

$\displaystyle 3( 2y - 21)+4y= 207$

$\displaystyle 6y-63+4y= 207$

$\displaystyle 10y-63= 207$

$\displaystyle 10y = 270$

$\displaystyle y = 27$

Backsolve:

$\displaystyle x = 2y - 21$

$\displaystyle x = 2 (27)- 21 = 54-21 = 33$

$\displaystyle x+y = 27+33 = 60$, which is the correct response.