Use the point-slope formula with \(\displaystyle m = -\frac{1}{5}, x_{1}= 1, y_{1} = 4\)

\(\displaystyle y -y _{1} = m (x -x _{1} )\)

\(\displaystyle y -4 = -\frac{1}{5} (x -1 )\)

\(\displaystyle y -\frac{20}{5} = -\frac{1}{5} x + \frac{1}{5}\)

\(\displaystyle y = -\frac{1}{5} x + \frac{21}{5}\)

Rewrite each in the slope-intercept form, \(\displaystyle y = mx + b\); \(\displaystyle m\) will be the slope.

\(\displaystyle 2x+y = 15\)

\(\displaystyle 2x+y -2x = 15-2x\)

\(\displaystyle y=15-2x\)

\(\displaystyle y = -2x+ 15\)

The slope of this line is \(\displaystyle m = -2\).

\(\displaystyle 3x+y = 15\)

\(\displaystyle 3x+y -3x = 15-3x\)

\(\displaystyle y=15-3x\)

\(\displaystyle y = -3x+ 15\)

The slope of this line is \(\displaystyle m = -3\).

Since \(\displaystyle -2 > -3\), (A) is greater.

Rewrite each in the slope-intercept form, \(\displaystyle y = mx + b\); \(\displaystyle m\) will be the slope.

\(\displaystyle 4x - 2y = 10\)

\(\displaystyle 4x - 2y -4x = 10-4x\)

\(\displaystyle - 2y = -4x+ 10\)

\(\displaystyle - 2y \div (-2) =\left ( -4x+ 10 \right )\div (-2)\)

\(\displaystyle y = 2x -5\)

The slope of the line of \(\displaystyle 4x - 2y = 10\) is \(\displaystyle m = 2\)

\(\displaystyle y - 2x = 7\)

\(\displaystyle y - 2x + 2x = 7+ 2x\)

\(\displaystyle y = 2x + 7\)

The slope of the line of \(\displaystyle y - 2x = 7\) is also \(\displaystyle m = 2\)

The slopes are equal.

The slope of a line through the points \(\displaystyle (A+ 1, B+ 1 )\) and \(\displaystyle (A-1, B- 1 )\) can be found by setting 

\(\displaystyle x_{1} = A-1,y_{1} = B - 1, x_{2} = A+1, y_{2} = B+ 1\)

in the slope formula:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle = \frac{ (B+ 1)- (B- 1)}{ (A+ 1)- (A- 1)}\)

\(\displaystyle = \frac{ B-B+ 1+1}{ A-A + 1 + 1}\)

\(\displaystyle = \frac{ 2}{ 2}\)

\(\displaystyle =1\)

The slope of a line through the points \(\displaystyle (B- 1, A- 1 )\) and \(\displaystyle (B+1, A+ 1 )\) can be found similarly:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle = \frac{(A+ 1)- (A- 1) }{ (B+ 1)- (B- 1) }\)

\(\displaystyle = \frac{A-A + 1+1}{B-B + 1 + 1}\)

\(\displaystyle = \frac{ 2}{ 2}\)

\(\displaystyle =1\)

The lines have the same slope.

The slope of a line through the points \(\displaystyle (A, B)\) and \(\displaystyle (-A, B )\), can be found by setting 

\(\displaystyle x_{1} = -A,y_{1} =B, x_{2} = A, y_{2} = B\):

in the slope formula:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\displaystyle m = \frac{B-B}{A - (-A)} = \frac{0}{A+A} = \frac{0}{2A} = 0\)

For this problem, simply plug in the values for the point (15,6) into the different equations (15 for the \(\displaystyle x\)-value and 6 for the \(\displaystyle y\)-value) to see which one fits.

\(\displaystyle (6)=\frac{1}{2}(15)-7\)          (NO)

\(\displaystyle (6)=\frac{2}{3}(15)-4\)          (YES!)

\(\displaystyle (6)=\frac{-1}{2}(15)+7\)       (NO)

\(\displaystyle (6)=\frac{-2}{3}(15)+4\)        (NO)  

At the intersection point of the two lines the \(\displaystyle x\)- and \(\displaystyle y\)- values for each equation will be the same. Thus, we can set the two equations as equal to each other:

\(\displaystyle \frac{3}{4}x-11=\frac{-5}{6}x+8\)

\(\displaystyle \frac{3}{4}x=\frac{-5}{6}x+19\)

\(\displaystyle \frac{9}{12}x=\frac{-10}{12}x+19\)

\(\displaystyle \frac{19}{12}x=19\)

\(\displaystyle (\frac{12}{19})(\frac{19}{12}x)=19 (\frac{12}{19})\)

\(\displaystyle \large x=12\)

\(\displaystyle y=\frac{3}{4}(12)-11= -2\)

point of intersection \(\displaystyle =\left ( 12,-2 \right )\)

In the formula \(\displaystyle y=mx+b\), the y-intercept is represented by \(\displaystyle b\) (because if you set \(\displaystyle x\) to zero, you are left with \(\displaystyle y=b\) ).

Thus, to find the \(\displaystyle x\)-intercept, set the \(\displaystyle y\) value to zero and solve for \(\displaystyle x\).

\(\displaystyle 0=\frac{7}{9}x-3\)

\(\displaystyle \frac{-7}{9}x=-3\)

\(\displaystyle (\frac{-9}{7})\frac{-7}{9}x=-3 (\frac{-9}{7})\)

\(\displaystyle x=\frac{27}{7}=3\frac{6}{7}\)

There are four quadrants in the coordinate plane. Quadrant I is the top right, and they are numbered counter-clockwise. Since the x-coordinate is \(\displaystyle -1\), you go to the left one unit (starting from the origin). Since the y-coordinate is \(\displaystyle 4\), you go upwards four units. Therefore, you are in Quadrant II.

Two angles that are supplementary add up to 180 degrees. They cannot both be acute, nor can they both be obtuse. Therefore, "Supplementary" is the correct answer.