Solution

In these problems we will make use of primarily two formulas: the slope formula and the Point-Slope Form of the equation of a line.\[ \begin{array}{rrcl} \text{Slope:}&m&=&\frac{y_2-y_1}{x_2-x_1}\\ \text{Point Slope Form:}&y&=&m(x-x_1)+y_1 \end{array} \]

A) First, what is the slope of the line which connects the two points given? \[ \begin{array}{{rcl}} m &=& \displaystyle\frac{-\frac{{11}}{{7}}-\left(-\frac{{17}}{{7}}\right)}{-2 - (-5)}\\ m&=&\displaystyle\frac{-\frac{{11}}{{7}}+\frac{{17}}{{7}}}{-2 +5}\\ m&=&\displaystyle\frac{\frac{{6}}{{7}}}{{3}}\\ m&=&\displaystyle\frac{{6}}{7\times 3}\\ m&=&\displaystyle\frac{{2}}{{7}} \end{array} \]

Since the line we want to create is parallel to the line with the above slope, we know the slopes are the same. Using the Point-Slope Form: \[ \begin{array}{rcl} y&=&\frac{{2}}{{7}}(x-(-5))-\frac{{24}}{{7}}\\ y&=&\frac{{2}}{{7}}(x+5)-\frac{{24}}{{7}}\\ \end{array} \]

What I particularly like about this form, is we can leave our answer right here! Unless the question demands a different form, it is fine to leave it in this particular form.

B) Similar to the previous problem, we need to identify the slope of the line connecting the points:\[ \begin{array}{rcl} m &=& \displaystyle\frac{-\frac{{9}}{{2}}-\left(-\frac{{3}}{{2}}\right)}{-5 - 1}\\ m&=&\displaystyle\frac{-\frac{{9}}{{2}}+\frac{{3}}{{2}}}{-6}\\ m&=&\displaystyle\frac{-\frac{{6}}{{2}}}{-6}\\ m&=&\displaystyle\frac{-6}{2\times -6}\\ m&=&\displaystyle\frac{{1}}{{2}} \end{array} \]

Since the lines are supposed to be perpendicular, the slope we need is the opposite reciprocal to the slope above.

\[ \begin{array}{rcl} y&=&-2(x-1)+6 \end{array} \]

As before, we do not need to simplify this any further.