Question

Find k so that the line containing the points (-5,k) and (4,6) is parallel to the line containing the points (7,4) and (3,-3) k=

Answer 1

For two lines to be parallel, the two lines must have the same slope. The slope of a line can be solved using the formula below:

$m=\frac{y_2-y_1}{x_2-x_1}$

For our first line, the points are (-5, k) and (4,6). The slope is:

$m=\frac{6-k}{4-(-5)}=\frac{6-k}{9}$

For our second line, the points are (7, 4) and (3, -3). The slope is:

$m=\frac{-3-4}{3-7}=\frac{-7}{-4}=\frac{7}{4}$

Since the slope of the two lines must be the same for them to be parallel, we can say that:

$\frac{6-k}{9}=\frac{7}{4}$

From the above equation, we can now solve for "k".

$\begin{gathered} \frac{6-k}{9}=\frac{7}{4} \\ 4(6-k)=7(9) \\ 24-4k=63 \\ 24-4k-24=63-24 \\ -4k=39 \\ \frac{-4k}{-4}=\frac{39}{-4} \\ k=-\frac{39}{4}or-9.75 \end{gathered}$

The value of k must be -39/4 or -9.75 for the two lines to be parallel.