# A line segment is bisected by a line with the equation  3 y - 8 x = 2 . If one end of the line segment is at (1 ,3 ), where is the other end?

Jan 17, 2017

The other end is at the point: $\left(\frac{57}{73} , \frac{225}{73}\right)$

#### Explanation:

When given a the equation of a line in standard form:

$a x + b y = c \text{ [1]}$

All lines perpendicular to the given line will be of the form:

$- b x + a y = k \text{ [2]}$

where k is an arbitrary constant.

Write the given equation for the bisecting line in standard form:

$8 x - 3 y = - 2 \text{ [3]}$

Write the equation for all lines perpendicular to equation [3]:

$3 x + 8 y = k \text{ [4]}$

The start of the bisected line segment, $\left({x}_{\text{start", y_"start}}\right)$, is given as the point $\left(1 , 3\right)$

Substitute 1 for x and 3 for y into equation [4] and solve for k:

$3 \left(1\right) + 8 \left(3\right) = k$

$k = 27$

The equation of the bisected line segment is:

$3 x + 8 y = 27 \text{ [5]}$

Add 8 times equation [3] to 3 times equation [5]:

$64 x - 24 y + 9 x + 24 y = - 16 + 81$

$73 x = 65$

${x}_{\text{intersect}} = \frac{65}{73}$

The x coordinate of the end can be found by adding the start coordinate to twice the change for the point of intersection:

x_"end" = x_"start" + 2(x_"intersect" - x_"start")

x_"end" = x_"start" + 2x_"intersect" - 2x_"start")

${x}_{\text{end" = 2x_"intersect" - x_"start}}$

${x}_{\text{end}} = 2 \frac{65}{73} - 1$

${x}_{\text{end}} = \frac{57}{73}$

Substitute $\frac{57}{73}$ for x in equation [5] and the solve for the value of ${y}_{\text{end}}$

$3 \left(\frac{57}{73}\right) + 8 {y}_{\text{end}} = 27$

${y}_{\text{end}} = \frac{27}{8} - \frac{3}{8} \left(\frac{57}{73}\right)$

${y}_{\text{end}} = \frac{225}{73}$

Here is a graph of, the two lines, the start point, and the end point.