# A line segment is bisected by a line with the equation  5 y -4 x = 1 . If one end of the line segment is at (3 ,4 ), where is the other end?

May 11, 2017

The other end is $\left(\frac{179}{41} , \frac{94}{41}\right)$

#### Explanation:

Let's rewrite the equation of the line

$5 y - 4 x = 1$

$5 y = 4 x + 1$

$y = \frac{4}{5} x + \frac{1}{5}$

The slope is

$m = \frac{4}{5}$

The slope of a line perpendicular is $m ' = - \frac{5}{4}$

as $m \cdot m ' = - 1$

The equation of the segment is

$y - 4 = - \frac{5}{4} \left(x - 3\right)$

$y = - \frac{5}{4} x + \frac{15}{4} + 4 = - \frac{5}{4} x + \frac{31}{4}$

The point of intersection of the lines

$y = \frac{4}{5} x + \frac{1}{5} = - \frac{5}{4} x + \frac{31}{4}$

$\left(\frac{4}{5} + \frac{5}{4}\right) x = - \frac{1}{5} + \frac{31}{4}$

$\frac{41}{20} x = \frac{161}{20}$

$x = \frac{151}{41}$

$y = \frac{4}{5} \cdot \frac{151}{41} + \frac{1}{5}$

$y = \frac{604 + 41}{205}$

$y = \frac{645}{205} = \frac{129}{41}$

The point of intersection is $\left(\frac{151}{41} , \frac{129}{41}\right)$

Let the other end of the segment be ${x}_{1} , {y}_{1}$

Then,

${x}_{1} - \frac{151}{41} = \frac{151}{41} - 3$

${x}_{1} = \frac{151}{41} + \frac{151}{41} - 3 = \frac{179}{41}$

and

${y}_{1} - \frac{129}{41} = \frac{129}{41} - 4$

${y}_{1} = \frac{129}{41} + \frac{129}{41} - 4 = \frac{94}{41}$

The other end is $\left(\frac{179}{41} , \frac{94}{41}\right)$
graph{(y-4/5x-1/5)(y+5/4x-31/4)((x-3)^2+(y-4)^2-0.01)((x-179/41)^2+(y-94/41)^2-0.01)=0 [-1.985, 9.116, -0.99, 4.56]}