# A line segment is bisected by a line with the equation  5 y + 2 x = 1 . If one end of the line segment is at (3 ,4 ), where is the other end?

the other is end is at $\left(- \frac{13}{29} , - \frac{134}{29}\right)$

#### Explanation:

The given line is $5 y + 2 x = 1$
The line perpendicular to this line and passing thru point $U \left(3 , 4\right)$ is
$y - 4 = \left(\frac{5}{2}\right) \left(x - 3\right)$ by the two-point form
or $5 x - 2 y = 7$
Simultaneous solution of
$2 x + 5 y = 1$ and $5 x - 2 y = 7$ yields point $I \left(\frac{37}{29} , - \frac{9}{29}\right)$

Let $v =$vertical distance from point $I \left(\frac{37}{29} , - \frac{9}{29}\right)$ to $U \left(3 , 4\right)$
$v = 4 - - \frac{9}{29} = \frac{125}{29}$
Let $h =$horizontal distance from point $I \left(\frac{37}{29} , - \frac{9}{29}\right)$ to $U \left(3 , 4\right)$
$h = 3 - \frac{37}{29} = \frac{50}{29}$
Let the other end point be $D \left({x}_{o} , {y}_{o}\right)$

${x}_{o} = \frac{37}{29} - h = \frac{37}{29} - \frac{50}{29} = - \frac{13}{29}$
${y}_{o} = - \frac{9}{29} - \frac{125}{29} = - \frac{134}{29}$

Check by distance formula from point $I \left(\frac{37}{29} , - \frac{9}{29}\right)$ to $U \left(3 , 4\right)$
${d}_{1} = \sqrt{{\left(3 - \frac{37}{29}\right)}^{2} + {\left(4 - - \frac{9}{29}\right)}^{2}} = \frac{25 \sqrt{29}}{29}$

Check by distance formula from point $I \left(\frac{37}{29} , - \frac{9}{29}\right)$ to $D \left(- \frac{13}{29} , - \frac{134}{29}\right)$
${d}_{2} = \sqrt{{\left(\frac{37}{29} - - \frac{13}{29}\right)}^{2} + {\left(- \frac{9}{29} - - \frac{134}{29}\right)}^{2}} = \frac{25 \sqrt{29}}{29}$

Therefore ${d}_{1} = {d}_{2}$

God bless....I hope the explanation is useful.