# A line segment is bisected by a line with the equation  - 3 y + 4 x = 6 . If one end of the line segment is at ( 3 , 1 ), where is the other end?

Jun 21, 2017

The other end is $= \left(\frac{51}{25} , \frac{129}{75}\right)$

#### Explanation:

Let the other end be $\left(x , y\right)$

Then the mid-point of the line is $= \left(\frac{x + 3}{2} , \frac{y + 1}{2}\right)$

The slope of the line $- 3 y + 4 x = 6$ ............$\left(1\right)$

is

$= \frac{4}{3}$

The slope of the line perpendicular is

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

The equation of the line is

$- \frac{3}{4} \left(x - 3\right) = \left(y - 1\right)$

$- 3 x + 9 = 4 y - 4$

$4 y + 3 x = 13$ .............$\left(2\right)$

Solving for $x$ and $y$, in equations $\left(1\right)$ and $\left(2\right)$

$4 x - 3 \cdot \frac{13 - 3 x}{4} = 6$

$16 x - 39 + 9 x = 24$

$25 x = 24 + 39 = 63$

$x = \frac{63}{25}$

$y = \frac{1}{3} \cdot \left(4 \cdot \frac{63}{25} - 6\right)$

$= \frac{102}{75}$

The mid-point is $= \left(\frac{63}{25} , \frac{102}{75}\right)$

So,

$\frac{x + 3}{2} = \frac{63}{25}$

$x = \frac{126}{25} - 3 = \frac{51}{25}$

$\frac{y + 1}{2} = \frac{102}{75}$

$y = \frac{204}{75} - 1 = \frac{129}{75}$

The other end is $= \left(\frac{51}{25} , \frac{129}{75}\right)$