# A line segment is bisected by a line with the equation  - 2 y + 3 x = 1 . If one end of the line segment is at (6 ,3 ), where is the other end?

May 14, 2018

$\left(\frac{12}{13} , \frac{83}{13}\right)$

#### Explanation:

$- 2 y + 3 x = 1 \implies y = \frac{3}{2} x - \frac{1}{2} \setminus \setminus \setminus \left[1\right]$

If the line segment was a line perpendicular to $\left[1\right]$. then its gradient would be the negative reciprocal of the gradient of $\left[1\right]$

$- \frac{2}{3}$

Forming the equation of a line for this:

$y - 3 = - \frac{2}{3} \left(x - 6\right)$

$y = - \frac{2}{3} x + 7 \setminus \setminus \setminus \setminus \left[2\right]$

Finding the intersection of $\left[1\right] \mathmr{and} \left[2\right]$

$- \frac{2}{3} x + 7 = \frac{3}{2} x - \frac{1}{2}$

$x = \frac{45}{13}$

Substituting in $\left[1\right]$

$y = \frac{3}{2} \left(\frac{45}{13}\right) - \frac{1}{2} = \frac{61}{13}$

The coordinates of the midpoint are given by:

$\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

We have $\left(6 , 3\right)$
$\therefore$

$\left(\frac{6 + {x}_{2}}{2} , \frac{3 + {y}_{2}}{2}\right) \implies \left(\frac{45}{13} , \frac{61}{13}\right)$

Hence:

$\frac{6 + x}{2} = \frac{45}{13} \implies x = \frac{12}{13}$

$\frac{3 + y}{2} = \frac{61}{13} \implies y = \frac{83}{13}$

Coordinates of the other end are:

$\left(\frac{12}{13} , \frac{83}{13}\right)$