# A line segment is bisected by line with the equation  6 y + 7 x = 4 . If one end of the line segment is at (2 ,4 ), where is the other end?

Oct 8, 2017

End point coordinates $\left(\left(\frac{6}{7}\right) , - 6\right)$

#### Explanation:

It is assumed that the intersecting line is a perpendicular bisector.
$6 y + 7 x = 4$
$y = - \left(\frac{7}{6}\right) x + \left(\frac{4}{6}\right) = - \left(\frac{7}{6}\right) x + \left(\frac{2}{3}\right)$
Slope $= - \left(\frac{7}{6}\right)$
Slope of the line segment $= - \frac{1}{-} \left(\frac{7}{6}\right) = \frac{6}{7}$
Equation of line segment is
$\left(y - 4\right) = \left(\frac{6}{7}\right) \left(x - 2\right)$
$7 y - 28 = 6 x - 12$
$7 x - 6 y = 16$
Solving the equations we will get the coordinates of the mid point (intersection point).
$14 x = 20$
$x = \left(\frac{10}{7}\right)$
$10 - 6 y = 16$
$y = - 1$
Mid point coordinates $\left(\left(\frac{10}{7}\right) , - 1\right)$
But mid point $\frac{4 + y 1}{2} = - 1$ & $\frac{2 + x 1}{2} = \frac{10}{7}$
$y 1 = - 6$ & $x 1 = \left(\frac{20}{7}\right) - 2 = \frac{6}{7}$
End point coordinates $\left(\left(\frac{6}{7}\right) , - 6\right)$

Oct 8, 2017

$\left(- \frac{18}{5} , - \frac{4}{5}\right)$

#### Explanation:

Let $L 1$ be the perpendicular bisector and $L 2$ (segment $A B$) be the bisected line, as shown in the figure.
Given that the equation of the bisector $L 1$ is $6 y + 7 x = 4$,
$\implies y = - \frac{7}{6} x + \frac{2}{3}$
Let ${m}_{1}$ be the slope of $L 1$, and ${m}_{2}$ the slope of $L 2$,
$\implies {m}_{1} = - \frac{7}{6}$
Recall that the product of the slopes of two perpendicular lines is $- 1$,
$\implies {m}_{2} \times {m}_{1} = - 1 , \implies {m}_{2} = \frac{6}{7}$
Given $A = \left(2 , 4\right)$,
$\implies$ equation of $L 2$ is : $y - 4 = \frac{6}{7} \left(x - 2\right)$
$\implies y = \frac{6}{7} x + \frac{16}{7}$
Set the equations of $L 1 \mathmr{and} L 2$ equal to each other to find the intersection point $M \left({x}_{m} , {y}_{m}\right)$, which is also the midpoint of $L 2$.
$\implies - \frac{7}{6} x + \frac{2}{3} = \frac{6}{7} x + \frac{16}{7}$
$\implies x = - \frac{4}{5}$
$\implies y = - \frac{7}{6} x + \frac{2}{3} = - \frac{7}{6} \times \left(- \frac{4}{5}\right) + \frac{2}{3} = \frac{8}{5}$
$\implies M \left({x}_{m} , {y}_{m}\right) = \left(- \frac{4}{5} , \frac{8}{5}\right)$
Let the other end point of $L 2$ be $B \left(x , y\right)$,
Since $M$ is the midpoint of $L 2$,
$\implies \left(\frac{x + 2}{2} , \frac{y + 4}{2}\right) = \left(- \frac{4}{5} , \frac{8}{5}\right)$
$\implies \left(x , y\right) = \left(- \frac{18}{5} , - \frac{4}{5}\right)$