# A line segment is bisected by a line with the equation  -7 y + 3 x = 2 . If one end of the line segment is at ( 2 , 4 ), where is the other end?

Jun 3, 2016

Infinite solutions: any of the points in the line $y = \left(\frac{3}{7}\right) x - \frac{26}{7}$, including the point $\left(\frac{110}{29} , - \frac{424}{203}\right)$ when the segment is perpendicularly bisected.

#### Explanation:

First we should notice that the point (2,4) isn't in the line $3 x - 7 y - 2 = 0$ since:
$3 \cdot 2 - 7 \cdot 4 - 2 = 6 - 28 - 2 = - 24$

But as the problem is stated it admits infinite solutions.

Calling AB the line segment with M the midpoint we have:
$A \left(2 , 4\right)$
$M \left(\frac{2 + {x}_{B}}{2} , \frac{4 + {y}_{B}}{2}\right)$
$B \left({x}_{B} , {y}_{B}\right)$

The orthogonal projection of A onto the line is also M when the segment is perpendicular bisected.
The slope of the line $y = \frac{3 x - 2}{7}$is
$k = \frac{3}{7}$
So the slope of line perpendicular to the aforementioned is
$p = - \frac{1}{k} = - \frac{7}{3}$
And the equation of such a perpendicular line passing through A is
$y - 4 = - \frac{7}{3} \left(x - 2\right)$ => $y = \frac{- 7 x + 14}{3} + 4$ => $y = \frac{- 7 x + 26}{3}$

Combining the equations of the 2 lines
$\frac{3 x - 2}{7} = \frac{- 7 x + 26}{3}$ => $9 x - 6 = - 49 + 162$ => $58 x = 168$ => $x = \frac{84}{29}$
$\to y = \frac{\frac{252}{29} - 2}{7}$ => $y = \frac{194}{203}$
And $M \left(\frac{84}{29} , \frac{194}{203}\right)$

Finding B
${x}_{M} = 1 + {x}_{B} / 2$ => $\frac{84}{29} = 1 + {x}_{B} / 2$=>${x}_{B} = \frac{55}{29} \cdot 2 = \frac{110}{29}$
${y}_{M} = 2 + {y}_{B} / 2$ =>$\frac{194}{203} = 2 + {y}_{B} / 2$ =>${y}_{B} = - \frac{212}{203} \cdot 2 = - \frac{424}{203}$
So when the segment AB is perpendicularly bisected by the line mentioned in the problem
$B \left(\frac{110}{29} , - \frac{424}{203}\right)$

But since it is not required that the bisection happens perpendicularly the set of possible points B lays in a line parallel to the bisecting line and passing through $\left(\frac{110}{29} , - \frac{424}{203}\right)$, itself one of the possible locations of B. Then the locus in which the point B can be located is
$y + \frac{424}{203} = \left(\frac{3}{7}\right) \left(x - \frac{110}{29}\right)$
$y = \left(\frac{3}{7}\right) x - \frac{330}{203} - \frac{424}{203}$
$y = \left(\frac{3}{7}\right) x - \frac{26}{7}$