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

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

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Answer 1 · 2016-06-03T00:42:31

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

Explanation:

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

But as the problem is stated it admits infinite solutions.

Calling AB the line segment with M the midpoint we have:
$A (2, 4)$ $A(2,4)$
$M (\frac{2 + x_{B}}{2}, \frac{4 + y_{B}}{2})$ $M((2+x_B)/2,(4+y_B)/2)$
$B (x_{B}, y_{B})$ $B(x_B,y_B)$

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

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

Finding B
$x_{M} = 1 + \frac{x_{B}}{2}$ $x_M=1+x_B/2$ => $\frac{84}{29} = 1 + \frac{x_{B}}{2}$ $84/29=1+x_B/2$ => $x_{B} = \frac{55}{29} \cdot 2 = \frac{110}{29}$ $x_B=55/29*2=110/29$
$y_{M} = 2 + \frac{y_{B}}{2}$ $y_M=2+y_B/2$ => $\frac{194}{203} = 2 + \frac{y_{B}}{2}$ $194/203=2+y_B/2$ => $y_{B} = - \frac{212}{203} \cdot 2 = - \frac{424}{203}$ $y_B=-212/203*2=-424/203$
So when the segment AB is perpendicularly bisected by the line mentioned in the problem
$B (\frac{110}{29}, - \frac{424}{203})$ $B(110/29,-424/203)$

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 $(\frac{110}{29}, - \frac{424}{203})$ $(110/29,-424/203)$ , itself one of the possible locations of B. Then the locus in which the point B can be located is
$y + \frac{424}{203} = (\frac{3}{7}) (x - \frac{110}{29})$ $y+424/203=(3/7)(x-110/29)$
$y = (\frac{3}{7}) x - \frac{330}{203} - \frac{424}{203}$ $y=(3/7)x-330/203-424/203$
$y = (\frac{3}{7}) x - \frac{26}{7}$ $y=(3/7)x-26/7$

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

1 Answer

Explanation:

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

1 Answer

Explanation:

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