A line segment is bisected by a line with the equation $- 6 y - x = 3$ $-6 y - x = 3$ . If one end of the line segment is at $(8, 3)$ $( 8 , 3 )$ , where is the other end?

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A line segment is bisected by a line with the equation $- 6 y - x = 3$ $-6 y - x = 3$ . If one end of the line segment is at $(8, 3)$ $( 8 , 3 )$ , where is the other end?

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Answer 1 · 2017-10-04T22:14:59

The other end point is $(\frac{238}{37}, - \frac{237}{37})$ $(238/37,-237/37)$

Explanation:

It is assumed that the line bisecting the line segment is assumed to be a perpendicular bisector.
$- 6 y - x = 3$ $-6y-x=3$ Eqn (1)
$- 6 y = x + 3$ $-6y=x+3$
$y = - (\frac{x}{6}) - (\frac{1}{2})$ $y=-(x/6)-(1/2)$
Slope of the equation $m 1 = - (\frac{1}{6})$ $m1=-(1/6)$
Slope the line segment $m 2 = - (\frac{1}{m 1}) = - (\frac{1}{- \frac{1}{6}}) = 6$ $m2=-(1/(m1))=-(1/(-1/6))=6$
Equation of the line segment is
$(y - 3) = 6 \cdot (x - 8)$ $(y-3)=6*(x-8)$
$y - 6 x = - 45$ $y-6x=-45$ Eqn (2)

Solving equations (1) & (2), we get the midpoint which is also the midpoint of the line segment.
$- x - 6 y = 3$ $-x-6y=3$ Eqn (1)
$- 36 x + 6 y = - 270$ $-36x+6y=-270$ Eqn (2) * 6; Adding,
$- 37 x = - 267$ $-37x=-267$
$x = \frac{267}{37}$ $x=267/37$
Substituting value of x in Eqn (1),
$- 6 y = 3 + (\frac{267}{37})$ $-6y=3+(267/37)$
$y = - \frac{111 + 267}{37 \cdot 6} = - (\frac{378}{222} = - (\frac{63}{3})$ $y=-(111+267)/(37*6)=-(378/(222)=-(63/3)$
Coordinate of midpoint $(\frac{267}{37}), - (\frac{63}{37})$ $(267/37),-(63/37)$
Let (x2,y2) be the other end point coordinates of the line segment.
$\frac{x 2 + 8}{2} = (\frac{267}{37})$ $(x2+8)/2=(267/37)$
$x 2 = (\frac{534}{37}) - 8 = \frac{534 - 296}{37} = \frac{238}{37}$ $x2=(534/37)-8=(534-296)/37=238/37$
$\frac{y 2 + 3}{2} = - (\frac{63}{37})$ $(y2+3)/2=-(63/37)$
$y 2 + 3 = - (\frac{126}{37})$ $y2+3=-(126/37)$ y2=-(126+111)/37=-(237/37)#

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A line segment is bisected by a line with the equation $- 6 y - x = 3$ $-6 y - x = 3$ . If one end of the line segment is at $(8, 3)$ $( 8 , 3 )$ , where is the other end?

1 Answer

Explanation:

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A line segment is bisected by a line with the equation −6y−x=3 -6 y - x = 3 . If one end of the line segment is at (8,3)( 8 , 3 ), where is the other end?

1 Answer

Explanation:

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A line segment is bisected by a line with the equation $- 6 y - x = 3$ $-6 y - x = 3$ . If one end of the line segment is at $(8, 3)$ $( 8 , 3 )$ , where is the other end?