A line segment is bisected by a line with the equation $- 3 y + 5 x = 8$ . If one end of the line segment is at $( 7 , 9 )$ , where is the other end?

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Answer 1 · 2017-12-31T09:07:49

Coordinates of the other end point $color(blue)(-61/17, -147/17)$

Assumption : Bisecting line is a perpendicular bisector

Standard form of equation $y=mx +c$
Slope of perpendicular bisector m is given by
$-3y + 5x= 8$
$y= (5/3)x - (8/3)$
$m= (5/3)$

Slope of line segment is
$y - 9 = -(1/m)(x-7)$
$y - 9= (-3/5)(x - 7)$
$5y - 45 = -3x + 21$

$-3y + 5x = 8 color (white)((aaaa)$ Eqn (1)
$5y + 3x = 66 color (white)((aaaa)$ Eqn (2)

Solving Eqns (1) & (2),

$x= color (purple)29/17$

$y= color (purple)3/17$
Mid point $color(purple)(29/17, 3/17)$

Let (x1,y1) the other end point.
$(7+x1)/2= 29/17$
$x1= (58/17) - 7 = color(red)( -61/17)$
$(9+y1)/2= 3/17$
$y1=(6/17) - 9 = color(red)(-147/17)$

Coordinates of other end point $color(red )(-61/17, -147/17)$

A line segment is bisected by a line with the equation - 3 y + 5 x = 8 - 3 y + 5 x = 8 . If one end of the line segment is at ( 7 , 9 )( 7 , 9 ), where is the other end?