A line segment is bisected by a line with the equation $- 2 y + 3 x = 1$ . If one end of the line segment is at $(6 ,3 )$ , where is the other end?

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Answer 1 · 2018-05-14T13:32:44

$(12/13,83/13)$

$-2y+3x=1=>y=3/2x-1/2 \ \ [1]$

If the line segment was a line perpendicular to $[1]$ . then its gradient would be the negative reciprocal of the gradient of $[1]$

$-2/3$

Forming the equation of a line for this:

$y-3=-2/3(x-6)$

$y=-2/3x+7\ \ \ \ [2]$

Finding the intersection of $[1] and [2]$

$-2/3x+7=3/2x-1/2$

$x=45/13$

Substituting in $[1]$

$y=3/2(45/13)-1/2=61/13$

The coordinates of the midpoint are given by:

$((x_1+x_2)/2,(y_1+y_2)/2)$

We have $(6,3)$
$:.$

$((6+x_2)/2,(3+y_2)/2)=>(45/13,61/13)$

Hence:

$(6+x)/2=45/13=>x=12/13$

$(3+y)/2=61/13=>y=83/13$

Coordinates of the other end are:

$(12/13,83/13)$

A line segment is bisected by a line with the equation - 2 y + 3 x = 1 - 2 y + 3 x = 1 . If one end of the line segment is at (6 ,3 )(6 ,3 ), where is the other end?