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A line segment is bisected by a line with the equation $- 5 y + 3 x = 1$ . If one end of the line segment is at $(6 ,4 )$ , where is the other end?

1 Answer

Apr 29, 2016

$-5y+3x=1$ is the bisector for a line segment between $(6,4)$ and any point on the line $color(blue)(-5y+3x=4)$

Explanation:

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If $(x,y)$ is the end point of a line segment with $(6,4)$ as its other end; and
if $-5y+3x=1$ bisects this line

Then for any point $(barx,bary)$ on $-5y+3x=1$
the $Deltax$ and $Deltay$ between $(6,4)$ and $(barx,bary)$
will be the same as $Deltax$ and $Deltay$ between $(barx,bary)$ and the corresponding target end point.

Since $Deltax=6-barx$ and $Deltay=4-bary$
for any point $(barx,bary)$ on $-5y+3x=1$
the corresponding target end point will be $(2barx-6,2bary-4)$

Specifically we can see that $(barx,bary)=(2,1)$ is a point on $-5y+3x=1$
with a corresponding target point of $(-2,-2)$

Note also that the target line is parallel to $-5y+3x=1$
and therefore has the same slope (namely $3/5$ ).

Using the slope-point form for the target line, we get
$color(white)("XXX")(y+2)=3/5(x+2)$
or
$color(white)("XXX")5y+10=3x+6$
or
$color(white)("XXX")-5y+3x=4$

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