A line segment is bisected by a line with the equation $7 y + x = 7$ . If one end of the line segment is at $(1 ,3 )$ , where is the other end?

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Answer 1 · 2017-07-09T06:44:34

The other end is $=(0.4,-1.2)$

Let's rewrite the equation of the line.

$7y+x=7$

$7y=-x+7$

$y=-1/7x+1$ ................................ $(1)$

The slope is $m=-1/7$

The slope of the perpendicular line is $m'=-1/m=-1/(-1/7)=7$

The equation of the perpendicular line is

$y-3=7(x-1)$

$y=7x-7+3=7x-4$ ....................... $(2)$

The point of intersection is obtained by solving for $x$ and $y$ in equations $(1)$ and $(2)$

$7x-4=-1/7x+1$

$7x+1/7x=4+1=5$

$50/7x=5$

$x=5*7/50=7/10=0.7$

$y=7*0.7-4=4.9-0.4=0.9$

The point of intersection is $=(0.7,0.9)$

Let the other end of the line segment be $=(a,b)$

Then,

$((a+1)/2,(b+3)/2)=(0.7,0.9)$

$a+1=2*0.7=1.4$ , $=>$ , $a=1.4-1=0.4$

$b+3=2*0.9=1.8$ , $=>$ , $b=1.8-3=-1.2$

The other end is $=(0.4,-1.2)$

A line segment is bisected by a line with the equation 7 y + x = 7 7 y + x = 7 . If one end of the line segment is at (1 ,3 )(1 ,3 ), where is the other end?