A line segment is bisected by a line with the equation $9 y + x = 5$ . If one end of the line segment is at $( 7 , 4 )$ , where is the other end?

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Answer 1 · 2017-06-15T18:29:50

The other end is: $( 249/41,-178/41)$
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Given: $x+9y=5" [1]"$

The family of perpendicular lines is:

$9x-y=k$

Use the point $(7,4)$ to find the value of x:

$9(7)-4=k$

$k = 59$

The equation of the bisected line is:

$9x-y=59" [2]"$

Multiply equation [2] by 9 and add to equation [1]:

$82x= 536$

$x_"midpoint" = 268/41$

Use the midpoint equation to find $x_"end"$

$x_"midpoint" = (x_"start"+x_"end")/2$

$268/41 = (7+x_"end")/2$

$x_"end" = 536/41-7$

$x_"end" = 249/41$

Find $y_"end"$ by substituting $x_"end"$ into equation [2]:

$9(249/41)-y_"end"=59$

$y_"end" = 9(249/41)-59$

$y_"end" = -178/41$

A line segment is bisected by a line with the equation 9 y + x = 5 9 y + x = 5 . If one end of the line segment is at ( 7 , 4 )( 7 , 4 ), where is the other end?