A line segment is bisected by a line with the equation $-6 y + 9 x = 2$ . If one end of the line segment is at $( 5 , 1 )$ , where is the other end?

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Answer 1 · 2016-10-22T01:19:59

The point is $(-9/13, 187/39)$

Write the given equation in slope-intercept form:

$y = 3/2x - 1/3$

The slope is $3/2$ any line perpendicular will have $-2/3$ slope.

Use the point-slope form of the equation of a line, to find the equation of the bisected line:

$y - 1 = -2/3(x - 5)$

$y = -2/3x + 13/3$

Subtract the second line from the first:

$0 = (3/2 + 2/3)x - 14/3$

$13/6x = 14/3$

$x = 28/13$

The change in x from $5$ to $28/13$ is:

$28/13 - 13/13(5) = -37/13$

Add twice that number to 5 to get the x coordinate of the other end of the line segment:

13/13(5) - 74/13 = -9/13

Substitute into the equation of the line for the y coordinate:

$y = -2/3(-9/13) + 13/3$

$y = 18/39 + 169/39$

$y = 187/39$

A line segment is bisected by a line with the equation -6 y + 9 x = 2 -6 y + 9 x = 2 . If one end of the line segment is at ( 5 , 1 )( 5 , 1 ), where is the other end?