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

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Answer 1 · 2016-10-26T18:05:19

The other end is a $(4.6, -4.8)$

Write the equation for the bisector in slope-intercept form:

$y = 1/3x - 1/3$ $[1]$

The slope is, $m = 1/3$

The slope, n, for the bisected line is, $n = -1/m = -1/(1/3) = -3$

Use the slope and the point, $(1, 6)$ into the slope, $-3$ , into the slope-intercept form of a line and then solve for b:

$6 = -3(1) + b$

$b = 9$

The equation for the bisected line is:

$y = -3x + 9$ $[2]$

Subtract equation $[2]$ from equation $[1]$

$y - y= 1/3x + 3x - 1/3 - 9$

$0 = 10/3x - 28/3$

The x coordinate of the point of intersection is:

$x = 2.8$

To go from 1 to 2.8, the x coordinate increased 1.8, therefore, to go to the other end of the line x coordinate must increase twice that much, 3.6.

$x = 1 + 3.6 = 4.6$

This is the x coordinate of the other end of the line.

To find the y coordinate, substitute 4.6 for x into equation $[2]$

#y = -3(4.6) + 9

$y = -4.8$

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