A line segment is bisected by a line with the equation # 3 y - 8 x = 2 #. If one end of the line segment is at #(1 ,3 )#, where is the other end?

1 Answer
Jan 17, 2017

The other end is at the point: #(57/73,225/73)#

Explanation:

When given a the equation of a line in standard form:

#ax + by = c" [1]"#

All lines perpendicular to the given line will be of the form:

#-bx + ay = k" [2]"#

where k is an arbitrary constant.

Write the given equation for the bisecting line in standard form:

#8x - 3y = -2" [3]"#

Write the equation for all lines perpendicular to equation [3]:

#3x + 8y = k" [4]"#

The start of the bisected line segment, #(x_"start", y_"start")#, is given as the point #(1, 3)#

Substitute 1 for x and 3 for y into equation [4] and solve for k:

#3(1) + 8(3) = k#

#k = 27#

The equation of the bisected line segment is:

#3x + 8y = 27" [5]"#

Add 8 times equation [3] to 3 times equation [5]:

#64x - 24y + 9x + 24y = -16 + 81#

#73x = 65#

#x_"intersect" = 65/73#

The x coordinate of the end can be found by adding the start coordinate to twice the change for the point of intersection:

#x_"end" = x_"start" + 2(x_"intersect" - x_"start")#

#x_"end" = x_"start" + 2x_"intersect" - 2x_"start")#

#x_"end" = 2x_"intersect" - x_"start"#

#x_"end" = 2(65)/73 - 1#

#x_"end" = 57/73#

Substitute #57/73# for x in equation [5] and the solve for the value of #y_"end"#

#3(57/73) + 8y_"end" = 27#

#y_"end" = 27/8 - 3/8(57/73)#

#y_"end" = 225/73#

Here is a graph of, the two lines, the start point, and the end point.

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