A line segment is bisected by a line with the equation $3 y - 7 x = 3$ . If one end of the line segment is at $(7 ,8 )$ , where is the other end?

Question

2059 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-02T16:01:40

The other point is at $(7/29, 10 26/29)$

Write the equation of the given line in standard form:

$3y = 7x +3 rArr y = (7x)/3+1$

The gradient is $7/3$

We will have to assume that this line is the perpendicular bisector rather than just a bisector, else the question cannot be answered.

The line segment has a gradient of $-3/7$
and passes through the point $(7,8)$

Find its equation using: $y-y_1 = m(x-x_1)$
$y - 8 = -3/7(x-7)$

$y = (-3x)/7 +3+8 rArr y = (-3x)/7 +11$

The midpoint of the line segment is where the lines intersect. Solve them simultaneously by equating the y's.

$y = (7x)/3+1 and y = (-3x)/7 +11$
$y=y rArr (7x)/3+1 =(-3x)/7 +11$

$(7x)/3 + (3x)/7 = 10" "xx21$

$49x+9x = 210$

$58x = 210 rArr x =105/29$ This gives y as $274/29$

This point is the midpoint of the line segment.

$(7 +x_2)/2 = 105/29 " and "(8+y_2)/2 = 274/29$

$x_2 = 210/29 -7 " and "y_2 = 548/29 -8$

The other point is at $(7/29, 10 26/29)$

A line segment is bisected by a line with the equation 3 y - 7 x = 3 3 y - 7 x = 3 . If one end of the line segment is at (7 ,8 )(7 ,8 ), where is the other end?