A line segment is bisected by a line with the equation $- 6 y + x = 3$ . If one end of the line segment is at $( 4 , 5 )$ , where is the other end?

Question

1484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-18T20:56:58

Midpoint $x = 9/37, y = -17/37$
Other end point $(-130/37, -219/37)$

Assumption : line segment is bisected by the line with equation $-6y+x = 3$ at right angle.

$x - 6y = 3 color (white)(aaa) Eqn (1)$

$y=(1/6)(x - 3)$ slope of line is $1/6$

Slope of perpendicular line is $-1/(1/6) = -6$

Equation of line segment is $(y - 5) = -6(x - 4)$

$6x + y = 1 color (white)(aaa)$ Eqn (2)

Solving Eqns (1) & (2) we get the midpoint.
$37x = 9, x = 9/37$
$y = -17/37$

Let the other end point be $(x_1, y_1)$

$(x_1+4)/2 = 9/37; x_1= —130/37$
$(y_1 + 5)/2 = -17/37; y_1 = -219/37$

A line segment is bisected by a line with the equation - 6 y + x = 3 - 6 y + x = 3 . If one end of the line segment is at ( 4 , 5 )( 4 , 5 ), where is the other end?