A line segment is bisected by a line with the equation $4 y + x = 8$ . If one end of the line segment is at $(5 ,2 )$ , where is the other end?

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Answer 1 · 2017-01-03T13:01:34

$B = (75/17, -6/17), AB ⊥ r, |AM| = |MB|$

$A = (5, 2)$

$r: y = 2 - 1/4 x$

$s ⊥ r Rightarrow s : y = a + 4x$

$A in s Rightarrow 2 = a + 4 * 5 Rightarrow a = -18$

$s cap r = M: 2 - 1/4 x = -18 + 4x$ // this is cohordinate x of M

$8 - x = -72 + 16x Rightarrow 80 = 17x$

$M = (80/17, 2 - 20/17) = (A + B)/2$ // middle point of A and B

$80/17 = (5 + x)/2 and 14/17 = (2 + y)/2$ // cohordinates of B

$160 = 85 + 17x and 28 = 34 + 17y$

$75 = 17x and -6 = 17y$

A line segment is bisected by a line with the equation 4 y + x = 8 4 y + x = 8 . If one end of the line segment is at (5 ,2 )(5 ,2 ), where is the other end?