A line segment is bisected by a line with the equation $4 y + 3 x = 4$ . If one end of the line segment is at $( 8 , 1 )$ , where is the other end?

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A line segment is bisected by a line with the equation $4 y + 3 x = 4$ . If one end of the line segment is at $( 8 , 1 )$ , where is the other end?

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Answer 1 · 2016-10-28T16:25:46

$P=(56/25;-167/25)$

Explanation:

Calculate the normal passing per $A=(8;1)$
$m=-3/4$
$m_(|--)=4/3$
Normal equation
$y-1=4/3(x-8)$
$4x-3y=29$
Calculate intersection I
${(3x+4y=4),(4x-3y=29):}$
Solving the system you get $I=(128/25;-71/25)$
Calculate $P$ so that $I$ is the medium point of $bar(PA)$
${((P_x+8)/2=128/25),((p_y+1)/2=-71/25):}$
$P_x=56/25$
$P_y=-167/25$

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A line segment is bisected by a line with the equation $4 y + 3 x = 4$ . If one end of the line segment is at $( 8 , 1 )$ , where is the other end?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

A line segment is bisected by a line with the equation 4 y + 3 x = 4 4 y + 3 x = 4 . If one end of the line segment is at ( 8 , 1 )( 8 , 1 ), where is the other end?

1 Answer

Explanation:

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Impact of this question

A line segment is bisected by a line with the equation $4 y + 3 x = 4$ . If one end of the line segment is at $( 8 , 1 )$ , where is the other end?