Equation of a straight line help me to solve this 1. the vertices of a rectangle are A(0,0), B(-8,0), C(-8,-6) and D(0.-6). write equations for the lines witch form its sides and diagonalsx = -8, 3x - 4y = 0x = 0, x - 4y = 02. the vertices of a rectangle are A(2,6), B(8,-4) and C(-3,-5). two lines are drawn through A. one parallel to and the other perpendicular to BC. find the equation of each of these linesx - 11y + 64 = 011x + y - 28 = 03. write the equation of the line that is the perpendicular bisector of the given segment AB. A(-2,4), B(4,0)3x - 2y + 1 = 0

Expert Answers

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1.

sides:x=0 x=-8

y=0, y=-6

 

Diagonal AC :`y=6/8 x=3/4x`

Diagonal BD: slope`=(-6-0)/(0-(-8))=-6/8=-3/4`

Slope point equation

`y-0=-(3/4)(x-(-8))`

Equation of BD: `y=-(3/4) x+8`

 

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