# Equation of a straight line help me to solve this1. 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...

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 diagonals

x = -8, 3x - 4y = 0

x = 0, x - 4y = 0

2. 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 lines

x - 11y + 64 = 0

11x + y - 28 = 0

3. 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

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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`

3.Let M be the slope of AB.

M=-2/3

for a line to be perpenducilar to this, it's slope (let's say K) must be 3/2, since K*M =-1 ALWAYS for perpendicular lines!

The midpont of AB is C(1,2) found with midpoint formula

[(x1+x2)/2 , (y1+y2)/2)

The line that passes through C(1,2) with a slope K=3/2 is:

y-2=K(x-1)

y-2=3x/2 -3/2

2y-3x = 4-3

3x-2y+1=0

this was done using the formula:

y-y1=K(x-x1) where x1 and y1 are the coordinates of the point through which it should pass and K is the slope.