# two sides of the rectangle are x-2y = 0 and x-2y 15 = 0 diagonal 7x + y-15 = 0.Coordinates A, B, C, D?Answers are A(2,1),B(4,2),C(1,8),D(-1,7).

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Determine if the given sides of the rectangle are parallel or perpendicular. To do so, express the equations in slope-intercept form to get the slope.

>> `x-2y=0`

`-2y= -x`

y=1/2x (Let this be EQ1.)` `

Hence, the slope of `x-2y=0` is `1/2` .

>> `x-2y+15=0`

`-2y=-x-15`

`y=1/2x +15/2` (Let this be EQ2.)

The slope of `x-2y+15=0` is `1/2` .

Since the two equations have the same slope, this means that the given sides are parallel. So, they do not intersect. But, these two lines would intersect the diagonal of the rectangle.

To determine the intersection of EQ1 and the diagonal, substitute EQ1 to 7x+y-15=0.

`7x+y-15=0 `

`7x+1/2x-15=0`

`15/2x=15`

`x=2`

Plug-in x=2 to EQ1.

`y=1/2x=1/2(2)=1`

Hence, the intersection of EQ1 and the diagonal is (2,1).

Next, substitute y=1/2x+15/2 to 7x+y-15=0.

`7x+y-15=0`

`7x+1/2x+15/2-15=0`

`15/2x-15/2=0`

`15/2x=15/2`

`x=1`

Plug-in x=1 to y=1/2x+15/2.

`y=1/2x+15/2=1/2*1+15/2=8`

Hence, EQ2 and the diagonal intersects at (1,8).

Next, determine the equations of the other two sides of the rectangle.

These two sides are both perpendicular the given sides. So their slopes are negative reciprocals of the slope of EQ1 and EQ2. Since EQ1 and EQ2 have the same slopes, then the other two sides have the same slope. Hence, they are parallel too.

The slopes of the other two sides are:

`m=-1/m_1=-1/(1/2)=-1*2/1=-2`

One of these two sides intersect EQ1 at (2,1). So, to determine its equation, apply the point-slope form of the line which is:

`y-y_1=m(x-x_1)`

Plug-in m=-2, x1=2 and y1=1.

`y-1=-2(x-2)`

`y-1=-2x+4`

`y=-2x+5 ` (Let this be EQ3.)

This is the equation of the third side of the rectangle.

Since the third side passes (2,1), the fourth side passes the other point (1,8). Take note that the 3rd and 4th sides of the rectangle do not contain the same points since they are parallel.

To determine the equation of the 4th side, use the point-slope form again. Plug-in m=-2, x1=1 and y1=8.

`y-8=-2(x-1)`

`y-8=-2x+2`

`y=-2x+10 ` (Let this be EQ.4).

Hence, the fourth side is y=-2x+10.

Since EQ3 and EQ4 are both perpendicular to the given sides, then the other end of EQ3 intersects with EQ2 and the other end of EQ4 intersects EQ1.

To determine the intersection of EQ2 and EQ3, substitute y=-2x+5 to y=1/2x+15/2.

`y=1/2x+15/2` `-4/2x-1/2x=15/2-10/2`

`-2x+5=1/2x+15/2 ` `-5/2x=5/2`

`-2x-1/2x=15/2-5 ` `x=-1`

Plug-in x=-1 to EQ3.

`y=-2x+5=-2(-1)+5=7`

Hence, EQ2 and EQ3 intersects at (-1,7).

And for the intersection of EQ1 and EQ4, substitute y=-2x+10 to y=1/2x.

`y=1/2x ` `-4/2x-1/2x=-10`

`-2x+10=1/2x ` `-5/2x=-10`

`-2x-1/2x=-10` `x=4`

Then, plug-in x=4 to EQ4.

`y=-2x+10=-2*4+10=2`

Hence, the intersection of EQ1 and EQ4 is (4,2).

Note that these four intersection points are the vertices's of the rectangle.

**Thus, the coordinates of the vertices's are (-1,7), (1,8) , (4,2) and (2,1).**