The coordinates of three vertices of a rectangle that are given below.Graph them.Then, find each fourth vertex

Question

1.) (2,0), (0,2), (-4,-2)

Answer 1

First, try to plot the points to scale so that you can see that the line containing (-4,-2) and (0,2) is perpendicular to the line containing (0,2) and (2,0).

The slope of the line containing (-4,-2) and (0,2) is 1 and the slope of the line containing (0,2) and (2,0) is -1.

Now, draw a line through (-4,-2) that is perpendicular to the line containing (-4,-2) and (0,2). Using point slope form, it has equation y+2=-1(x+4).

Next, draw a line through (2,0) that is perpendicular to the line containing (0,2) and (2,0). Using point slope form, it has equation y-0=1(x-2).

Simplifying a bit, we see that to find the intersection point of the two lines we solve {y+x=-6,y-x=-2}. Adding the equations together gives 2y=-8 so y=-4. Substituting into either equation yields x=-2. Thus, the fourth vertex is (-2,-4).

Note that we have used the fact that two lines are perpendicular if and only if their slopes are negative reciprocals: if l_1 has slope m_1 and l_2 has slope m_2 then l_1 and l_2 are perpendicular if and only if m_1=-1/m_2.

Answer 2

(2,0) and (0,2)
D = sqrt((2 - 0)^2 + (0 - 2)^2)
D = sqrt(2^2 + (-2)^2)
D = sqrt(4 + 4)
D = sqrt(8)
D = sqrt(4 * 2)
D = 2sqrt(2)

(0,2) and (-4,-2)
D = sqrt((-2 - 2)^2 + (-4 - 0)^2)
D = sqrt((-2 + (-2))^2 + (-4)^2)
D = sqrt((-4)^2 + 16)
D = sqrt(16 + 16)
D = sqrt(16 * 2)
D = 4sqrt(2)

(-4,-2) and (x,y)
sqrt(8) = sqrt((x - (-4))^2 + (y - (-2))^2)
8 = (x + 4)^2 + (y + 2)^2

(2,0) and (x,y)
sqrt(16) = sqrt((x - 2)^2 + (y - 0)^2)
16 = (x - 2)^2
-4 = x - 2
-2 = x

8 = (x + 4)^2 + (y + 2)^2
8 = (-2 + 4)^2 + (y + 2)^2
8 = (2)^2 + (y + 2)^2
8 = 4 + (y + 2)^2
4 = (y + 2)^2
-2 = y + 2
y = -4

ANS : (-2,-4)

As you may notice, all you had to do is which the values of x and y

Answer 3

(-2,-4)

Answer 4

Lotta is right.