Answer:
See Explanation
Step-by-step explanation:
The question is incomplete, as the coordinates of the three vertices.
I will answer your question using the following illustration.
Assume that the square is ABCD are the given coordinates are:
[tex]A = (3,1)[/tex]
[tex]B = (-1,5)[/tex]
[tex]C = (-5,1)[/tex]
Required
Find D
Let the coordinates of D be:
[tex]D = (x,y)[/tex]
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Calculate the slope of each side.
AB, BC, CD and DA using:
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
AB:
[tex]A = (3,1)[/tex] -- [tex](x_1,y_1)[/tex]
[tex]B = (-1,5)[/tex] -- [tex](x_2,y_2)[/tex]
So:
[tex]m_1 = \frac{5 - 1}{-1 - 3} = \frac{4}{-4} = -1[/tex]
BC:
[tex]B = (-1,5)[/tex] -- [tex](x_1,y_1)[/tex]
[tex]C = (-5,1)[/tex] -- [tex](x_2,y_2)[/tex]
So:
[tex]m_2 = \frac{1-5}{-5--1} = \frac{-4}{-4} = 1[/tex]
CD:
[tex]C = (-5,1)[/tex] -- [tex](x_1,y_1)[/tex]
[tex]D = (x,y)[/tex] -- [tex](x_2,y_2)[/tex]
So:
[tex]m_3 = \frac{y - 1}{x --5} = \frac{y-1}{x+5}[/tex]
DA
[tex]D = (x,y)[/tex] -- [tex](x_1,y_1)[/tex]
[tex]A = (3,1)[/tex] -- [tex](x_2,y_2)[/tex]
[tex]m_4 = \frac{1-y}{3-x}[/tex]
AB and CD are parallel sides. So, they have the same slope
i.e.
[tex]m_1 = m_3[/tex]
[tex]\frac{y-1}{x+5} = -1[/tex]
Solve:
[tex]y-1 =-x-5[/tex]
Make y the subject
[tex]y =-x-5+1[/tex]
[tex]y =-x-4[/tex] ---- (1)
BC and DA are parallel sides. So, they have the same slope
i.e.
[tex]m_2 = m_4[/tex]
[tex]1 = \frac{1-y}{3-x}[/tex]
Solve:
[tex]1-y =3-x[/tex]
Make y the subject
[tex]y =1-3+x[/tex]
[tex]y =x-2[/tex] ---- (2)
So, we have:
[tex]y =-x-4[/tex] and [tex]y =x-2[/tex]
Equate both:
[tex]y=y[/tex]
[tex]-x-4=x-2[/tex]
Collect like terms
[tex]-x-x=4-2[/tex]
[tex]-2x=2[/tex]
Solve for x
[tex]x = -\frac{2}{2}[/tex]
[tex]x = -1[/tex]
Substitute [tex]x = -1[/tex] in [tex]y =x-2[/tex]
[tex]y = -1-2[/tex]
[tex]y = -3[/tex]
So, the missing coordinate is:
[tex]D = (-1,-3)[/tex]
