Answer:
[tex]k =(2,1)[/tex]
[tex]JK = 2[/tex]
Step-by-step explanation:
Given
[tex]J = (0,1)[/tex] ---- [tex](x_1,y_1)[/tex]
[tex]H = (1,-2)[/tex] --- [tex](x_2,y_2)[/tex]
[tex]I = (3,-2)[/tex] --- [tex](x_3,y_3)[/tex]
See attachment for grid
Solving (a): The coordinates of K
The parallelogram has the following diagonals: IJ and HK
Diagonals bisect one another. So:
Midpoint of IJ = Midpoint of HK
This gives:
[tex]\frac{1}{2}(I + J) = \frac{1}{2}(H+K)[/tex]
[tex]\frac{1}{2}(x_3+x_1,y_3+y_1) = \frac{1}{2}(x_2+x,y_2+y)[/tex]
[tex]\frac{1}{2}(3+0,-2+1) = \frac{1}{2}(1+x,-2+y)[/tex]
[tex]\frac{1}{2}(3,-1) = \frac{1}{2}(1+x,-2+y)[/tex]
Multiply through by 2
[tex](3,-1) = (1+x,-2+y)[/tex]
By comparison:
[tex]1 + x = 3[/tex]
[tex]-2 + y = -1[/tex]
Solve for x and y
[tex]x = 3 - 1 =2[/tex]
[tex]y = -1 +2 = 1[/tex]
So, the coordinates of k is:
[tex]k =(2,1)[/tex]
The length of JK is calculated using distance (d) formula
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2[/tex]
[tex]J = (0,1)[/tex] ---- [tex](x_1,y_1)[/tex]
[tex]k =(2,1)[/tex] ---- [tex](x_2,y_2)[/tex]
So:
[tex]d = \sqrt{(0 - 2)^2 + (1 - 1)^2[/tex]
[tex]d = \sqrt{(- 2)^2 + (0)^2[/tex]
[tex]d = \sqrt{4 + 0[/tex]
[tex]d = \sqrt{4[/tex]
[tex]d = 2[/tex]
Hence:
[tex]JK = 2[/tex]
