<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B28%20-%20%20%5Csqrt%7B7%7D%20%7D%20" id="TexFormula1" title=" \sqrt{28

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BigorU [14]

3 years ago

} } " alt=" \sqrt{28 - \sqrt{7} } " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

kogti [31]3 years ago

6 0

The answer is 5.0353

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wolverine [178]

Answer:

$k =(2,1)$

$JK = 2$

Step-by-step explanation:

Given

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$H = (1,-2)$ --- $(x_2,y_2)$

$I = (3,-2)$ --- $(x_3,y_3)$

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:

$\frac{1}{2}(I + J) = \frac{1}{2}(H+K)$

$\frac{1}{2}(x_3+x_1,y_3+y_1) = \frac{1}{2}(x_2+x,y_2+y)$

$\frac{1}{2}(3+0,-2+1) = \frac{1}{2}(1+x,-2+y)$

$\frac{1}{2}(3,-1) = \frac{1}{2}(1+x,-2+y)$

Multiply through by 2

$(3,-1) = (1+x,-2+y)$

By comparison:

$1 + x = 3$

$-2 + y = -1$

Solve for x and y

$x = 3 - 1 =2$

$y = -1 +2 = 1$

So, the coordinates of k is:

$k =(2,1)$

The length of JK is calculated using distance (d) formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

$J = (0,1)$ ---- $(x_1,y_1)$

$k =(2,1)$ ---- $(x_2,y_2)$

So:

$d = \sqrt{(0 - 2)^2 + (1 - 1)^2$

$d = \sqrt{(- 2)^2 + (0)^2$

$d = \sqrt{4 + 0$

$d = \sqrt{4$

$d = 2$

Hence:

$JK = 2$

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