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kogti [31]
3 years ago
5

What could be the coordinates of the third vertex, z, of triangle xyz so that it would have a hypotenuse with a length of square

root 45 units
Mathematics
1 answer:
ICE Princess25 [194]3 years ago
8 0

Answer:

z = (6,-1)

Step-by-step explanation:

The missing parameters are:

x = (3,-1)

y = (3,5)

yz= \sqrt{45 --- Hypotenuse

Required

The coordinate of Z

First, calculate the distance xy

xy = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

xy = \sqrt{(3 - 3)^2 + (-1 - 5)^2}

xy = \sqrt{0 + 36}

xy = \sqrt{36}

xy = 6

By Pythagoras theorem, distance xz is:

yz^2 = xz^2 + xy^2

(\sqrt 45)^2 = xz^2 + 6^2

45 = xz^2 + 36

Collect like terms

xz^2 =45-36

xz^2 =9

xz = \sqrt{9}

xz = 3

This means that z is 3 units to the right of x

We have:

x = (3,-1)

The rule to determine z is:

(x,y) \to (x + 3, y)

So, we have:

z = (3 + 3,-1)

z = (6,-1)

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To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}

The angle, from E to N, can be calculated as:

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