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umka2103 [35]
3 years ago
11

Find the coordinate point for C that would make ABCD a rhombus. *tia

Mathematics
2 answers:
yKpoI14uk [10]3 years ago
4 0

Answer:

(5,1)

Step-by-step explanation:

alexandr1967 [171]3 years ago
3 0

Answer:

The correct answer is 3,5

Step-by-step explanation:

What is a Rhombus ?

a parallelogram with opposite equal acute angles, opposite equal obtuse angles, and four equal sides.

Alright cool

Wait what is a acute and obtuse angle.

Acute : An acute angle is an angle that measures between 90° and 0°

Oh its just small.

Obtuse : an obtuse angle is an angle that is greater than 90° and less than 180°.

And obtuse is big.

So if we use our 3,5 point we can make this diamond looking thing.

The 3,5 point is basically just a mirror image of the point A by the way it makes the shape symmetrical on the x axis which is what we want for a rhombus.

So if we were to draw this diamond. (I am a professional artist luckily // check the bottom of the page.)

We have 2 equal (ish) acute angles and two equal (ish) obtuse angles. And we know this b is symmetrical we went up from the middle 2 units just like the A point is down 2 units. So it all is even.

This is a rhombus.

Thank you

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4 years ago
The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:
deff fn [24]

Step-by-step explanation:

Given equation of circle is,

{x}^{2}  +  {y}^{2}  - 6y - 12 = 0 \\ by \: compairing \: it \: with \:   \\ {x}^{2}  +  {y}^{2}   + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f=  - 6 \\ or \: f=  - 3 \\ c =  - 12

again the another form of circle is,

{x}^{2}  +  {(y - k)}^{2}  = 21 \\ or \:  {x}^{2}  + {y}^{2}  - 2ky +  {k}^{2}  - 21 = 0 \\ by \: compairing \:it \: with \:  \\ {x}^{2}  +  {y}^{2}   + 2gx + 2fy = 0 \: we \: get  \\ g = 0 \\f =  - k \\ c =  {k}^{2}  - 21

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