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3 years ago

∆ABC rotates 90° clockwise about point P to form ∆A′B′C′.

1 answer:

6 0

Assume P(xp,yp), A(xa,ya), etc.
We know that rotation rule of 90<span>° clockwise about the origin is
R_-90(x,y) -> (y,-x)
For example, rotating A about the origin 90</span><span>° clockwise is
(xa,ya) -> (ya, -xa)
or for a point at H(5,2), after rotation, H'(2,-5), etc.

To rotate about P, we need to translate the point to the origin, rotate, then translate back. The rule for translation is
T_(dx,dy) (x,y) -> (x+dx, y+dy)

So with the translation set at the coordinates of P, and combining the rotation with the translations, the complete rule is:
T_(xp,yp) R_(-90) T_(-xp,-yp) (x,y)
-> </span>T_(xp,yp) R_(-90) (x-xp, y-yp)
-> T_(xp,yp) (y-yp, -(x-xp))
-> (y-yp+xp, -x+xp+yp)

Example: rotate point A(7,3) about point P(4,2)
=> x=7, y=3, xp=4, yp=2
=> A'(3-2+4, -7+4+2) => A'(5,-1)

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BabaBlast [244]

$\underline{\bf{Given \:equation:-}}$

$\\ \sf{:}\dashrightarrow ax^2+by+c=0$

$\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.$

$\sf We\:know,$

$\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}$

$\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}$

$\underline{\large{\bf Identities\:used:-}}$

$\boxed{\sf (a+b)^2=a^2+2ab+b^2}$

$\boxed{\sf (√a)^2=a}$

$\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}$

$\boxed{\sf \sqrt{\sqrt{a}}=a}$

$\underline{\bf Final\: Solution:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}$

$\bull\sf Apply\: Squares$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}$

$\bull\sf Put\:values$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}$

$\bull\sf Simplify$

$\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}$

$\underline{\bf More\: simplification:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}$

$\underline{\Large{\bf Simplified\: Answer:-}}$

$\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}$

5 0

2 years ago

Read 2 more answers

Evaluate.

Daniel [21]

Answer:

d) 192

Step-by-step explanation:

Your calculator can do this for you.

Or, you can simplify to

$12\cdot\dfrac{4^{-2}}{4^{-4}}=12(4^{-2+4})=12(4^2)=192$

3 0

3 years ago

What is the derivative of y=(4x+1)(1-x)^3

Rudik [331]

I am not so sure of this but I feel it would serve as hint

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3 years ago

Three cards are dealt from a shuffled standard deck of playing cards. Find the probability that the first card dealt is red, the

harina [27]

There are 52 cards in a deck, half the cards aware red and half the cards are red.

First card being red is 26/52 = 1/2

After the first card is dealt there are 51 cards left and 26 are black so dealing a black card = 26/51

After 2 cards are dealt there are 50 cards left and 25 red ones left. The probability would be 25/50 = 1/2

The probability of all 3 happening = 1/2 x 26/51 x 1/2 = 13/102

Answer: 13/102

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Anarel [89]

Answer:

i think this is right i hope it helped :)

Step-by-step explanation:

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6 0

2 years ago

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