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kifflom [539]
1 year ago
9

Triangle abc is located on a coordinate plane angle c is at (1,2) the triangle is rotated counter clockwise 90° about the origin

enter coordinates for c’
Mathematics
1 answer:
sertanlavr [38]1 year ago
6 0

After rotating a point C 90° counterclockwise about the origin the coordinates of C' would be (-2, 1)

In this question, we have been given a triangle ABC.

A point C is at (1, 2)

The triangle is rotated counter clockwise 90° about the origin.

We need to find the coordinates of C' which is image of vertex C  after rotation.

We know that, if we rotate a point 90° counterclockwise about the origin a point (x, y) becomes (-y, x).

Here C(1, 2) is rotated 90 degrees counterclockwise about the origin.

So, the coordinates of C' would be,

C' = (-2, 1)

Therefore, after rotating a point C 90° counterclockwise about the origin the coordinates of C' are (-2, 1)

Learn more about the rotation here:

brainly.com/question/2763408

#SPJ1

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Step-by-step explanation:  

see the attached figure with letters to better understand the problem

<em>In the right triangle BCD find BC</em>

Applying the Pythagoras Theorem

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Need help!<br> A. y= -x+2<br> B. y=2x+2<br> C. y=2x-1<br> D. y=5x+2
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\large\boxed{y=5x+2}

Step-by-step explanation:

\text{Let:}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\\text{then}\\\\k\ ||\ l\iff m_1=m_2\\\\k \perp l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\==================================

\text{We have the equation of a line:}\\\\y=-\dfrac{1}{5}x+4\to m_1=-\dfrac{1}{5}\\\\\text{Therefore}\\\\m_2=-\dfrac{1}{-\frac{1}{5}}=-\left(-\dfrac{5}{1}\right)=5\\\\\text{Initially, our equation has the form}\\\\y=5x+b

\text{The line should pass through point (-1, -3).}\\\text{We substitute the coordinates of the point into the equation:}\\\\-3=5(-1)+b\\-3=-5+b\qquad\text{add 5 to both sides}\\-3+5=-5+5+b\\2=b\to b=2\\\\\text{Finally:}\\\\y=5x+2

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