The vertices of a triangle are A(0,3) B(-2,-4) and C(1,5) find the new vertices

Mathematics

1 answer:

worty [1.4K]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Using the translation rule (x, y ) → (x - 2, y + 4 )

Subtract 2 from the original x- coordinate and add 4 to the original y- coordinate, thus

A(0, 3 ) → A'(0 - 2, 3 + 4 ) → A'(- 2, 7 )

B(- 2, - 4 ) → B'(- 2 - 2, - 4 + 4 ) → B'(- 4, 0 )

C(1, 5 ) → C'(1 - 2, 5 + 4 ) → C'(- 1, 9 )

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The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

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Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.