A triangle has corners at #(5, 8 )#, #( 3, 1 )#, and #( 9 , 2 )#. If the triangle is dilated by # 3 x# around #(4, 2)#, what will the new coordinates of its corners be?

Answer 1

New coordinates are

#A’ ((7),(20)), B’ ((1),(-1)), C’ ((19),(2))#

#A (5,8), B(3,1), C(9,2)#

Dilated around D(4,2) by a factor of 3.

#vec(AD) = 3 * vec(AB)#
#A’ = 3A - 2D = 3((5),(8)) - 2((4),(2)) = ((15),(24)) - ((8),(4)) = ((7),(20))#
#B’ = 3B - 2D = 3((3),(1)) - 2((4),(2)) = ((9),(3)) - ((8),(4)) = ((1),(-1))#
#C’ = 3C - 2D = 3((9),(2)) - 2((4),(2)) = ((27),(6)) - ((8),(4)) = ((19),(2))#
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Answer 2

To dilate a triangle by a factor of (3) around the point ((4, 2)), you apply the dilation formula to each corner of the triangle. The formula for dilation from a point ((x_0, y_0)) by a scale factor (k) is given by:

[ \begin{align*} x' &= x_0 + k(x - x_0) \ y' &= y_0 + k(y - y_0) \end{align*} ]

where ((x, y)) are the original coordinates, and ((x', y')) are the dilated coordinates.

Applying this formula to each corner of the triangle:

  1. For the corner at ((5, 8)):

[ \begin{align*} x' &= 4 + 3(5 - 4) = 4 + 3(1) = 4 + 3 = 7 \ y' &= 2 + 3(8 - 2) = 2 + 3(6) = 2 + 18 = 20 \end{align*} ]

So, the new coordinates are ((7, 20)).

  1. For the corner at ((3, 1)):

[ \begin{align*} x' &= 4 + 3(3 - 4) = 4 - 3 = 1 \ y' &= 2 + 3(1 - 2) = 2 - 3 = -1 \end{align*} ]

So, the new coordinates are ((1, -1)).

  1. For the corner at ((9, 2)):

[ \begin{align*} x' &= 4 + 3(9 - 4) = 4 + 3(5) = 4 + 15 = 19 \ y' &= 2 + 3(2 - 2) = 2 + 0 = 2 \end{align*} ]

So, the new coordinates are ((19, 2)).

Therefore, after the dilation by a factor of (3) around the point ((4, 2)), the new coordinates of the triangle's corners are ((7, 20)), ((1, -1)), and ((19, 2)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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