# A triangle has corners at (2, 7 ), ( 6, 3 ), and ( 2 , 5 ). If the triangle is dilated by  2 x around (2, 5), what will the new coordinates of its corners be?

Oct 11, 2016

$\left(2 , 5\right) , \left(10 , 1\right) , \left(2 , 9\right)$

#### Explanation:

$A B C$ is the original triangle. $A B ' C '$ is the dilated triangle.

Given: $A \left(2 , 5\right) , B \left(6 , 3\right) , C \left(2 , 7\right)$

1) Given the 2X dilation is around Pt $A \left(2 , 5\right) , A ' s$ coordinates remain unchanged at $\left(2 , 5\right)$

2) From Pt $A \left(2 , 5\right)$ to Pt $B \left(6 , 3\right)$
$\Delta x = {x}_{2} - {x}_{1} = 6 - 2 = 4$
$\Delta y = {y}_{2} - {y}_{1} = 3 - 5 = - 2$
Given scaling factor $= 2 X , \implies A B ' = 2 A B$
$\implies 2 \Delta x = 2 \cdot 4 = 8$,
$\implies 2 \Delta y = 2 \cdot \left(- 2\right) = - 4$

$\implies$ coordinates of $B ' = \left(2 + 8 , 5 - 4\right) = \left(10 , 1\right)$

3) From Pt $A \left(2 , 5\right)$ to Pt $C \left(2 , 7\right)$
$\Delta x = {x}_{2} - {x}_{1} = 2 - 2 = 0$ (vertical line)
$\Delta y = {y}_{2} - {y}_{1} = 7 - 5 = 2$
Given scaling factor $= 2 X , \implies A C ' = 2 A C$
$\implies 2 \Delta x = 2 \cdot 0 = 0$,
$\implies 2 \Delta y = 2 \cdot 2 = 4$

$\implies$ coordinates of $C ' = \left(2 + 0 , 5 + 4\right) = \left(2 , 9\right)$

Hence, the new coordinates of the dilated triangle will be:
$A \left(2 , 5\right) , B ' \left(10 , 1\right) , C ' \left(2 , 9\right)$