# A triangle has corners at (2, 7 ), ( 8, 3 ), and ( 4 , 8 ). If the triangle is dilated by  7 x around (1, 3), what will the new coordinates of its corners be?

Mar 14, 2017

The new coordinates are $\left(8 , 31\right)$, $\left(50 , 0\right)$ and $\left(22 , 38\right)$

#### Explanation:

We can work this with vectors

Let the points be

$A = \left(2 , 7\right)$

$B = \left(8 , 3\right)$

$C = \left(4 , 8\right)$

$D = \left(1 , 3\right)$

$\vec{D A} = < 2 - 1 , 7 - 3 > = < 1 , 4 >$

$\vec{D B} = < 8 - 1 , 3 - 3 > = < 7 , 0 >$

$\vec{D C} = < 4 - 1 , 8 - 3 > = < 3 , 5 >$

Let $A '$ be the new point

Then

$\vec{D A '} = 7 \vec{D A} = 7 + < 1 , 4 > = < 7 , 28 >$

So the coordinates of $A ' = \left(1 + 7 , 3 + 28\right) = \left(8 , 31\right)$

Let $B '$ be another new point

Then,

$\vec{D B '} = 7 \vec{D B} = 7 \cdot < 7 , 0 > = < 49 , 0 >$

The coordinates of $B ' = \left(1 + 49 , 3 + 0\right) = \left(50 , 0\right)$

Let $C '$ be another new point

Then,

$\vec{D C '} = 7 \vec{D C} = 7 \cdot < 3 , 5 > = < 21 , 35 >$

The coordinates of $C ' = \left(1 + 21 , 3 + 35\right) = \left(22 , 38\right)$