Question

Quadrilateral ABCD with vertices A(0, 6), B(-3, -6), C(-9, -6), and D(-12, -3): a) dilation with scale factor of 1/3 centered at the origin b) translation along the vector <-5,-1>

Answer 1

a) The points of the new quadrillateral are $A'(x,y) = (0, 2)$, $B'(x,y) = (-1, -2)$, $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$, respectively.

b) The points of the new quadrillateral are $A'(x,y) = (-5, 5)$, $B'(x,y) = (-8,-7)$, $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$, respectively.

How to perform transformations with points

a) A dillation centered at the origin is defined by following operation:

$P'(x,y) = k\cdot P(x,y)$ (1)

Where:

• $P(x,y)$ - Original point
• $P'(x,y)$ - Dilated point.

If we know that $k = \frac{1}{3}$, $A(x,y) = (0,6)$, $B(x,y) = (-3,-6)$, $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$, then the new points of the quadrilateral are:

$A'(x,y) = \frac{1}{3}\cdot (0,6)$

$A'(x,y) = (0, 2)$

$B'(x,y) = \frac{1}{3} \cdot (-3,-6)$

$B'(x,y) = (-1, -2)$

$C'(x,y) = \frac{1}{3}\cdot (-9,-6)$

$C'(x,y) = \left(-3,-2\right)$

$D'(x,y) = \frac{1}{3}\cdot (-12,-3)$

$D'(x,y) = (-4, -1)$

The points of the new quadrillateral are $A'(x,y) = (0, 2)$, $B'(x,y) = (-1, -2)$, $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$, respectively. $\blacksquare$

b) A translation along a vector is defined by following operation:

$P'(x,y) = P(x,y) +T(x,y)$ (2)

Where $T(x,y)$ is the transformation vector.

If we know that $T(x,y) = (-5,-1)$, $A(x,y) = (0,6)$, $B(x,y) = (-3,-6)$, $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$,

$A'(x,y) = (0,6) + (-5, -1)$

$A'(x,y) = (-5, 5)$

$B'(x,y) = (-3, -6) + (-5, -1)$

$B'(x,y) = (-8,-7)$

$C'(x,y) = (-9, -6) + (-5, -1)$

$C'(x,y) = (-13, -7)$

$D'(x,y) = (-12,-3)+(-5,-1)$

$D'(x,y) = (-17, -4)$

The points of the new quadrillateral are $A'(x,y) = (-5, 5)$, $B'(x,y) = (-8,-7)$, $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$, respectively. $\blacksquare$

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