# Find the line of reflection that maps the trapezoid, formed with vertices at A(-3,1), B(-1,1), C(0,4) and D(-4,4), onto itself?

##### 1 Answer
Dec 19, 2016

$x + 2 = 0$

#### Explanation:

The trapezoid is formed with its vertices at $A \left(- 3 , 1\right)$, $B \left(- 1 , 1\right)$, $C \left(0 , 4\right)$ and $D \left(- 4 , 4\right)$. These have been named to add clarity to answer.

As ordinates of $A$ and $B$ are same, as also ordinates of $C$and $D$, it the lines $A B$ and $C D$ are parallel to $x$-axiss and hence it is apparent that $A B$||$C D$.

Further $A D = \sqrt{{\left(- 3 - \left(- 4\right)\right)}^{2} + {\left(1 - 4\right)}^{2}} = \sqrt{10}$ and
$B C = \sqrt{{\left(- 1 - 0\right)}^{2} + {\left(1 - 4\right)}^{2}} = \sqrt{10}$ and as such $A D = B C$ and it is an isosceles trapezium.

Hence the line joining the midpoints of $A B$ and $C D$ i.e.$\left(- 2 , 1\right)$ and $\left(- 2 , 4\right)$ is the line of reflection that maps the trapezoid onto itself, which is parallel to $y$-axis.

and its equation is $x = - 2$ or $x + 2 = 0$.