144,555 views
30 votes
30 votes
Quadrilateral FORT is reflected over the line y=1/2x+2, which draw below

Quadrilateral FORT is reflected over the line y=1/2x+2, which draw below-example-1
User Serhii Shynkarenko
by
3.0k points

1 Answer

17 votes
17 votes
Step-by-step explanation

Let's recall that to reflect a polygon it is sufficient to reflect its vertices. After reflecting them, we only need to join the resulting points by segments to get the reflection of the total quadrilateral.

With this in mind, let's reflect the four vertices of our figure over the given line.

Let's start with


O=(1,0)\text{.}

Firstly, we need to find the perpendicular line passing through O. It has the equation


y_p=m\cdot x_p+b,

where m is the slope, and b is the y-intercept. For we want the line to be perpendicular to the given one, we have


m_p=-(1)/(((1)/(2)))=-2.

Then, the equation above becomes


y_p=-2x_p+b\text{.}

Now, O must lie in the line; then, ("we can evaluate the equation for O")


0=-2(1)+b\text{.}

Solving it for b, we get


\begin{gathered} 0=-2(1)+b, \\ 0=-2+b, \\ 2=b, \\ b=2. \end{gathered}

Thus, the equation for the line perpendicular to the given one passing through O is


y_p=-2x_p+2.

Let's graph our current situation:

The green line represents the given line, and the blue line represents the perpendicular line, which equation we just found. Point B is the point where both lines intersect. The reflection of O over the provided line is point C, the point on the perpendicular line which is "as far from B as O".

Now, How to calculate C? Note that B is the middle point between C and O. Then, if


C=(c_1,c_2),
(0,2)=B=((1+c_1)/(2),(0+c_2)/(2))\text{.}

That is


\begin{cases}(1+c_1)/(2)=0, \\ \\ (0+c_2)/(2)=2.\end{cases}

Solving both equations, we get


\begin{gathered} (1+c_1)/(2)=0,\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.\text{ }(0+c_2)/(2)=2,\text{ } \\ 1+c_1=0,.\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..(c_2)/(2)=2,_{} \\ {\textcolor{green}{c}_{\textcolor{green}{1}}\textcolor{green}{=-1}}.\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots...{\textcolor{green}{c}_{\textcolor{green}{2}}\textcolor{green}{=4}}. \end{gathered}

Thus


C=(-1,4)\text{.}

Repeating this process to the remaining vertices, we get that the reflection of

F is (-8,3), the reflection of T is (-5,7), and the reflection of R is (4,9).

Let's graph what we have obtained:

It only remains to apply the idea given in the first paragraph.

Answer

Quadrilateral FORT is reflected over the line y=1/2x+2, which draw below-example-1
Quadrilateral FORT is reflected over the line y=1/2x+2, which draw below-example-2
Quadrilateral FORT is reflected over the line y=1/2x+2, which draw below-example-3
User Lord Spectre
by
2.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.